Math Problems Vault

Problem V.3.1

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

The matrix

A=\left\| \begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right\| .

is given.

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(a)

Find its rank and some basis submatrix.

(b)

Find the coefficients of the expansion of the non-basis row in the basis rows and of the non-basis column in the basis columns.

(c)

Add the second row of the matrix to the first and verify that the linear dependence among the columns has remained the same.

(d)

How many basis submatrices are there in this matrix in all?

Problem V.3.2

A square matrix of order $n$ has a zero submatrix of order $n-1$ . Estimate the rank of the matrix.

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Problem V.3.3

A matrix $A$ of order $n$ contains a zero submatrix of size $m \times k$ , where $m+k>n$ . Prove that $A$ is singular.

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Problem V.3.4

(a)(b)

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(a)

By means of elementary row transformations reduce to the reduced form the matrix

\left\| \begin{array}{lllll} 1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 & 7 \\ 5 & 6 & 7 & 8 & 9 \end{array}\right\| .

(b)

Expand the fifth column of this matrix in the first and second columns.

Problem V.3.5

Let $A$ be a matrix with entries $a_{i j}, i=1, \ldots , m$ ; $j=1, \ldots , n$ and $\operatorname {Rg} A=1$ . Prove that there exist a nonzero column $\mathbf{a}=\left\| \alpha_{1}, \ldots , \alpha_{m}\right\|$ and a nonzero row $\mathbf{b}=\left\| \beta_{1}, \ldots , \beta_{n}\right\|$ , such that $a_{i j}=\alpha_{i} \beta_{j}$ for all $i$ and $j$ .

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Problem V.3.6

$B$ a matrix of rank $r$ there are marked $r$ linearly independent rows and $r$ linearly independent columns. Prove that at their intersection there stands a minor different from zero. Show by an example that the statement is false if the number of marked rows is less than $r$ .

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Problem V.3.7

Prove that for any matrices $A$ and $B$ of the same size the rank of the sum is not greater than the sum of the ranks.

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Problem V.3.8

Let the columns of a matrix $A$ of size $m \times n$ be linearly independent and let the product $A B$ be defined. Prove that $\operatorname {Rg} A B=\operatorname {Rg} B$ .

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