3.10 Invariant subspace

Suppose $k$ field, $V$ is $k$-vector space, $f:V\to V$ is $k$-linear map

Definition

Given a subspace $S\subseteq V$. We will say that $S$ is $f$-invariant (or invariant under $f$) if $f(S)\subseteq S$

Example

$S=\{0_V\}$ is $f$-invariant $\forall f$
$\ker f,Im f$ is $f$-invariant $\forall f$
If $f=Id_{V_1}$ all subspace are $f$-invariant
$f:V\to V$, $\dim_kV=m$, $\mathcal{B}=\{v_1,...,v_m\}$ and $[f]_{\mathcal{B}}= \begin{pmatrix} a_{11} & \ldots & a_{1m} \\ & \ddots & \vdots \\ & & a_{\operatorname{\mathrm{mm}}} \end{pmatrix}$

$S_1=\lang v_1\rang$, $f(v_1)=a_{11}v_1\in S_1$ which is $f$-invariant

Also $S_j=\lang v_1,...,v_j\rang$ is $f$-invariant
$f=0,S$ is $f$-invariant $\forall S$
$f:k^2\to k^2$ where $f(e_1)=e_2$ and $f(e_2)=e_1$

$\lang e_1\rang$ is not $f$-invariant but $\lang e_1+e_2\rang$ is $f$-invariant
$f:V\to V$ is $k$-linear and $g:V\to V$ is $k$-linear such that $f\circ g=g\circ f$

Let $W=Img,U=\ker g$, then $W$ and $U$ are $f$-invariant

Let $w\in W,\exists v\in V:w=g(v)$, then $f(w)=f(g(v))=f\circ g(v)=g\circ f(v)=g(f(w))\in W$

Let $v\in U=\ker g$, then $g\left(f\left(v\right)\right)=g\circ f\left(v\right)=f\circ g\left(v\right)=0_{V} \Rightarrow f\left(v\right)\in U$
$V=\R^2$ and $\R$ vector space. Fix $0\leq \alpha<2\pi$

$A= \begin{pmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{pmatrix}$, $f_A:\R^2\to \R^2$ rotation with any $\alpha$ to the left

(a subspace is a line, rotate 0 or $\pi$ is invariant since $-v$ in it)

Clearly, $\{0\}$ and $\R^2$ are $f$-invariant. Suppose that $S\subseteq \R^2$ subspace of $\dim=1$ such that $S$ is invariant

Then $S=\lang v_1\rang$ $,v_{1}\neq 0_{V}$ and $f_A(v_1)=\lambda v_1$ for some $\lambda\in \R$

Then assume $v_{1}= \begin{pmatrix} x_{1} \\ y_{1} \end{pmatrix}$ and $\lambda$ is an eigenvalue of $f_A$ with eigenvector $v_1$

Let's calculate $\chi_{f_{A}}(x)=\chi_{A}\left(x\right)=\det \begin{pmatrix} x-\cos\alpha & \sin\alpha \\ -\sin\alpha & x-\cos\alpha \end{pmatrix}=x^{2}-2x\cos\alpha+1$

Thus the roots are $\cos\alpha\pm\sqrt{\cos^2\alpha-1}$, since we have an eigenvalue, thus $\alpha = 0$ or $\pi$

Restriction linear map

Suppose $V$-finite dimension, $f:V\to V,k$-linear. Let $S$ be an $f$-invariant

Let $\mathcal{B}_{1}=\left\lbrace v_{1},\ldots,v_{s}\right\rbrace$ be basis of $S$, we can extend it $\{\mathcal{B}_{1},v_{s+1},\ldots,v_{n}\}$ of $V$

Then $[f]_{\mathcal{B}}=$ where $M=[f|_{S}]_{\mathcal{B_1}}$

Consider the restriction $f|_S$ of $f$ to $S$. $f|_s:S\to S$ is linear map.

If $D$ is invariant, then it is upper triangular

Lemma

Let $W\subseteq V$ be an $f$-invariant subspace, then $\chi_{f|_{W}}(x)\mid\chi_{f}(x)$ and $m_{f|_{W}}(x)\mid m_{f}(x)$

Proof

Main idea: Consider the matrix of restriction linear map, remember definition of $\chi_f$ is about matrix.

Let $\mathcal{B}$ be a basis of $V$ such that

So $\chi_{f|_{W}}(x)\mid\chi_{f}(x)$

Main idea: Remember $m(x)$ is a polynomial, so it can be written as $m(x)=\sum^n_{i=0}a_ix^i$

For all $i\in \N$,

Since $m_{f}(x)=m_{[f]_\mathcal{B}}(x)$ and $m_{[f]_\mathcal{B}}([f]_{\mathcal{B}})=0$, then $m_{[f]_{\mathcal{B}}}(x)=\sum_{j=0}^{r}\alpha_{j}x^{j}$ and $0=m_{[f]_\mathcal{B}}([f]_{\mathcal{B}})=\sum_{j=0}^{r}\alpha_{j}\left\lbrack f\right\rbrack _{\mathcal{B}}^{j}=\sum_{j=0}^{r}\alpha_{j}$($)^j$=

Thus $\sum_{j=0}^{r}\alpha_{i}M^{j}=0$, then $m_{[f]_\mathcal{B}}([f|_W]_\mathcal{B_1})=0$ since representation of matrix

Thus by definition of $m_{f|_W}(x)$, we get that $m_{f|_{W}}(x)\mid m_{f}(x)$

Example

$f: V \to V$, $\dim V = n$. Let $W = \langle \text{all the eigenvectors of } f \rangle$. Assume $W \neq \{0\}$.
Let $c_1, \dots, c_t$ be the different eigenvalues of $f$. Let $B_i$ be a basis of $V_{c_i}$ ($c_i$ eigenspace), $B' = B_1 \cup \dots \cup B_r$ is also basis for $W$.

Given $w \in W$, $w=\sum_{j=1}^{t}\alpha_{j}v_{j}\quad\text{(eigenvectors)}$, then $f(w)=\sum_{j=1}^{t}\alpha_{j}f(v_{j})\quad\text{(since }f(v_{j})=c_{j}v_{j}\text{)} \in W$

So $W$ is $f$-invariant.

$\chi_f(x)=\chi_{f|_w}(x)\cdot P(x)$ some polynomial $=(x-c_{1})^{m_1}\cdot\ldots\cdot(x-c_{r})^{m_{r}}\cdot P\left(x\right)$ where $m_1+...+m_r=\dim W$ and $m_1=\dim V_{c_1}$ .... $m_r=\dim V_{c_r}$

Definition

Let $W\subseteq V$ be an $f$-invariant subspace, let $v\in V\text{ is finite dimension}$, $m_f(f)(v)=0\in W $

Consider $m_{f}(x)\in\underbrace{\left.\{p(x\right)\in k[x]:p(f)(v)\in W\}}_{S_f(v,W)}\neq \left\lbrace0 \right\rbrace$, this is a set of polynomials that send $v\in V\to w\in W$ via $p(x)$ evaluating on $f$-invariant

Remark: If $w=\{0_V\}$, then $S_{f}(v,\{0\})=\{p\left(x\right)\in k[x]:p(f)(v)=0\}$

Lemma

Suppose that $\dim_k V<\infty$, let $f:V\to V$ be a $k$-linear map. Let $W$ be an $f$-invariant subspace, then $W$ is $p(f)$-invariant, $\forall p(x)\in k[x]$

Proof

Let $p(x)=\sum_{i=0}^{t}a_{i}x^{i}\in k[x]$, then $p(f)=\sum_{i=0}^{t}a_{i}f^{i}$. Given $v\in W$, $p(f)(v)=\sum_{i=0}^{t}a_{i}f^{i}\left(v\right)$

Note that if $W$ is $f$-invariant and $v\in W$, then $\forall i\in \N$, $f^i(v)\in W$ by induction

Thus $v\in p(f)(v)$

Corollary

$\forall v\in V,S_f(v,W)$ is closed under sums and by multiplication by any polynomial in $k[x]$

Proof

Suppose $p_1(x),p_2(x)\in S_f(v,W)$, then $p_{1}\left(x\right)=\sum_{i=0}^{t}a_{i}x^{i},p_{2}\left(x\right)=\sum_{j=0}^{t}b _{j}x^{j}$

We know that $p_1(f)(v)\in W$ and $p_2(f)(v)\in W$, then $p_1(f)(v)+p_2(f)(v)\in W$

Then $(p_1+p_2)(f)(v)\in W$, thus $p_1(x)+p_2(x)\in S_f(v,W)$

Let $p(x)\in S_f(v,W)$, that is $p(f)(v)\in W$ and $h(x)\in k[x]$

Let's prove that $h(x)\cdot p(x)\in S_f(v,W)$

Since $(h\cdot p)(f)=h(f)\circ p(f)$, then $(h\cdot p)(f)(v)=\underbrace{h(f)(\underbrace{p(f)(v)}_{\in W})}_{\in W\text{ use lemma}}$

Remark: we have proven that $S_f(v,W)$ is an ideal in the ring $k[x]$

As we did for the definition of $m_f(x)$, we can prove that if we take a monic polynomial $p(x)$ of minimal degree in $S_f(v,W)$ then it divides all other polynomials in $S_f(v,W)$ and it is unique

Proof

Assume $p(x)$ is unique monic polynomial of minimal degree, then $p(f)(v)\in W$ and $h(x)\in k[x]:h(f)(v)\in W$, we are going to prove $p(x)\mid h(x)$

Let's $h(x)=p(x)q(x)+r(x),r(x)=0\text{ or }dgr(x)<dgp(x)$\, then $h(f)(v)=p(f)q(f)(v)+r(f)(v)$

Then $h(f)(v)\in W,q\left(f)\left(v\right)=v^{\prime}\in V\Rightarrow\left.p(f\right)( v^{\prime}\right)\in W$ by lemma

Thus $r(f)(v)\in W$. Here $r(f)(v)=0$. If not it contradicts to $p$ is the minimality, then $p(x)\mid h(x)$

We call this polynomial the $f$-conductor of $v$ into $W$. Note that $p(x)\mid m_f(x)$ since $m_{f}(x)\in S(f,v)$

Lemma

Suppose that $\dim_k V<\infty$, let $f:V\to V$ be a $k$-linear map.

Suppose $m_{f}(x)=(x-c_{1})^{m_1}\cdot....\cdot\left(x-c_{r}\right)^{m_r}$ with $c_1,...,c_r\in k$ and $c_i\neq c_j$ if $i\neq j$ and $m_i\geq 0$

Let $W\subsetneq V$ be an $f$-invariant subspace, then $\exists v\in V:$ 1. $v\notin W$ 2. $(f-c\cdot Id)(v)\in W$ for some $c$ eigenvalue of $f$

Proof

Main idea: Recall $S(v,W)$ because it can send $v\in V$ to a smaller subspace $W$ via $f$-invariant and an important property: $p(x)\mid m_f(x)$ which allowed us to find $x-c$

In other words, we want to prove that $\exists v\notin W:S_f(v,W)$ contains a polynomial of degree 1 such that $x-c$ where $c$ is an eigenvalue of $f$ ($(x-c)(f)(v)\in W$)

Let $z\in V\setminus W$. Let $g(x)$ be the $f$-conduct of $z$ into $W$. Then $g(x)\mid m_f(x)$.

Since $z\notin W$, then $dgg(x)>0$, therefore $g(x)=(x-c_{1})^{e_1}...(x-c_{r})^{e_r}$ with $e_i\leq m_i,\forall i,1\leq i\leq r$ and $e_1+...+e_r>0$

Choose $j$, $1\leq j\leq r$ such that $e_j>0$. Then $x-c_{j}\mid g(x)$

Let $l(x)\in k[x]$ such that $g(x)=(x-c_j)\cdot l(x)$

Since $dgl(x)<dgg(x)$, then we know that $l(x)\notin S_f(z,W)$ that is $\underbrace{l(f)(z)}_{v}\notin W$

So $v\notin W$ and $g(f)(z)\in W$, then $(f-c_jId)(\underbrace{l(f)(z)}_{v})\in W$

Then $(f-c_{j}Id)(v)\in W$