Math Problems Vault

Problem 18.1

Let $\mathbf{R}$ represent an $n \times n$ matrix, $\mathbf{S}$ an $n \times m$ matrix, $\mathbf{T}$ an $m \times m$ matrix, and $\mathbf{U}$ an $m \times n$ matrix. Derive (for the special case where $\mathbf{R}$ and $\mathbf{T}$ are nonsingular), the formula

\left|\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}\right| = \left|\mathbf{R}\right| \left|\mathbf{T} + \mathbf{T}\mathbf{U}\mathbf{R}^{-1}\mathbf{S}\mathbf{T}\right|/\left|\mathbf{T}\right|.

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Remark.

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Do so by making two applications of the formula

\begin{vmatrix} \mathbf{T} & \mathbf{U} \\ \mathbf{V} & \mathbf{W} \end{vmatrix} = \begin{vmatrix} \mathbf{W} & \mathbf{V} \\ \mathbf{U} & \mathbf{T} \end{vmatrix} = \left|\mathbf{T}\right| \left|\mathbf{W} - \mathbf{V}\mathbf{T}^{-1}\mathbf{U}\right|. \tag {*}

(in which $\mathbf{V}$ is an $n \times m$ matrix and $\mathbf{W}$ an $n \times n$ matrix and in which $\mathbf{T}$ is assumed to be nonsingular) to the partitioned matrix $\begin{vmatrix} \mathbf{R} & -\mathbf{S}\mathbf{T} \\ \mathbf{T}\mathbf{U} & \mathbf{T} \end{vmatrix}$ -- one with $\mathbf{W}$ set equal to $\mathbf{R}$ , and the other with $\mathbf{T}$ set equal to $\mathbf{R}$ .

Problem 18.2

Let $\mathbf{R}$ represent an $n \times n$ matrix, $\mathbf{S}$ an $n \times m$ matrix, $\mathbf{T}$ an $m \times m$ matrix, and $\mathbf{U}$ an $m \times n$ matrix. Show that if $\mathbf{R}$ is nonsingular, then

\left|\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}\right| = \left|\mathbf{R}\right| \left|\mathbf{I}_{m} + \mathbf{U}\mathbf{R}^{-1}\mathbf{S}\mathbf{T}\right| = \left|\mathbf{R}\right| \left|\mathbf{I}_{m} + \mathbf{T}\mathbf{U}\mathbf{R}^{-1}\mathbf{S}\right|.

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Remark.

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Do so by using the formula

\left|\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}\right| = \left|\mathbf{R}\right| \left|\mathbf{T}\right| \left|\mathbf{T}^{-1} + \mathbf{U}\mathbf{R}^{-1}\mathbf{S}\right|, \tag {*}

(in which $\mathbf{R}$ and $\mathbf{T}$ are assumed to be nonsingular), or alternatively the formula $\left|\mathbf{I}_{n} + \mathbf{S}\mathbf{U}\right| = \left|\mathbf{I}_{m} + \mathbf{U}\mathbf{S}\right|$ or the formula $\left|\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}\right| = \left|\mathbf{R}\right| \left|\mathbf{T} + \mathbf{T}\mathbf{U}\mathbf{R}^{-1}\mathbf{S}\mathbf{T}\right|/\left|\mathbf{T}\right|$ .

Problem 18.3

Let $\mathbf{A}$ represent an $n \times n$ symmetric nonnegative definite matrix. Show that if $\mathbf{I} - \mathbf{A}$ is nonnegative definite and if $\left|\mathbf{A}\right| = 1$ , then $\mathbf{A} = \mathbf{I}$ .

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Problem 18.4

Show that, for any $n \times n$ symmetric nonnegative definite matrix $\mathbf{B}$ and for any $n \times n$ symmetric matrix $\mathbf{C}$ such that $\mathbf{C} - \mathbf{B}$ is nonnegative definite,

\left|\mathbf{C}\right| \geq \left|\mathbf{C} - \mathbf{B}\right|,

with equality holding if and only if $\mathbf{C}$ is singular or $\mathbf{B} = \mathbf{0}$ .

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Problem 18.5

(a)(b)(c)

Let $\mathbf{A}$ represent a symmetric nonnegative definite matrix that has been partitioned as

\mathbf{A} = \begin{pmatrix} \mathbf{T} & \mathbf{U} \\ \mathbf{U}' & \mathbf{W} \end{pmatrix},

where $\mathbf{T}$ is of dimensions $m \times m$ and $\mathbf{W}$ of dimensions $n \times n$ (and where $\mathbf{U}$ is of dimensions $m \times n$ ). And, define $\mathbf{Q} = \mathbf{W} - \mathbf{U}'\mathbf{T}^{-}\mathbf{U}$ (which is the Schur complement of $\mathbf{T}$ ).

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(a)

Using the result that the symmetry and nonnegative definiteness of $\mathbf{A}$ imply the nonnegative definiteness of $\mathbf{Q}$ and the result of Exercise 14.33 (or otherwise), show that

\left|\mathbf{W}\right| \geq \left|\mathbf{U}'\mathbf{T}^{-}\mathbf{U}\right|,

with equality holding if and only if $\mathbf{W}$ is singular or $\mathbf{Q} = \mathbf{0}$ .

(b)

Suppose that $n = m$ and that $\mathbf{T}$ is nonsingular. Show that

\left|\mathbf{W}\right| \left|\mathbf{T}\right| \geq \left|\mathbf{U}\right|^{2},

with equality holding if and only if $\mathbf{W}$ is singular or $\operatorname {rank}(\mathbf{A}) = m$ .

(c)

Suppose that $n = m$ and that $\mathbf{A}$ is positive definite. Show that

\left|\mathbf{W}\right| \left|\mathbf{T}\right| > \left|\mathbf{U}\right|^{2}.

Problem 18.6

Show that, for any $n \times p$ matrix $\mathbf{X}$ and any symmetric positive definite matrix $\mathbf{W}$ ,

\left|\mathbf{X}'\mathbf{W}\mathbf{X}\right| \, \left|\mathbf{X}'\mathbf{W}^{-1}\mathbf{X}\right| \geq \left|\mathbf{X}'\mathbf{X}\right|^{2}. \tag {E.1}

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Remark.

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Begin by showing that the matrices $\mathbf{X}'\mathbf{X}(\mathbf{X}'\mathbf{W}\mathbf{X})^{-}\mathbf{X}'\mathbf{X}$ and $\mathbf{X}'\mathbf{W}^{-1}\mathbf{X} - \mathbf{X}'\mathbf{X}(\mathbf{X}'\mathbf{W}\mathbf{X})^{-}\mathbf{X}'\mathbf{X}$ are symmetric and nonnegative definite.

Problem 18.7

(a)(b)

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(a)

Show that, for any $n \times n$ skew-symmetric matrix $\mathbf{C}$ ,

\left|\mathbf{I}_{n} + \mathbf{C}\right| \geq 1,

with equality holding if and only if $\mathbf{C} = \mathbf{0}$ .

(b)

Generalize the result of Part (a) by showing that, for any $n \times n$ symmetric positive definite matrix $\mathbf{A}$ and any $n \times n$ skew-symmetric matrix $\mathbf{B}$ ,

\left|\mathbf{A} + \mathbf{B}\right| \geq \left|\mathbf{A}\right|,

with equality holding if and only if $\mathbf{B} = \mathbf{0}$ .

Problem 18.8

(a)(b)

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(a)

Let $\mathbf{R}$ represent an $n \times n$ nonsingular matrix, and let $\mathbf{B}$ represent an $n \times n$ matrix of rank one. Show that $\mathbf{R} + \mathbf{B}$ is nonsingular if and only if $\operatorname {tr}(\mathbf{R}^{-1}\mathbf{B}) \neq -1$ , in which case

(\mathbf{R} + \mathbf{B})^{-1} = \mathbf{R}^{-1} - [1 + \operatorname {tr}(\mathbf{R}^{-1}\mathbf{B})]^{-1}\mathbf{R}^{-1}\mathbf{B}\mathbf{R}^{-1}.

(b)

To what does the result of Part (a) simplify in the special case where $\mathbf{R} = \mathbf{I}_{n}$ ?

Problem 18.9

(a)(b)

Let $\mathbf{R}$ represent an $n \times n$ matrix, $\mathbf{S}$ an $n \times m$ matrix, $\mathbf{T}$ an $m \times m$ matrix, and $\mathbf{U}$ an $m \times n$ matrix. Suppose that $\mathbf{R}$ is nonsingular. Show

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(a)

that $\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}$ is nonsingular if and only if $\mathbf{I}_{m} + \mathbf{U}\mathbf{R}^{-1}\mathbf{S}\mathbf{T}$ is nonsingular, in which case

(\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U})^{-1} = \mathbf{R}^{-1} - \mathbf{R}^{-1}\mathbf{S}\mathbf{T}(\mathbf{I}_{m} + \mathbf{U}\mathbf{R}^{-1}\mathbf{S}\mathbf{T})^{-1}\mathbf{U}\mathbf{R}^{-1},

and

(b)

that $\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}$ is nonsingular if and only if $\mathbf{I}_{m} + \mathbf{T}\mathbf{U}\mathbf{R}^{-1}\mathbf{S}$ is nonsingular, in which case

(\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U})^{-1} = \mathbf{R}^{-1} - \mathbf{R}^{-1}\mathbf{S}(\mathbf{I}_{m} + \mathbf{T}\mathbf{U}\mathbf{R}^{-1}\mathbf{S})^{-1}\mathbf{T}\mathbf{U}\mathbf{R}^{-1}.

Remark.

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Do so by using the result -- applicable when $\mathbf{T}$ (as well as $\mathbf{R}$ ) is nonsingular -- that $\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}$ is nonsingular if and only if $\mathbf{T}^{-1} + \mathbf{U}\mathbf{R}^{-1}\mathbf{S}$ is nonsingular, or equivalently if and only if $\mathbf{T} + \mathbf{T}\mathbf{U}\mathbf{R}^{-1}\mathbf{S}\mathbf{T}$ is nonsingular, in which case

\begin{aligned} (\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U})^{-1} & = \mathbf{R}^{-1} - \mathbf{R}^{-1}\mathbf{S}(\mathbf{T}^{-1} + \mathbf{U}\mathbf{R}^{-1}\mathbf{S})^{-1}\mathbf{U}\mathbf{R}^{-1} \\ & = \mathbf{R}^{-1} - \mathbf{R}^{-1}\mathbf{S}\mathbf{T}(\mathbf{T} + \mathbf{T}\mathbf{U}\mathbf{R}^{-1}\mathbf{S}\mathbf{T})^{-1}\mathbf{T}\mathbf{U}\mathbf{R}^{-1}. \end{aligned}

Problem 18.10

Let $\mathbf{R}$ represent an $n \times q$ matrix, $\mathbf{S}$ an $n \times m$ matrix, $\mathbf{T}$ an $m \times p$ matrix, and $\mathbf{U}$ a $p \times q$ matrix. Extend the results of Exercise 9 by showing that if $\mathcal{R}(\mathbf{S}\mathbf{T}\mathbf{U}) \subset \mathcal{R}(\mathbf{R})$ and $C(\mathbf{S}\mathbf{T}\mathbf{U}) \subset C(\mathbf{R})$ , then the matrix

\mathbf{R}^{-} - \mathbf{R}^{-}\mathbf{S}\mathbf{T}(\mathbf{I}_{p} + \mathbf{U}\mathbf{R}^{-}\mathbf{S}\mathbf{T})^{-}\mathbf{U}\mathbf{R}^{-}

and the matrix

\mathbf{R}^{-} - \mathbf{R}^{-}\mathbf{S}(\mathbf{I}_{m} + \mathbf{T}\mathbf{U}\mathbf{R}^{-}\mathbf{S})^{-}\mathbf{T}\mathbf{U}\mathbf{R}^{-}

are both generalized inverses of the matrix $\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}$ .

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Problem 18.11

(a)(b)(c)

Let $\mathbf{R}$ represent an $n \times q$ matrix, $\mathbf{S}$ an $n \times m$ matrix, $\mathbf{T}$ an $m \times p$ matrix, and $\mathbf{U}$ a $p \times q$ matrix.

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(a)

Take $\mathbf{G}$ to be a generalized inverse of the partitioned matrix $\begin{pmatrix} \mathbf{R} & -\mathbf{S}\mathbf{T} \\ \mathbf{T}\mathbf{U} & \mathbf{T} \end{pmatrix}$ , and partition $\mathbf{G}$ as $\mathbf{G} = \begin{pmatrix} \mathbf{G}_{11} & \mathbf{G}_{12} \\ \mathbf{G}_{21} & \mathbf{G}_{22} \end{pmatrix}$ (where $\mathbf{G}_{11}$ is of dimensions $q \times n$ ). Show that $\mathbf{G}_{11}$ is a generalized inverse of the matrix $\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}$ . Do so by using the result that, for any partitioned matrix $\mathbf{A} = \begin{pmatrix} \mathbf{A}_{11} & \mathbf{A}_{12} \\ \mathbf{A}_{21} & \mathbf{A}_{22} \end{pmatrix}$ such that $C(\mathbf{A}_{21}) \subset C(\mathbf{A}_{22})$ and $\mathcal{R}(\mathbf{A}_{12}) \subset \mathcal{R}(\mathbf{A}_{22})$ and for any generalized inverse $\begin{pmatrix} \mathbf{C}_{11} & \mathbf{C}_{12} \\ \mathbf{C}_{21} & \mathbf{C}_{22} \end{pmatrix}$ of $\mathbf{A}$ (where $\mathbf{C}_{11}$ is of the same dimensions as $\mathbf{A}_{11}'$ ), $\mathbf{C}_{11}$ is a generalized inverse of the matrix $\mathbf{A}_{11} - \mathbf{A}_{12}\mathbf{A}_{22}^{-}\mathbf{A}_{21}$ .

(b)

Let $\mathbf{E}_{\mathbf{R}} = \mathbf{I} - \mathbf{R}\mathbf{R}^{-}$ , $\mathbf{F}_{\mathbf{R}} = \mathbf{I} - \mathbf{R}^{-}\mathbf{R}$ , $\mathbf{X} = \mathbf{E}_{\mathbf{R}}\mathbf{S}\mathbf{T}$ , $\mathbf{Y} = \mathbf{T}\mathbf{U}\mathbf{F}_{\mathbf{R}}$ , $\mathbf{E}_{\mathbf{Y}} = \mathbf{I} - \mathbf{Y}\mathbf{Y}^{-}$ , $\mathbf{F}_{\mathbf{X}} = \mathbf{I} - \mathbf{X}^{-}\mathbf{X}$ , $\mathbf{Q} = \mathbf{T} + \mathbf{T}\mathbf{U}\mathbf{R}^{-}\mathbf{S}\mathbf{T}$ , $\mathbf{Z} = \mathbf{E}_{\mathbf{Y}}\mathbf{Q}\mathbf{F}_{\mathbf{X}}$ , and $\mathbf{Q}^{*} = \mathbf{F}_{\mathbf{X}}\mathbf{Z}^{-}\mathbf{E}_{\mathbf{Y}}$ . Use the result of Part (a) of Exercise 10.10 to show that the matrix

\begin{aligned} & \mathbf{R}^{-} - \mathbf{R}^{-}\mathbf{S}\mathbf{T}\mathbf{Q}^{*}\mathbf{T}\mathbf{U}\mathbf{R}^{-} - \mathbf{R}^{-}\mathbf{S}\mathbf{T}(\mathbf{I} - \mathbf{Q}^{*}\mathbf{Q})\mathbf{X}^{-}\mathbf{E}_{\mathbf{R}} \\ & \qquad - \mathbf{F}_{\mathbf{R}}\mathbf{Y}^{-}(\mathbf{I} - \mathbf{Q}\mathbf{Q}^{*})\mathbf{T}\mathbf{U}\mathbf{R}^{-} + \mathbf{F}_{\mathbf{R}}\mathbf{Y}^{-}(\mathbf{I} - \mathbf{Q}\mathbf{Q}^{*})\mathbf{Q}\mathbf{X}^{-}\mathbf{E}_{\mathbf{R}} \end{aligned} \tag {E.2}

is a generalized inverse of the matrix $\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}$ .

(c)

Show that if $\mathcal{R}(\mathbf{T}\mathbf{U}) \subset \mathcal{R}(\mathbf{R})$ and $C(\mathbf{S}\mathbf{T}) \subset C(\mathbf{R})$ , then the formula

\mathbf{R}^{-} - \mathbf{R}^{-}\mathbf{S}\mathbf{T}\mathbf{Q}^{-}\mathbf{T}\mathbf{U}\mathbf{R}^{-} \tag {*}

for a generalized inverse of $\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}$ can be obtained as a special case of formula (E.2).

Problem 18.12

(a)(b)

Let $\mathbf{A}_{1}, \mathbf{A}_{2}, \ldots$ represent a sequence of $m \times n$ matrices, and let $\mathbf{A}$ represent another $m \times n$ matrix.

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(a)

Using the result of Exercise 6.1 (i.e., the triangle inequality), show that if $\left\| \mathbf{A}_{k} - \mathbf{A}\right\| \to 0$ , then $\left\| \mathbf{A}_{k}\right\| \to \left\| \mathbf{A}\right\|$ .

(b)

Show that if $\mathbf{A}_{k} \to \mathbf{A}$ , then $\left\| \mathbf{A}_{k}\right\| \to \left\| \mathbf{A}\right\|$ (where the norms are the usual norms).

Problem 18.13

Let $\mathbf{A}$ represent an $n \times n$ matrix. Using the results of Exercise 6.1 and of Part (b) of Exercise 12, show that if $\left\| \mathbf{A}\right\| < 1$ , then (for $k = 0, 1, 2, \ldots$ )

\left\| (\mathbf{I} - \mathbf{A})^{-1} - (\mathbf{I} + \mathbf{A} + \mathbf{A}^{2} + \cdots + \mathbf{A}^{k})\right\| \leq \left\| \mathbf{A}\right\| ^{k+1}/(1 - \left\| \mathbf{A}\right\| )

(where the norms are the usual norms). (Note. If $\left\| \mathbf{A}\right\| < 1$ , then $\mathbf{I} - \mathbf{A}$ is nonsingular.)

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Problem 18.14

Let $\mathbf{A}$ and $\mathbf{B}$ represent $n \times n$ matrices. Suppose that $\mathbf{B}$ is nonsingular, and define $\mathbf{F} = \mathbf{B}^{-1}\mathbf{A}$ . Using the result of Exercise 13, show that if $\left\| \mathbf{F}\right\| < 1$ , then (for $k = 0, 1, 2, \ldots$ )

\left\| (\mathbf{B} - \mathbf{A})^{-1} - (\mathbf{B}^{-1} + \mathbf{F}\mathbf{B}^{-1} + \mathbf{F}^{2}\mathbf{B}^{-1} + \cdots + \mathbf{F}^{k}\mathbf{B}^{-1})\right\| \leq \left\| \mathbf{B}^{-1}\right\| \left\| \mathbf{F}\right\| ^{k+1}/(1 - \left\| \mathbf{F}\right\| )

(where the norms are the usual norms). (Note. If $\left\| \mathbf{F}\right\| < 1$ , then $\mathbf{B} - \mathbf{A}$ is nonsingular.)

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Problem 18.15

Let $\mathbf{A}$ represent an $n \times n$ symmetric nonnegative definite matrix, and let $\mathbf{B}$ represent an $n \times n$ matrix. Show that if $\mathbf{B} - \mathbf{A}$ is nonnegative definite (in which case $\mathbf{B}$ is also nonnegative definite), then $\mathcal{R}(\mathbf{A}) \subset \mathcal{R}(\mathbf{B})$ and $C(\mathbf{A}) \subset C(\mathbf{B})$ .

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Problem 18.16

Let $\mathbf{A}$ represent an $n \times n$ symmetric idempotent matrix, and let $\mathbf{B}$ represent an $n \times n$ symmetric nonnegative definite matrix. Show that if $\mathbf{I} - \mathbf{A} - \mathbf{B}$ is nonnegative definite, then $\mathbf{B}\mathbf{A} = \mathbf{A}\mathbf{B} = \mathbf{0}$ .

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Remark.

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Show that $\mathbf{A}'(\mathbf{I} - \mathbf{A} - \mathbf{B})\mathbf{A} = -\mathbf{A}'\mathbf{B}\mathbf{A}$ , and then consider the implications of this equality.

Problem 18.17

Let $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ represent $n \times n$ symmetric matrices, and define $\mathbf{A} = \mathbf{A}_{1} + \cdots + \mathbf{A}_{k}$ . Suppose that $\mathbf{A}$ is idempotent. Suppose further that $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k-1}$ are idempotent and that $\mathbf{A}_{k}$ is nonnegative definite. Using the result of Exercise 16 (or otherwise), show that $\mathbf{A}_{i}\mathbf{A}_{j} = \mathbf{0}$ (for $j \neq i = 1, \ldots , k$ ), that $\mathbf{A}_{k}$ is idempotent, and that $\operatorname {rank}(\mathbf{A}_{k}) = \operatorname {rank}(\mathbf{A}) - \sum_{i=1}^{k-1} \operatorname {rank}(\mathbf{A}_{i})$ .

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Problem 18.18

Let $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ represent $n \times n$ symmetric matrices, and define $\mathbf{A} = \mathbf{A}_{1} + \cdots + \mathbf{A}_{k}$ . Suppose that $\mathbf{A}$ is idempotent. Show that if $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ are nonnegative definite and if $\operatorname {tr}(\mathbf{A}) \leq \sum_{i=1}^{k} \operatorname {tr}(\mathbf{A}_{i})$ , then $\mathbf{A}_{i}\mathbf{A}_{j} = \mathbf{0}$ (for $j \neq i = 1, \ldots , k$ ) and $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ are idempotent.

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Remark.

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Show that $\sum_{i, j \neq i} \operatorname {tr}(\mathbf{A}_{i}\mathbf{A}_{j}) \leq 0$ and then make use of the result that, for any two symmetric nonnegative definite matrices $\mathbf{B}$ and $\mathbf{C}$ (of the same order), $\operatorname {tr}(\mathbf{B}\mathbf{C}) \geq 0$ , with equality holding if and only if $\mathbf{B}\mathbf{C} = \mathbf{0}$ .

Problem 18.19

Let $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ represent $n \times n$ symmetric matrices such that $\mathbf{A}_{1} + \cdots + \mathbf{A}_{k} = \mathbf{I}$ . Show that if $\operatorname {rank}(\mathbf{A}_{1}) + \cdots + \operatorname {rank}(\mathbf{A}_{k}) = n$ , then, for any (strictly) positive scalars $c_{1}, \ldots , c_{k}$ , the matrix $c_{1}\mathbf{A}_{1} + \cdots + c_{k}\mathbf{A}_{k}$ is positive definite.

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Problem 18.20

Let $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ represent $n \times n$ symmetric idempotent matrices such that $\mathbf{A}_{i}\mathbf{A}_{j} = \mathbf{0}$ for $j \neq i = 1, \ldots , k$ . Show that, for any (strictly) positive scalar $c_{0}$ and any nonnegative scalars $c_{1}, \ldots , c_{k}$ , the matrix $c_{0}\mathbf{I} + \sum_{i=1}^{k} c_{i}\mathbf{A}_{i}$ is positive definite (and hence nonsingular), and

\left(c_{0}\mathbf{I} + \sum _{i=1}^{k} c_{i}\mathbf{A}_{i}\right)^{-1} = d_{0}\mathbf{I} + \sum _{i=1}^{k} d_{i}\mathbf{A}_{i},

where $d_{0} = 1/c_{0}$ and (for $i = 1, \ldots , k$ ) $d_{i} = -c_{i}/[c_{0}(c_{0}+c_{i})]$ .

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Problem 18.21

Let $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ represent $n \times n$ symmetric idempotent matrices such that (for $j \neq i = 1, \ldots , k$ ) $\mathbf{A}_{i}\mathbf{A}_{j} = \mathbf{0}$ , and let $\mathbf{A}$ represent an $n \times n$ symmetric idempotent matrix such that (for $i = 1, \ldots , k$ ) $C(\mathbf{A}_{i}) \subset C(\mathbf{A})$ . Show that if $\operatorname {rank}(\mathbf{A}_{1}) + \cdots + \operatorname {rank}(\mathbf{A}_{k}) = \operatorname {rank}(\mathbf{A})$ , then $\mathbf{A}_{1} + \cdots + \mathbf{A}_{k} = \mathbf{A}$ .

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Problem 18.22

Let $\mathbf{A}$ represent an $m \times n$ matrix and $\mathbf{B}$ an $n \times m$ matrix. If $\mathbf{B}$ is a generalized inverse of $\mathbf{A}$ , then $\operatorname {rank}(\mathbf{I} - \mathbf{B}\mathbf{A}) = n - \operatorname {rank}(\mathbf{A})$ . Show that the converse is also true; that is, show that if $\operatorname {rank}(\mathbf{I} - \mathbf{B}\mathbf{A}) = n - \operatorname {rank}(\mathbf{A})$ , then $\mathbf{B}$ is a generalized inverse of $\mathbf{A}$ .

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Problem 18.23

(a)(b)

Let $\mathbf{A}$ represent the ( $n \times n$ ) projection matrix for a subspace $\mathcal{U}$ of $\mathbb {R}^{n \times 1}$ along a subspace $\mathcal{V}$ of $\mathbb {R}^{n \times 1}$ (where $\mathcal{U} \oplus \mathcal{V} = \mathbb {R}^{n \times 1}$ ), and let $\mathbf{B}$ represent the ( $n \times n$ ) projection matrix for a subspace $\mathcal{W}$ of $\mathbb {R}^{n \times 1}$ along a subspace $\mathcal{X}$ of $\mathbb {R}^{n \times 1}$ (where $\mathcal{W} \oplus \mathcal{X} = \mathbb {R}^{n \times 1}$ ).

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(a)

Show that $\mathbf{A} + \mathbf{B}$ is the projection matrix for some subspace $\mathcal{C}$ of $\mathbb {R}^{n \times 1}$ along some subspace $\mathcal{M}$ of $\mathbb {R}^{n \times 1}$ (where $\mathcal{C} \oplus \mathcal{M} = \mathbb {R}^{n \times 1}$ ) if and only if $\mathbf{B}\mathbf{A} = \mathbf{A}\mathbf{B} = \mathbf{0}$ , in which case $\mathcal{C} = \mathcal{U} \oplus \mathcal{W}$ and $\mathcal{M} = \mathcal{V} \cap \mathcal{X}$ .

(b)

Show that $\mathbf{A} - \mathbf{B}$ is the projection matrix for some subspace $\mathcal{C}$ of $\mathbb {R}^{n \times 1}$ along some subspace $\mathcal{M}$ of $\mathbb {R}^{n \times 1}$ (where $\mathcal{C} \oplus \mathcal{M} = \mathbb {R}^{n \times 1}$ ) if and only if $\mathbf{B}\mathbf{A} = \mathbf{A}\mathbf{B} = \mathbf{B}$ , in which case $\mathcal{C} = \mathcal{U} \cap \mathcal{X}$ and $\mathcal{M} = \mathcal{V} \oplus \mathcal{W}$ .

Remark.

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for Part (b). Observe (in light of the result that a matrix is a projection matrix for one subspace along another if and only if it is idempotent and the result that a matrix, say $\mathbf{K}$ , is idempotent if and only if $\mathbf{I} - \mathbf{K}$ is idempotent) that $\mathbf{A} - \mathbf{B}$ is the projection matrix for some subspace $\mathcal{C}$ along some subspace $\mathcal{M}$ if and only if $\mathbf{I} - (\mathbf{A} - \mathbf{B}) = (\mathbf{I} - \mathbf{A}) + \mathbf{B}$ is the projection matrix for some subspace $\mathcal{C}^{*}$ along some subspace $\mathcal{M}^{*}$ , and then make use of Part (a) and the result that the projection matrix for a subspace $\mathcal{M}^{*}$ along a subspace $\mathcal{C}^{*}$ (where $\mathcal{M}^{*} \oplus \mathcal{C}^{*} = \mathbb {R}^{n \times 1}$ ) equals $\mathbf{I} - \mathbf{H}$ , where $\mathbf{H}$ is the projection matrix for $\mathcal{C}^{*}$ along $\mathcal{M}^{*}$ .

Problem 18.24

(a)(b)

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(a)

Let $\mathbf{B}$ represent an $n \times n$ symmetric matrix, and let $\mathbf{W}$ represent an $n \times n$ symmetric nonnegative definite matrix. Show that

\mathbf{W}\mathbf{B}\mathbf{W}\mathbf{B}\mathbf{W} = \mathbf{W}\mathbf{B}\mathbf{W} \Leftrightarrow (\mathbf{B}\mathbf{W})^{3} = (\mathbf{B}\mathbf{W})^{2} \Leftrightarrow \operatorname {tr}[(\mathbf{B}\mathbf{W})^{2}] = \operatorname {tr}[(\mathbf{B}\mathbf{W})^{3}] = \operatorname {tr}[(\mathbf{B}\mathbf{W})^{4}].

(b)

Let $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ represent $n \times n$ matrices, let $\mathbf{V}$ represent an $n \times n$ symmetric nonnegative definite matrix, and define $\mathbf{A} = \mathbf{A}_{1} + \cdots + \mathbf{A}_{k}$ . If $\mathbf{V}\mathbf{A}_{i}\mathbf{V}\mathbf{A}_{i}\mathbf{V} = \mathbf{V}\mathbf{A}_{i}\mathbf{V}$ for all $i$ and if $\mathbf{V}\mathbf{A}_{i}\mathbf{V}\mathbf{A}_{j}\mathbf{V} = \mathbf{0}$ for all $i$ and $j \neq i$ , then $\mathbf{V}\mathbf{A}\mathbf{V}\mathbf{A}\mathbf{V} = \mathbf{V}\mathbf{A}\mathbf{V}$ and $\operatorname {rank}(\mathbf{V}\mathbf{A}_{1}\mathbf{V}) + \cdots + \operatorname {rank}(\mathbf{V}\mathbf{A}_{k}\mathbf{V}) = \operatorname {rank}(\mathbf{V}\mathbf{A}\mathbf{V})$ . Conversely, if $\mathbf{V}\mathbf{A}\mathbf{V}\mathbf{A}\mathbf{V} = \mathbf{V}\mathbf{A}\mathbf{V}$ , then each of the following three conditions implies the other two: (1) $\mathbf{V}\mathbf{A}_{i}\mathbf{V}\mathbf{A}_{j}\mathbf{V} = \mathbf{0}$ (for $j \neq i = 1, \ldots , k$ ) and $\operatorname {rank}(\mathbf{V}\mathbf{A}_{i}\mathbf{V}\mathbf{A}_{i}\mathbf{V}) = \operatorname {rank}(\mathbf{V}\mathbf{A}_{i}\mathbf{V})$ (for $i = 1, \ldots , k$ ); (2) $\mathbf{V}\mathbf{A}_{i}\mathbf{V}\mathbf{A}_{i}\mathbf{V} = \mathbf{V}\mathbf{A}_{i}\mathbf{V}$ (for $i = 1, \ldots , k$ ); (3) $\operatorname {rank}(\mathbf{V}\mathbf{A}_{1}\mathbf{V}) + \cdots + \operatorname {rank}(\mathbf{V}\mathbf{A}_{k}\mathbf{V}) = \operatorname {rank}(\mathbf{V}\mathbf{A}\mathbf{V})$ . Indicate how, in the special case where $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ are symmetric, the conditions $\mathbf{V}\mathbf{A}\mathbf{V}\mathbf{A}\mathbf{V} = \mathbf{V}\mathbf{A}\mathbf{V}$ and $\mathbf{V}\mathbf{A}_{i}\mathbf{V}\mathbf{A}_{i}\mathbf{V} = \mathbf{V}\mathbf{A}_{i}\mathbf{V}$ can be reexpressed by applying the results of Part (a) (of the current exercise).

Problem 18.25

(a)(b)

Let $\mathbf{R}$ represent an $n \times q$ matrix, $\mathbf{S}$ an $n \times m$ matrix, $\mathbf{T}$ an $m \times p$ matrix, and $\mathbf{U}$ a $p \times q$ matrix.

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(a)

Show that

\operatorname {rank}(\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}) = \operatorname {rank}\begin{pmatrix} \mathbf{R} & -\mathbf{S}\mathbf{T} \\ \mathbf{T}\mathbf{U} & \mathbf{T} \end{pmatrix} - \operatorname {rank}(\mathbf{T}). \tag {E.3}

(b)

Let $\mathbf{E}_{\mathbf{R}} = \mathbf{I} - \mathbf{R}\mathbf{R}^{-}$ , $\mathbf{F}_{\mathbf{R}} = \mathbf{I} - \mathbf{R}^{-}\mathbf{R}$ , $\mathbf{X} = \mathbf{E}_{\mathbf{R}}\mathbf{S}\mathbf{T}$ , $\mathbf{Y} = \mathbf{T}\mathbf{U}\mathbf{F}_{\mathbf{R}}$ , $\mathbf{E}_{\mathbf{Y}} = \mathbf{I} - \mathbf{Y}\mathbf{Y}^{-}$ , $\mathbf{F}_{\mathbf{X}} = \mathbf{I} - \mathbf{X}^{-}\mathbf{X}$ , $\mathbf{Q} = \mathbf{T} + \mathbf{T}\mathbf{U}\mathbf{R}^{-}\mathbf{S}\mathbf{T}$ , and $\mathbf{Z} = \mathbf{E}_{\mathbf{Y}}\mathbf{Q}\mathbf{F}_{\mathbf{X}}$ . Use the result of Part (b) of Exercise 10.10 to show that

\operatorname {rank}(\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}) = \operatorname {rank}(\mathbf{R}) + \operatorname {rank}(\mathbf{X}) + \operatorname {rank}(\mathbf{Y}) + \operatorname {rank}(\mathbf{Z}) - \operatorname {rank}(\mathbf{T}).

Problem 18.26

Show that, for any $m \times n$ matrices $\mathbf{A}$ and $\mathbf{B}$ ,

\operatorname {rank}(\mathbf{A} + \mathbf{B}) \geq \left|\operatorname {rank}(\mathbf{A}) - \operatorname {rank}(\mathbf{B})\right|.

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Problem 18.27

Show that, for any $n \times n$ symmetric nonnegative definite matrices $\mathbf{A}$ and $\mathbf{B}$ ,

C(\mathbf{A} + \mathbf{B}) = C(\mathbf{A}, \mathbf{B}), \qquad \mathcal{R}(\mathbf{A} + \mathbf{B}) = \mathcal{R}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix},

\operatorname {rank}(\mathbf{A} + \mathbf{B}) = \operatorname {rank}(\mathbf{A}, \mathbf{B}) = \operatorname {rank}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix}.

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Problem 18.28

(a)(b)

Let $\mathbf{A}$ and $\mathbf{B}$ represent $m \times n$ matrices.

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(a)

Show that (1) $C(\mathbf{A}) \subset C(\mathbf{A} + \mathbf{B})$ if and only if $\operatorname {rank}(\mathbf{A}, \mathbf{B}) = \operatorname {rank}(\mathbf{A} + \mathbf{B})$ and (2) $\mathcal{R}(\mathbf{A}) \subset \mathcal{R}(\mathbf{A} + \mathbf{B})$ if and only if $\operatorname {rank}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix} = \operatorname {rank}(\mathbf{A} + \mathbf{B})$ .

(b)

Show that (1) if $\mathcal{R}(\mathbf{A})$ and $\mathcal{R}(\mathbf{B})$ are essentially disjoint, then $C(\mathbf{A}) \subset C(\mathbf{A} + \mathbf{B})$ and (2) if $C(\mathbf{A})$ and $C(\mathbf{B})$ are essentially disjoint, then $\mathcal{R}(\mathbf{A}) \subset \mathcal{R}(\mathbf{A} + \mathbf{B})$ .

Problem 18.29

Let $\mathbf{A}$ and $\mathbf{B}$ represent $m \times n$ matrices. Show that each of the following five conditions is necessary and sufficient for rank additivity [i.e., for $\operatorname {rank}(\mathbf{A} + \mathbf{B}) = \operatorname {rank}(\mathbf{A}) + \operatorname {rank}(\mathbf{B})$ ]:

?

(a)

$\operatorname {rank}(\mathbf{A}, \mathbf{B}) = \operatorname {rank}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix} = \operatorname {rank}(\mathbf{A}) + \operatorname {rank}(\mathbf{B})$ ;

(b)

$\operatorname {rank}(\mathbf{A}) = \operatorname {rank}[\mathbf{A}(\mathbf{I} - \mathbf{B}^{-}\mathbf{B})] = \operatorname {rank}[(\mathbf{I} - \mathbf{B}\mathbf{B}^{-})\mathbf{A}]$ ;

(c)

$\operatorname {rank}(\mathbf{B}) = \operatorname {rank}[\mathbf{B}(\mathbf{I} - \mathbf{A}^{-}\mathbf{A})] = \operatorname {rank}[(\mathbf{I} - \mathbf{A}\mathbf{A}^{-})\mathbf{B}]$ ;

(d)

$\operatorname {rank}(\mathbf{A}) = \operatorname {rank}[\mathbf{A}(\mathbf{I} - \mathbf{B}^{-}\mathbf{B})]$ and $\operatorname {rank}(\mathbf{B}) = \operatorname {rank}[(\mathbf{I} - \mathbf{A}\mathbf{A}^{-})\mathbf{B}]$ ;

(e)

$\operatorname {rank}(\mathbf{A}) = \operatorname {rank}[(\mathbf{I} - \mathbf{B}\mathbf{B}^{-})\mathbf{A}]$ and $\operatorname {rank}(\mathbf{B}) = \operatorname {rank}[\mathbf{B}(\mathbf{I} - \mathbf{A}^{-}\mathbf{A})]$ .

Problem 18.30

Let $\mathbf{A}$ and $\mathbf{B}$ represent $m \times n$ matrices. And, let

\mathbf{H} = \left[\mathbf{I} - \begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix}^{-}\right] \begin{pmatrix} \mathbf{A} & \mathbf{0} \\ \mathbf{0} & -\mathbf{B} \end{pmatrix} [\mathbf{I} - (\mathbf{A}, \mathbf{B})^{-}(\mathbf{A}, \mathbf{B})].

?

(a)

Show that

\begin{aligned} \operatorname {rank}(\mathbf{A} - \mathbf{B}) & = \operatorname {rank}(\mathbf{A}) - \operatorname {rank}(\mathbf{B}) + [\operatorname {rank}(\mathbf{A}, \mathbf{B}) - \operatorname {rank}(\mathbf{A})] \\ & \qquad + \left[\operatorname {rank}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix} - \operatorname {rank}(\mathbf{A})\right] + \operatorname {rank}(\mathbf{H}). \end{aligned}

(b)

Show that $\mathbf{A}$ and $\mathbf{B}$ are rank subtractive [in the sense that $\operatorname {rank}(\mathbf{A} - \mathbf{B}) = \operatorname {rank}(\mathbf{A}) - \operatorname {rank}(\mathbf{B})$ ] if and only if $\operatorname {rank}(\mathbf{A}, \mathbf{B}) = \operatorname {rank}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix} = \operatorname {rank}(\mathbf{A})$ and $\mathbf{H} = \mathbf{0}$ .

(c)

Show that if $\operatorname {rank}(\mathbf{A}, \mathbf{B}) = \operatorname {rank}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix} = \operatorname {rank}(\mathbf{A})$ , then (1) $(\mathbf{A}^{-}, \mathbf{0})$ and $\begin{pmatrix} \mathbf{A}^{-} \\ \mathbf{0} \end{pmatrix}$ are generalized inverses of $\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix}$ and $(\mathbf{A}, \mathbf{B})$ , respectively, and (2) for $\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix}^{-} = (\mathbf{A}^{-}, \mathbf{0})$ and $(\mathbf{A}, \mathbf{B})^{-} = \begin{pmatrix} \mathbf{A}^{-} \\ \mathbf{0} \end{pmatrix}$ , $\mathbf{H} = \begin{pmatrix} \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{B}\mathbf{A}^{-}\mathbf{B} - \mathbf{B} \end{pmatrix}$ .

(d)

Show that each of the following three conditions is necessary and sufficient for rank subtractivity [i.e., for $\operatorname {rank}(\mathbf{A} - \mathbf{B}) = \operatorname {rank}(\mathbf{A}) - \operatorname {rank}(\mathbf{B})$ ]:

$\operatorname {rank}(\mathbf{A}, \mathbf{B}) = \operatorname {rank}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix} = \operatorname {rank}(\mathbf{A})$ and $\mathbf{B}\mathbf{A}^{-}\mathbf{B} = \mathbf{B}$ ;
$C(\mathbf{B}) \subset C(\mathbf{A})$ , $\mathcal{R}(\mathbf{B}) \subset \mathcal{R}(\mathbf{A})$ , and $\mathbf{B}\mathbf{A}^{-}\mathbf{B} = \mathbf{B}$ ;
$\mathbf{A}\mathbf{A}^{-}\mathbf{B} = \mathbf{B}\mathbf{A}^{-}\mathbf{A} = \mathbf{B}\mathbf{A}^{-}\mathbf{B} = \mathbf{B}$ .

(e)

Using the result of Exercise 29 (or otherwise), show that $\operatorname {rank}(\mathbf{A} - \mathbf{B}) = \operatorname {rank}(\mathbf{A}) - \operatorname {rank}(\mathbf{B})$ if and only if $\operatorname {rank}(\mathbf{A} - \mathbf{B}) = \operatorname {rank}[\mathbf{A}(\mathbf{I} - \mathbf{B}^{-}\mathbf{B})] = \operatorname {rank}[(\mathbf{I} - \mathbf{B}\mathbf{B}^{-})\mathbf{A}]$ .

Remark.

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Do so by applying (with $-\mathbf{B}$ in place of $\mathbf{B}$ ) the formula

\operatorname {rank}(\mathbf{A} + \mathbf{B}) = \operatorname {rank}(\mathbf{A}, \mathbf{B}) + \operatorname {rank}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix} - \operatorname {rank}(\mathbf{A}) - \operatorname {rank}(\mathbf{B}) + \operatorname {rank}(\mathbf{K}), \tag {*}

where

\mathbf{K} = \left[\mathbf{I} - \begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix}\begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix}^{-}\right] \begin{pmatrix} \mathbf{A} & \mathbf{0} \\ \mathbf{0} & \mathbf{B} \end{pmatrix} [\mathbf{I} - (\mathbf{A}, \mathbf{B})^{-}(\mathbf{A}, \mathbf{B})].

Problem 18.31

Let $\mathbf{A}_{1}, \ldots , \mathbf{A}_{k}$ represent $m \times n$ matrices. Adopting the terminology of Exercise 17.6, use Part (a) of that exercise to show that if $\mathcal{R}(\mathbf{A}_{1}), \ldots , \mathcal{R}(\mathbf{A}_{k})$ are independent and $C(\mathbf{A}_{1}), \ldots , C(\mathbf{A}_{k})$ are independent, then $\operatorname {rank}(\mathbf{A}_{1} + \cdots + \mathbf{A}_{k}) = \operatorname {rank}(\mathbf{A}_{1}) + \cdots + \operatorname {rank}(\mathbf{A}_{k})$ .

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Problem 18.32

(a)(b)(c)(d)

Let $\mathbf{T}$ represent an $m \times p$ matrix, $\mathbf{U}$ an $m \times q$ matrix, $\mathbf{V}$ an $n \times p$ matrix, and $\mathbf{W}$ an $n \times q$ matrix, and define $\mathbf{Q} = \mathbf{W} - \mathbf{V}\mathbf{T}^{-}\mathbf{U}$ . Further, let $\mathbf{E}_{\mathbf{T}} = \mathbf{I} - \mathbf{T}\mathbf{T}^{-}$ , $\mathbf{F}_{\mathbf{T}} = \mathbf{I} - \mathbf{T}^{-}\mathbf{T}$ , $\mathbf{X} = \mathbf{E}_{\mathbf{T}}\mathbf{U}$ , and $\mathbf{Y} = \mathbf{V}\mathbf{F}_{\mathbf{T}}$ .

?

(a)

Show that

\operatorname {rank}\begin{pmatrix} \mathbf{T} & \mathbf{U} \\ \mathbf{V} & \mathbf{W} \end{pmatrix} = \operatorname {rank}(\mathbf{T}) + \operatorname {rank}\begin{pmatrix} \mathbf{0} & \mathbf{X} \\ \mathbf{Y} & \mathbf{Q} \end{pmatrix}. \tag {E.4}

(b)

Show that

\operatorname {rank}\begin{pmatrix} \mathbf{0} & \mathbf{V} \\ \mathbf{U} & \mathbf{T} \end{pmatrix} = \operatorname {rank}(\mathbf{T}) + \operatorname {rank}\begin{pmatrix} -\mathbf{V}\mathbf{T}^{-}\mathbf{U} & \mathbf{Y} \\ \mathbf{X} & \mathbf{0} \end{pmatrix}.

(c)

Show that

\operatorname {rank}\begin{pmatrix} \mathbf{0} & \mathbf{V} \\ \mathbf{U} & \mathbf{T} \end{pmatrix} = \operatorname {rank}(\mathbf{T}) + \operatorname {rank}(\mathbf{X}) + \operatorname {rank}(\mathbf{Y}) + \operatorname {rank}(\mathbf{E}_{\mathbf{Y}}\mathbf{V}\mathbf{T}^{-}\mathbf{U}\mathbf{F}_{\mathbf{X}}),

where $\mathbf{E}_{\mathbf{Y}} = \mathbf{I} - \mathbf{Y}\mathbf{Y}^{-}$ and $\mathbf{F}_{\mathbf{X}} = \mathbf{I} - \mathbf{X}^{-}\mathbf{X}$ .

(d)

Show that

\begin{aligned} \operatorname {rank}\begin{pmatrix} \mathbf{T} & \mathbf{U} \\ \mathbf{V} & \mathbf{W} \end{pmatrix} & = \operatorname {rank}(\mathbf{T}) + \operatorname {rank}(\mathbf{Q}) + \operatorname {rank}(\mathbf{A}) + \operatorname {rank}(\mathbf{B}) \\ & \qquad + \operatorname {rank}[(\mathbf{I} - \mathbf{A}\mathbf{A}^{-})\mathbf{X}\mathbf{Q}^{-}\mathbf{Y}(\mathbf{I} - \mathbf{B}^{-}\mathbf{B})], \end{aligned}

where $\mathbf{A} = \mathbf{X}(\mathbf{I} - \mathbf{Q}^{-}\mathbf{Q})$ and $\mathbf{B} = (\mathbf{I} - \mathbf{Q}\mathbf{Q}^{-})\mathbf{Y}$ .

Remark.

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for Part (b). Observe that (since the rank of a matrix is not affected by a permutation of rows or columns) $\operatorname {rank}\begin{pmatrix} \mathbf{0} & \mathbf{V} \\ \mathbf{U} & \mathbf{T} \end{pmatrix} = \operatorname {rank}\begin{pmatrix} \mathbf{T} & \mathbf{U} \\ \mathbf{V} & \mathbf{0} \end{pmatrix}$ , and make use of Part (a).

for Part (c). Use Part (b) in combination with the result that, for any $r \times s$ matrix $\mathbf{A}$ , $r \times v$ matrix $\mathbf{B}$ , and $u \times s$ matrix $\mathbf{C}$ ,

\operatorname {rank}\begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{0} \end{pmatrix} = \operatorname {rank}(\mathbf{B}) + \operatorname {rank}(\mathbf{C}) + \operatorname {rank}[(\mathbf{I} - \mathbf{B}\mathbf{B}^{-})\mathbf{A}(\mathbf{I} - \mathbf{C}^{-}\mathbf{C})]. \tag {*}

for Part (d). Use Part (c) in combination with Part (a).

Problem 18.33

Let $\mathbf{R}$ represent an $n \times q$ matrix, $\mathbf{S}$ an $n \times m$ matrix, $\mathbf{T}$ an $m \times p$ matrix, and $\mathbf{U}$ a $p \times q$ matrix, and define $\mathbf{Q} = \mathbf{T} + \mathbf{T}\mathbf{U}\mathbf{R}^{-}\mathbf{S}\mathbf{T}$ . Further, let $\mathbf{E}_{\mathbf{R}} = \mathbf{I} - \mathbf{R}\mathbf{R}^{-}$ , $\mathbf{F}_{\mathbf{R}} = \mathbf{I} - \mathbf{R}^{-}\mathbf{R}$ , $\mathbf{X} = \mathbf{E}_{\mathbf{R}}\mathbf{S}\mathbf{T}$ , $\mathbf{Y} = \mathbf{T}\mathbf{U}\mathbf{F}_{\mathbf{R}}$ , $\mathbf{A} = \mathbf{X}(\mathbf{I} - \mathbf{Q}^{-}\mathbf{Q})$ , $\mathbf{B} = (\mathbf{I} - \mathbf{Q}\mathbf{Q}^{-})\mathbf{Y}$ . Use the result of Part (d) of Exercise 32 in combination with the result of Part (a) of Exercise 25 to show that

\begin{aligned} \operatorname {rank}(\mathbf{R} + \mathbf{S}\mathbf{T}\mathbf{U}) & = \operatorname {rank}(\mathbf{R}) + \operatorname {rank}(\mathbf{Q}) \\ & \qquad - \operatorname {rank}(\mathbf{T}) + \operatorname {rank}(\mathbf{A}) + \operatorname {rank}(\mathbf{B}) \\ & \qquad + \operatorname {rank}[(\mathbf{I} - \mathbf{A}\mathbf{A}^{-})\mathbf{X}\mathbf{Q}^{-}\mathbf{Y}(\mathbf{I} - \mathbf{B}^{-}\mathbf{B})]. \end{aligned}

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