Triangular form and invariant flags

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Let TT represent a linear transformation from an nn-dimensional linear space V\mathcal{V} into V\mathcal{V}.

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(a)

Let U\mathcal{U} represent an rr-dimensional subspace of V\mathcal{V}, and suppose that U\mathcal{U} is invariant relative to TT. Show that there exists a basis BB for V\mathcal{V} such that the matrix representation of TT with respect to BB and BB is of the (upper block-triangular) form (EF0H)\begin{pmatrix} \mathbf{E} & \mathbf{F} \\ \mathbf{0} & \mathbf{H} \end{pmatrix} (where E\mathbf{E} is of dimensions r×rr \times r).

(b)

Let U\mathcal{U} and W\mathcal{W} represent subspaces of V\mathcal{V} such that U⊕W=V\mathcal{U} \oplus \mathcal{W} = \mathcal{V} (i.e., essentially disjoint subspaces of V\mathcal{V} whose sum is V\mathcal{V}). Suppose that both U\mathcal{U} and W\mathcal{W} are invariant relative to TT. Show that there exists a basis BB for V\mathcal{V} such that the matrix representation of TT with respect to BB and BB is of the (block-diagonal) form \diag(E,H)\diag (\mathbf{E}, \mathbf{H}) [where the dimensions of E\mathbf{E} equal dim⁡(W)\dim (\mathcal{W})].

Suppose F=C,V\mathbf{F} = \mathbb {C}, V is finite-dimensional, and T∈L(V)T \in \mathcal{L}(V). Prove that if k∈{1,…,dim⁡V}k \in \left\{ 1, \ldots , \operatorname {dim} V\right\}, then VV has a kk-dimensional subspace invariant under TT.

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Suppose T∈L(V)T \in \mathcal{L}(V) and v1,…,vnv_{1}, \ldots , v_{n} is a basis of VV. Show that the following are equivalent.

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(a)

The matrix of TT with respect to v1,…,vnv_{1}, \ldots , v_{n} is lower triangular.

(b)

span⁡(vk,…,vn)\operatorname {span}\left(v_{k}, \ldots , v_{n}\right) is invariant under TT for each k=1,…,nk = 1, \ldots , n.

(c)

Tvk∈span⁡(vk,…,vn)T v_{k} \in \operatorname {span}\left(v_{k}, \ldots , v_{n}\right) for each k=1,…,nk = 1, \ldots , n.

Remark.
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A square matrix is called lower triangular if all entries above the diagonal are 00.

What form does the matrix of a linear transformation of an nn-dimensional vector space have, if the basis of an invariant subspace is formed by:

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(1)

the first kk basis vectors

(2)

the last n−kn - k basis vectors?

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(1)

Let a linear transformation φ\varphi of an nn-dimensional vector space L\mathcal{L} possess a chain of pairwise distinct invariant subspaces nested in one another: L1⊂L2⊂…⊂Ln=L\mathcal{L}_{1} \subset \mathcal{L}_{2} \subset \ldots \subset \mathcal{L}_{n} = \mathcal{L}. Prove that there exists in L\mathcal{L} a basis in which the matrix of the transformation φ\varphi is upper triangular.

(2)

Let, in the basis e1,…,ene_{1}, \ldots , e_{n}, the matrix of a linear transformation φ\varphi of the space L\mathcal{L} be upper triangular. Prove that the subspaces Lk=L{e1,…,ek}(k=1,…,n)\mathcal{L}_{k} = \mathcal{L}\left\{ e_{1}, \ldots , e_{k}\right\} \quad (k = 1, \ldots , n) are invariant with respect to φ\varphi and Lk⊂Lk+1(k=1,…,n−1)\mathcal{L}_{k} \subset \mathcal{L}_{k + 1} \quad (k = 1, \ldots , n - 1).

A linear transformation of the space R3\mathcal{R}_{3} is given by the matrix AA in the standard basis. Reduce the matrix of the transformation to triangular form, if:

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(1)

A=A241A = \mathbf{A}_{241}. A241=[6−115−554−99]\mathbf{A}_{241} = \left[\begin{smallmatrix} 6 & -1 & 1 \\ 5 & -5 & 5 \\ 4 & -9 & 9 \end{smallmatrix}\right].

(2)

A=A222A = \mathbf{A}_{222}. A222=[111222−3−3−3]\mathbf{A}_{222} = \left[\begin{smallmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ -3 & -3 & -3 \end{smallmatrix}\right].

(3)

A=A283A = \mathbf{A}_{283}. A283=[251−1−30−2−3−2]\mathbf{A}_{283} = \left[\begin{smallmatrix} 2 & 5 & 1 \\ -1 & -3 & 0 \\ -2 & -3 & -2 \end{smallmatrix}\right].

(4)

A=A263A = \mathbf{A}_{263}. A263=[3−313−22−120]\mathbf{A}_{263} = \left[\begin{smallmatrix} 3 & -3 & 1 \\ 3 & -2 & 2 \\ -1 & 2 & 0 \end{smallmatrix}\right].

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(1)

Let L1⊂L2⊂…⊂Lr=L\mathcal{L}_{1} \subset \mathcal{L}_{2} \subset \ldots \subset \mathcal{L}_{r} = \mathcal{L} be a chain of subspaces of a vector space L\mathcal{L}, invariant with respect to a linear transformation φ,dim⁡Li=ni(n1<n2<……<nr=n)\varphi , \operatorname {dim} \mathcal{L}_{i} = n_{i}\left(n_{1} < n_{2} < \ldots \ldots < n_{r} = n\right). Suppose the basis e1,…,ene_{1}, \ldots , e_{n} is chosen so that the vectors e1,…,enie_{1}, \ldots , e_{n_{i}} belong to Li(i=1,…,r)\mathcal{L}_{i}(i = 1, \ldots , r). Show that the matrix AφA_{\varphi } is upper block-triangular with diagonal blocks of size ki×kik_{i} \times k_{i}, where ki=ni−ni−1(i=2,…,r)k_{i} = n_{i} - n_{i - 1}(i = 2, \ldots , r), k1=n1k_{1} = n_{1}.

(2)

Let, in some basis, the matrix of a linear transformation be upper block-triangular. Prove that the transformation possesses a chain of invariant subspaces. Express their dimensions in terms of the orders of the diagonal blocks.

Prove that in an nn-dimensional complex space every linear operator has an invariant subspace of dimension n−1n-1.

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Let A\mathcal{A} be a linear operator on a finite-dimensional real space VV. Prove that VV has a basis in which the matrix of the operator A\mathcal{A} has the block upper-triangular form

[A1∗A2⋱0An], \left[\begin{smallmatrix} \boxed {A_{1}} & & & * \\ & \boxed {A_{2}} & & \\ & & \ddots & \\ 0 & & & \boxed {A_{n}} \end{smallmatrix}\right],

where the square blocks A1,…,AnA_{1}, \ldots , A_{n} have size at most 2.

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Let GG be a connected solvable topological group and ρ\rho a continuous homomorphism of GG into the group of nonsingular linear operators on a finite-dimensional complex space VV.

Prove that:

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(a)

VV has a nonzero vector that is an eigenvector for all the operators ρ(g),g∈G\rho (g), g \in G;

(b)

VV has a basis e1,…,ene_{1}, \ldots , e_{n} such that all the matrices ρ(g)\rho (g), g∈Gg \in G, are upper triangular in this basis.

Let FF be an algebraically closed field and GG a solvable group of nonsingular linear operators on a finite-dimensional vector space VV over FF. Prove that there exist a basis e1,…,ene_{1}, \ldots , e_{n} of VV and a normal subgroup NN of GG of finite index (depending only on nn) such that NN consists of upper triangular matrices.

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Let T\mathbf{T} and Q\mathbf{Q} be the matrices

T=(−2−1−5−2−90−8−2231153−5−13−7)andQ=(100−1113−4−20103−1−43). \mathbf{T} = \begin{pmatrix} -2 & -1 & -5 & -2 \\ -9 & 0 & -8 & -2 \\ 2 & 3 & 11 & 5 \\ 3 & -5 & -13 & -7 \end{pmatrix} \quad \text{and} \quad \mathbf{Q} = \begin{pmatrix} 1 & 0 & 0 & -1 \\ 1 & 1 & 3 & -4 \\ -2 & 0 & 1 & 0 \\ 3 & -1 & -4 & 3 \end{pmatrix}.
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(a)

Explain why the columns of Q\mathbf{Q} are a basis for R4\mathbb {R}^{4}.

(b)

Verify that X=span⁡{Q∗1,Q∗2}\mathcal{X} = \operatorname {span}\left\{ \mathbf{Q}_{*1}, \mathbf{Q}_{*2}\right\} and Y=span⁡{Q∗3,Q∗4}\mathcal{Y} = \operatorname {span}\left\{ \mathbf{Q}_{*3}, \mathbf{Q}_{*4}\right\} are each invariant subspaces under T\mathbf{T}.

(c)

Describe the structure of Q−1TQ\mathbf{Q}^{-1}\mathbf{T}\mathbf{Q} without doing any computation.

(d)

Now compute the product Q−1TQ\mathbf{Q}^{-1}\mathbf{T}\mathbf{Q} to determine [T/X]{Q∗1,Q∗2}[\mathbf{T}_{/\mathcal{X}}]_{\left\{ \mathbf{Q}_{*1}, \mathbf{Q}_{*2}\right\} } and [T/Y]{Q∗3,Q∗4}[\mathbf{T}_{/\mathcal{Y}}]_{\left\{ \mathbf{Q}_{*3}, \mathbf{Q}_{*4}\right\} }.

Let T\mathbf{T} be a linear operator on a space V\mathcal{V}, and suppose that

B={u→1,…,u→r,  w→1,…,w→q} \mathcal{B} = \left\{ \overrightarrow {u}_1, \ldots , \overrightarrow {u}_r, \; \overrightarrow {w}_1, \ldots , \overrightarrow {w}_q\right\}

is a basis for V\mathcal{V} such that [T]B[\mathbf{T}]_{\mathcal{B}} has the block-diagonal form

[T]B=(Ar×r00Cq×q). [\mathbf{T}]_{\mathcal{B}} = \begin{pmatrix} \mathbf{A}_{r \times r} & \mathbf{0} \\ \mathbf{0} & \mathbf{C}_{q \times q} \end{pmatrix}.

Explain why U=span⁡{u→1,…,u→r}\mathcal{U} = \operatorname {span}\left\{ \overrightarrow {u}_1, \ldots , \overrightarrow {u}_r\right\} and W=span⁡{w→1,…,w→q}\mathcal{W} = \operatorname {span}\left\{ \overrightarrow {w}_1, \ldots , \overrightarrow {w}_q\right\} must each be invariant subspaces under T\mathbf{T}.

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If Tn×n\mathbf{T}_{n \times n} and Pn×n\mathbf{P}_{n \times n} are matrices such that

P−1TP=(Ar×r00Cq×q), \mathbf{P}^{-1}\mathbf{T}\mathbf{P} = \begin{pmatrix} \mathbf{A}_{r \times r} & \mathbf{0} \\ \mathbf{0} & \mathbf{C}_{q \times q} \end{pmatrix},

explain why U=span⁡{P∗1,…,P∗r}\mathcal{U} = \operatorname {span}\left\{ \mathbf{P}_{*1}, \ldots , \mathbf{P}_{*r}\right\} and W=span⁡{P∗r+1,…,P∗n}\mathcal{W} = \operatorname {span}\left\{ \mathbf{P}_{*r+1}, \ldots , \mathbf{P}_{*n}\right\} are each invariant subspaces under T\mathbf{T}.

Recall (Example 4.8.3): if T\mathbf{T} is regarded as the linear operator on Rn×1\mathbb {R}^{n \times 1} defined by matrix multiplication T(x→)=Tx→\mathbf{T}(\overrightarrow {x}) = \mathbf{T}\overrightarrow {x}, then with respect to the standard basis S\mathcal{S}, [T]S=T[\mathbf{T}]_{\mathcal{S}} = \mathbf{T}.

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For complementary subspaces XX and YY of Rn\mathbb {R}^{n}, let PP be the projector onto XX along YY, and let Q=[X∣Y]Q = \left[X\mid Y\right], where the columns of XX and YY constitute bases for XX and YY, respectively. Prove that if Q−1An×nQQ^{-1}A_{n\times n}Q is partitioned as Q−1AQ=(A11A12A21A22)Q^{-1}AQ = \begin{pmatrix} A_{11}& A_{12} \\ A_{21}& A_{22} \end{pmatrix}, then

Q(A11000)Q−1=PAP,Q(0A1200)Q−1=PA(I−P),Q(00A210)Q−1=(I−P)AP,Q(000A22)Q−1=(I−P)A(I−P). Q\begin{pmatrix} A_{11}& 0 \\ 0& 0 \end{pmatrix}Q^{-1} = PAP, \quad Q\begin{pmatrix} 0& A_{12} \\ 0& 0 \end{pmatrix}Q^{-1} = PA(I-P), \quad Q\begin{pmatrix} 0& 0 \\ A_{21}& 0 \end{pmatrix}Q^{-1} = (I-P)AP, \quad Q\begin{pmatrix} 0& 0 \\ 0& A_{22} \end{pmatrix}Q^{-1} = (I-P)A(I-P).
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Remark.
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This means that if AA is a linear operator on Rn\mathbb {R}^{n}, and B=BX∪BYB = B_{X}\cup B_{Y} (the bases making up the columns of X,YX,Y), then the matrix representation of AA with respect to BB is [A]B=(A11A12A21A22)[A]_{B} = \begin{pmatrix} A_{11}& A_{12}\\ A_{21}& A_{22} \end{pmatrix}, where A11=[PAP/X]BXA_{11} = \left[PA P/X\right]_{B_{X}}, A12=[PA(I−P)/Y]BYBXA_{12} = \left[PA(I-P)/Y\right]_{B_{Y}B_{X}}, A21=[(I−P)AP/X]BXBYA_{21} = \left[(I-P)AP/X\right]_{B_{X}B_{Y}}, A22=[(I−P)A(I−P)/Y]BYA_{22} = \left[(I-P)A(I-P)/Y\right]_{B_{Y}} (each a matrix representation of a restricted operator).

Characterize the matrices of operators А in the basis e1,…,ene_{1}, \ldots , e_{n} such that

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(1)

the subspace ⟨e1,…,ek⟩\left\langle e_{1}, \ldots , e_{k}\right\rangle is invariant under A\mathbf{A};

(2)

the subspace ⟨ek+1,…,en⟩\left\langle e_{k+1}, \ldots , e_{n}\right\rangle is invariant under A\mathbf{A}.

Prove that a linear operator on a space VV of dimension nn over an algebraically closed field always has an invariant subspace of dimension n−1n-1.

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Show that in a space VV of dimension nn over an algebraically closed field every operator А has invariant subspaces of all dimensions from 0 to nn.

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References

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