How to find all the invariant subspaces in relation to Matrix

Question

Let $$A=\begin{bmatrix}1 & 1 & 0 & 0\\ -1 & 1 & 2 & 1\\ 0 & 0 & 3 & 1\\ 0 & 0 & -1 & 1 \end{bmatrix}\in M_4(\mathbb C)$$ I need to find all the $A$-invariant subspaces of $\mathbb{C}^{4}$.

I proved a theorem that says: every subspace which is $J_{k}\left(\lambda\right) $ invariant when $\lambda$ is eigenvalue of A is of the form $Span\left(e_{i},\ldots,e_{k}\right)$, and I know how to find the Canonical Jordan form of A, but I don't understand how to use the theorem or if there is another way to find all the invariant subspaces of A.

Answer 1

Too long for a comment:

We know that: a subspace $U$ is invariant under the matrix $A\in M_{n}(F)$ if, and only if, $$\forall u\in U:\quad Au\in U$$ However, as far as I have learned, finding them all is not always an easy task, but if the matrix $A$ is diagonalizable, everything is usually easier.

Notice that, the characteristic polynomial for $A$ is given by $$p_{A}(\lambda)=\lambda^{4}-6\lambda^{3}+14\lambda^{2}-16\lambda+8$$ The polynomial can be factorable over $\mathbb{C}$ but by hypothesis $A\in M_{4}(\mathbb{C})$ so no problem.

$$p_{A}(\lambda)=(\lambda-2)^{2}(\lambda-2\lambda+2)$$ Then the characteristic equation is given by $$p_{A}(\lambda)=0\implies \lambda_{1}=2,\quad \lambda_{2}=1+i,\quad \lambda_{3}=1-i$$ With algebraic multiplicity $2,1,1$ to $2,1+i, 1-i$ respectively. We know $A$ is diagonalizable if, and only if, the algebraic multiplicity of the eigenvalue $\lambda_{i}$ is the same which the geometric multiplicity of the eigenvalue $\lambda_{i}$ for each $i$.

However solving $(A-\lambda_{i}I)=\vec{0}$ for each eigenvalues we get

$$\lambda_{1}=2\longrightarrow v_{1}=\begin{pmatrix}-1/2\\-1/2\\-1\\1\end{pmatrix}\implies 2\not=1$$ $$\lambda_{2}=1+i\longrightarrow v_{2}=\begin{pmatrix} -i\\1\\0\\0\end{pmatrix}\implies 1=1$$ $$\lambda_{3}=1-i\longrightarrow v_{3}=\begin{pmatrix}i\\1\\0\\0\end{pmatrix}\implies 1=1$$ So $A$ is not diagonalizable since the algebraic multiplicity of $\lambda_{1}=2$ is not the same of geometric multiplicity.

Hence in this problem you have to work step-by-step using the definition given above and your theorem also should be useful because give a characterisation for the invariant subspace with the canonical Jordan form $J=\begin{pmatrix} 1-i&0&0&0\\0&1+i&0&0\\0&0&2&1\\0&0&0&2\end{pmatrix}$ where $A=PJP^{-1}$ with $P$ be a invertible matrix.

By other hand,notice for example the subspaces $\{\vec{0}\}$ and $\mathbb{C}^{4}$ they're invariant subspaces. Indeed, $A\vec{0}=\vec{0}\in\{\vec{0}\}$ and $Au=u\in \mathbb{C}^{4}$ for all $u\in\mathbb{C}^{4}$. Now see in the columns span, rows span, etc.

If $M_{4}(\mathbb{C})$ is working over the field $\mathbb{C}$ so $\dim_{\mathbb{C}}\left(M_{4}(\mathbb{C})\right)=4^{2}$ and it should be useful.

Answer 2

If $\ J=P^{-1}AP\ $ is the Jordan canonical form of an $\ n\times n\ $ matrix $\ A\ $, then $$ Av=v\Leftrightarrow JP^{-1}v= P^{-1}v\, $$ so $\ V\ $ is an invariant subspace of $\ A\ $ if and only if $\ P^{-1}V\ $ is an invariant subspace of $ J\ $.

Let $$ J=\pmatrix{J_1&0&\dots&\dots&\dots&0\\ 0&J_2&\dots&\dots&\dots&0\\ \vdots&&\ddots&\vdots&\vdots&\vdots\\ \vdots&&&J_i&&\vdots\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&\dots&\dots&J_k}\ , $$ where each $\ J_i\ $ is a Jordan block of size $\ s_i\ $ corresponding to an eigenvalue $\ \lambda_i\ $. If $\ e_{ij}\ $ is the $\ j^{\,\text{th}}\ $ column of the $\ s_i\times s_i\ $ identity matrix, then $\ \big(J_i-\lambda_iI\big)e_{i1}=0\ $ and $\ \big(J_i-\lambda_iI\big)e_{ir}=e_{i\,r-1}\ $ for $\ r=2,3,\dots,s_i\ $. The matrix $\ J_i\ $ therefore has $\ s_i+1\ $ distinct invariant subspaces—namely $\ \{0\}\ $ and $$ Span\left(\big\{e_{ij}\,\big|\,j=1,2,\dots,r\,\big\}\right) $$ for $\ r=1,2,\dots,s_i\ $. Assume $\ \lambda_i\ne\lambda_j\ $ for $\ i\ne j\ $. The invariant subspaces of $\ J\ $ are then all subspaces of $\ \mathbb{C}^n\ $ of the form $$ W_1\times W_2\times\dots\times W_k\ $$ where each $\ W_i\ $ is an invariant subspace of $\ J_i\ $. There are thus $\ \prod_\limits{i=1}^k\big(s_i+1\big)\ $ distinct invariant subspaces of $\ J\ $, obtained by allowing each $\ W_i\ $ in the above Cartesian product to take on each of its $\ s_i+1\ $ possible values.

If $\ t_1=1\ $ and $\ t_i=1+\sum_\limits{j=1}^{i-1}s_j\ $ for $\ i=2,3,\dots,k\ $ and $\ \pi_{ir}\ $ is the $\ \big(t_i+r-1\big)^\text{th}\ $ column of the matrix $\ P\ $, then $\ \pi_{i1},\pi_{i2},\dots,\pi_{is_i}\ $ are the columns of $\ P\ $ corresponding to the Jordan block $\ J_i\ $. For those columns, we have $\ \big(A-\lambda_iI\big)\pi_{i1}=0\ $ and $\ \big(A-\lambda_iI\big)\pi_{ir}=\pi_{i\,r-1}\ $ for $\ r=2,3,\dots,s_i\ $, and for each invariant subspace of $\ J_i\ $ we get a corresponding invariant subspace $\ \{0\}\ $ or $$ Span\left(\big\{\pi_{ij}\,\big|\,j=1,2,\dots,r\,\big\}\right) $$ of $\ A\ $. The invariant subspaces of $\ A\ $ are all subspaces of $\ \mathbb{C}^n\ $ of the form $$ V_1\oplus V_2\oplus\dots\oplus V_k $$ where each $\ V_i\ $ is an invariant subspace of $\ A\ $ corresponding to the Jordan block $\ J_i\ $.

Event Horizon's answer has given the Jordan canonical form for your $\ A\ $, with three Jordan blocks of sizes $\ 1, 1\ $ and $\ 2\ $ and three of the columns of $\ P\ $: $$ \pi_{11}=\pmatrix{i\\1\\0\\0},\ \pi_{21}=\pmatrix{-i\\1\\0\\0},\ \pi_{31}=\pmatrix{1/2\\1/2\\1\\-1} $$ (where I have inverted the sign of the third column). The fourth column can be any solution, $\ v\ $, of the equation $\ \big(A-2I\big)v=\pi_{31}\ $. I'll take it to be $$ \pi_{32}=\pmatrix{\frac{1}{2}\\1\\1\\0}\ . $$ Using the notation $\ \big\langle v_1,v_2,\dots, v_n\big\rangle\triangleq Span\big(v_1,v_2,\dots, v_n\big)\ $, we can list the twelve invariant subspaces of your matrix $\ A\ $ as \begin{array}{ll} \text{Dim }0&\{0\}\\ \text{Dim }1&\big\langle\pi_{11}\big\rangle,\ \big\langle\pi_{21}\big\rangle,\ \big\langle\pi_{31}\big\rangle\\ \text{Dim }2&\big\langle\pi_{11},\pi_{21}\big\rangle,\ \big\langle\pi_{11},\pi_{31}\big\rangle,\ \big\langle\pi_{21},\pi_{31}\big\rangle ,\ \big\langle\pi_{31},\pi_{32}\big\rangle\\ \text{Dim }3&\big\langle\pi_{11},\pi_{21},\pi_{31}\big\rangle,\ \big\langle\pi_{11},\pi_{31},\pi_{32}\big\rangle,\ \big\langle\pi_{21},\pi_{31},\pi_{32}\big\rangle\\ \text{Dim }4&\big\langle\pi_{11},\pi_{21},\pi_{31},\pi_{32}\big\rangle=\mathbb{C}^4\ . \end{array}