# M216: Exercise sheet 6

## Warmup questions

• 1. Let $$f:X\to X$$ be a map of sets. Show that if $$f$$ is injective then $$f^k$$ is injective for each $$k\in \N$$.

• 2. Let $$U_1\leq U_2\leq \dots \leq V$$ be an increasing sequence of subspaces of $$V$$, so that $$U_m\leq U_n$$ whenever $$m\leq n$$.

Show that $$\bigcup _{n\in \N }U_n\leq V$$.

## Homework

• 3. Let $$\phi =\phi _A\in L(\C ^3)$$ where $$A$$ is given by

\begin{equation*} \begin{pmatrix*}[r] 0&1&-1\\-10&-2&5\\-6&2&1 \end {pmatrix*}. \end{equation*}

• (a) Compute the characteristic and minimum polynomials of $$\phi$$.

• (b) Find bases for the eigenspaces and generalised eigenspaces of $$\phi$$.

• 4. Let $$\phi =\phi _A\in L(\C ^3)$$ where $$A$$ is given by

\begin{equation*} \begin{pmatrix*} 0&0&0\\4&0&0\\0&0&5 \end {pmatrix*}. \end{equation*}

• (a) Compute the characteristic and minimum polynomials of $$\phi$$.

• (b) Find bases for the eigenspaces and generalised eigenspaces of $$\phi$$.

## Extra questions

• 5. Let $$\phi \in L(V)$$ be an invertible linear operator on a finite-dimensional vector space and $$\lambda$$ an eigenvalue of $$\phi$$. Show that $$G_{\phi }(\lambda )=G_{\phi ^{-1}}(\lambda ^{-1})$$.

• 6. Let $$\lambda \in \F$$ and define $$J(\lambda ,n)\in M_n(\F )$$ by

\begin{equation*} J(\lambda ,n)= \begin{pmatrix} \lambda &1&0&\dots &0\\ &\ddots &\ddots &\ddots &\vdots \\ &&\ddots &\ddots &0\\ &&&\ddots &1\\ 0&&&&\lambda \end {pmatrix}. \end{equation*}

Set $$J_n:=J(0,n)$$.

Prove:

• (a) $$\ker J_n^k=\Span {\lst {e}1k}$$.

• (b) $$\im J_n^k=\Span {\lst {e}1{n-k}}$$.

• (c) $$m_{J(\lambda ,n)}=\pm \Delta _{J(\lambda ,n)}=(x-\lambda )^n$$.

• (d) $$\lambda$$ is the only eigenvalue of $$J(\lambda ,n)$$ and $$E_{J(\lambda ,n)}(\lambda )=\Span {e_1}$$, $$G_{J(\lambda ,n)}(\lambda )=\F ^{n}$$.

Please hand in at 4W level 1 by NOON on Friday 17th November

# M216: Exercise sheet 6—Solutions

• 1. Let $$x,y\in X$$ be such that $$f^k(x)=f^k(y)$$. Thus $$f(f^{k-1}(x))=f(f^{k-1}(y))$$ whence, since $$f$$ is injective, $$f^{k-1}(x)=f^{k-1}(y)$$. Repeat the argument to eventually conclude that $$x=y$$ so that $$f^k$$ is injective. For a more formal argument, induct on $$k$$.

• 2. Let $$U=\bigcup _{n\in \N }U_n$$ and let $$v,w\in U$$. Then there are $$n,m\in \N$$ with $$v\in U_n$$ and $$w\in U_m$$. Without loss of generality, assume that $$m\leq n$$ so that $$U_m\leq U_n$$ whence $$v,w\in U_n$$. Since $$U_n$$ is a subspace, $$v+\lambda w\in U_n\subseteq U$$, for any $$\lambda \in \F$$, so that $$U$$ is indeed a subspace.

• 3.

• (a) We compute the characteristic polynomial: $$\Delta _{\phi }=\Delta _A=-x^{3}-x^{2}+8x+12=(3-x)(x+2)^{2}$$. Consequently, $$m_{\phi }$$ is either $$(x-3)(x+2)^{2}$$ or $$(x-3)(x+2)$$. We try the latter:

\begin{align*} A-3I_3&= \begin{pmatrix*}[r] -3&1&-1\\-10&-5&5\\-6&2&-2 \end {pmatrix*}& A+2I_{3}&= \begin{pmatrix*}[r] 2&1&-1\\-10&0&5\\-6&2&3 \end {pmatrix*} \end{align*} so that

\begin{equation*} (A-3I_3)(A+2I_3)= \begin{pmatrix*}[r] -10&-5&5\\0&0&0\\-20&-10&10 \end {pmatrix*}\neq 0. \end{equation*}

Thus $$m_{\phi }=m_A=(x-3)(x+2)^2$$.

• (b) We deduce that $$G_{\phi }(3)=E_{\phi }(3)=\ker (A-3I_3)$$ while $$E_{\phi }(-2)=\ker (A+2I_3)$$ and $$G_{\phi }(-2)=\ker (A+2I_3)^2$$. We compute these: an eigenvector $$x$$ with eigenvalue 3 solves

\begin{align*} -3x_1+x_3-x_3&=0\\ -2x_1-x_2+x_3&=0 \end{align*} which rapidly yields $$x_1=0$$ and $$x_2=x_3$$. Thus the $$3$$-eigenspace is spanned by $$(0,1,1)$$.

An eigenvector $$x$$ with eigenvalue $$2$$ solves

\begin{align*} 2x_1+x_2-x_3&=0\\ -2x_1+0x_2+x_3&=0 \end{align*} giving $$x_2=0$$ and $$2x_1=x_3$$ so the eigenspace is spanned by $$(1,0,2)$$.

Finally,

\begin{equation*} (A+2I_3)^{2}= \begin{pmatrix*}[r] 0&0&0\\-50&0&25\\-50&0&25 \end {pmatrix*} \end{equation*}

with kernel spanned by $$(1,0,2)$$ and $$(0,1,0)$$.

To summarise:

\begin{align*} E_{\phi }(3)=G_{\phi }(3)&=\Span {(0,1,1)}\\ E_{\phi }(-2)&=\Span {(1,0,2)}\\ G_{\phi }(-2)&=\Span {(1,0,2),(0,1,0)}. \end{align*}

• 4.

• (a) Since $$A$$ is lower triangular, we immediately see that $$\Delta _{\phi }=\Delta _A=x^{2}(x-5)$$. So the only possibilities for $$m_{\phi }=x(x-5)$$ and $$x^2(x-5)$$. However

\begin{equation*} A-5I_3= \begin{pmatrix*}[r] -5&0&0\\4&-5&0\\0&0&0 \end {pmatrix*} \end{equation*}

so that

\begin{equation*} A(A-5I_3)= \begin{pmatrix*}[r] 0&0&0\\-20&0&0\\0&0&0 \end {pmatrix*}\neq 0. \end{equation*}

We conclude that $$m_{\phi }=x^2(x-5)$$.

Alternatively, $$A$$ is block diagonal:

\begin{equation*} A= \begin{pmatrix} 0&0\\4&0 \end {pmatrix}\oplus \begin{pmatrix} 5 \end {pmatrix} \end{equation*}

and the summands clearly have minimum polynomials $$x^2$$ and $$x-5$$ respectively. It follows from a previous sheet that $$m_{\phi }=x^2(x-5)$$.

• (b) We have $$E_{\phi }(5)=G_{\phi }(5)=\Span {(0,0,1)}$$, $$E_{\phi }(0)=\ker A=\Span {(0,1,0)}$$ and finally $$G_{\phi }(0)=\ker A^2=\Span {(1,0,0),(0,1,0)}$$ since

\begin{equation*} A^{2}= \begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&0&25 \end {pmatrix}. \end{equation*}

• 5. Note that $$(\phi -\lambda \id _V)^{n}(v)=0$$ if and only if $$\lambda ^{-n}\phi ^{-n}(\phi -\lambda ^n\id _V)^n(v)=0$$, that is $$(\lambda ^{-1}\id _V-\phi )^n(v)=0$$. Thus $$G_{\phi }(\lambda )=G_{\phi ^{-1}}(\lambda ^{-1})$$.

Here, of course, we need $$\lambda \neq 0$$ but, since $$\phi$$ is invertible, zero is not an eigenvalue.

• 6. Note that $$\phi _{J_n}(x)=(x_2,\dots ,x_n,0)$$ so that $$\phi _{J_n}^k(x)=(x_{k+1},\dots ,x_{n},0,\dots ,0)$$, $$k<n$$ and $$\phi _{J_n}^{n}=0$$.

• (a) It is clear from the above that $$\ker J_n^k=\set {x\in \F ^{n}\st x_{k+1}=\dots =x_n=0}=\Span {\lst {e}1k}$$.

• (b) Similarly, $$\im J_n^k=\set {y\in \F ^n\st y_{n-k+1}=\dots =y_n=0}=\Span {\lst {e}1{n-k}}$$.

• (c) $$J(\lambda ,n)$$ is upper triangular so that $$\Delta _{J(\lambda ,n)}=(\lambda -x)^n$$. Therefore $$m_{J(\lambda ,n)}=(x-\lambda )^{s}$$, for some $$s\leq n$$. However $$(J(\lambda ,n)-\lambda I_n)^{k}=J_n^k\neq 0$$, for $$k<n$$, so that $$m_{J(\lambda ,n)}=(x-\lambda )^{n}$$.

• (d) Finally, it is clear that $$\lambda$$ is the only eigenvalue and the eigenspace is $$\ker (J(\lambda ,n)-\lambda I_{n})=\ker J_n=\Span {e_1}$$ by part (a). Similarly, $$G_{J(\lambda ,n)}(\lambda )=\ker J_n^n=\F ^n$$.