# M216: Exercise sheet 7

## Warmup questions

• 1. Let $$\lst {v}1n$$ be a basis for a vector space and $$\phi \in L(V)$$. Show that the following are equivalent:

• (1) $$\phi (v_1)=0$$ and $$\phi (v_i)=v_{i-1}$$, for $$\bw 2in$$.

• (2) $$v_i=\phi ^{n-i}(v_n)$$ and $$\phi ^n(v_n)=0$$.

• 2. Let $$\phi \in L(V)$$ be a nilpotent linear operator on a finite-dimensional vector space $$V$$ with Jordan normal form $$J_{n_1}\oplus \dots \oplus J_{n_k}$$.

Show that

\begin{equation*} \#\set {i\st n_i= s}=2\dim \ker \phi ^s-\dim \ker \phi ^{s-1}-\dim \ker \phi ^{s+1}. \end{equation*}

## Homework

• 3. Let $$\phi \in L(V)$$ be a linear operator on a finite-dimensional complex vector space $$V$$ with distinct eigenvalues $$\lst {\lambda }1k$$.

Show that $$\phi$$ is diagonalisable if and only if $$m_{\phi }=\prod _{i=1}^k(x-\lambda _i)$$.

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

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

Find the JNF and a Jordan basis for $$\phi$$.

(You have studied $$\phi$$ before in question 3 of sheet 6.)

## Extra questions

• 5. Let $$\phi \in L(V)$$ be a nilpotent linear operator on a finite-dimensional vector space $$V$$ with Jordan normal form $$J_{n_1}\oplus \dots \oplus J_{n_k}$$.

Show that $$m_{\phi }=x^s$$ where $$s=\max \set {\lst {n}1k}$$.

Hint: use question 6 on sheet 5.

• 6. 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*}

Find the JNF and a Jordan basis for $$\phi$$.

(You have studied $$\phi$$ before in question 4 of sheet 6.)

Please hand in at 4W level 1 by NOON on Friday 24th November

# M216: Exercise sheet 7—Solutions

• 1. Assume (1). Then $$v_{n-1}=\phi (v_n)$$ and induction gives $$v_i=\phi ^{n-i}(v_n)$$. In particular, $$v_1=\phi ^{n-1}(v_n)$$ so that $$\phi (v_1)$$ gives $$\phi ^n(v_n)=0$$. This establishes (2).

Assume (2). Then $$0=\phi ^n(v_n)=\phi (\phi ^{n-1}(v_n))=\phi (v_1)$$. Moreover $$\phi (v_i)=\phi (\phi ^{n-i}(v_n))=\phi ^{n-(i-1)}(v_n)=v_{i-1}$$, for $$\bw 2in$$. Thus we have (1).

• 2. From lectures, we know that, for $$s\geq 1$$,

\begin{equation*} \#\set {i\st n_i\geq s}=\dim \ker \phi ^s-\dim \ker \phi ^{s-1}. \end{equation*}

Now

\begin{align*} \#\set {i\st n_i=s}&=\#\set {i\st n_i\geq s}-\#\set {i\st n_i\geq s+1}\\ &=\dim \ker \phi ^s-\dim \ker \phi ^{s-1}-(\dim \ker \phi ^{s+1}-\dim \ker \phi ^{s})\\ &= 2\dim \ker \phi ^s-\dim \ker \phi ^{s-1}-\dim \ker \phi ^{s+1}. \end{align*}

• 3. Let $$p=\prod _{i=1}^k(x-\lambda _i)\in \C [x]$$.

If $$\phi$$ is diagonalisable then $$p(\phi )=0$$ since $$\phi -\lambda _i\id _{V}=0$$ on $$E_{\phi }(\lambda _i)$$ and $$V$$ is the direct sum of these eigenspaces.

Conversely, if $$m_{\phi }=\prod _{i=1}^k(x-\lambda _i)$$, a result from lectures tells us that all Jordan blocks in the Jordan normal form of $$\phi$$ have size $$1$$. Otherwise said, the JNF is diagonal and the Jordan basis is an eigenbasis. So $$\phi$$ is diagonalisable.

• 4. From question 3 of sheet 6, we know that $$m_{\phi }=(x-3)(x+2)^2$$ so that the JNF of $$\phi$$ must be $$J(3,1)\oplus J(-2,2)$$. A Jordan basis of $$G_{\phi }(3)=E_{\phi }(3)$$ is an arbitrary basis and one is given by $$(0,1,1)$$ as we found out in question 3 of sheet 6.

For $$G_{\phi }(-2)$$, we want $$(\phi +2\id _{\C ^3})(v),v$$ with $$(\phi +2\id _{\C ^3})^{2}(v)=0$$ so work backwards from an eigenvector $$w$$ with eigenvalue $$-2$$ and then solve $$(A+2I_3)\mathbf {v}=\mathbf {w}$$ to get $$v$$. We know from sheet 6 that we can take $$w=(1,0,2)$$ and then

\begin{equation*} (A+2I_{3})\mathbf {v}= \begin{pmatrix*}[r] 2&1&-1\\-10&0&5\\-6&2&3 \end {pmatrix*}\mathbf {v}= \begin{pmatrix} 1\\0\\2 \end {pmatrix} \end{equation*}

is clearly solved by $$(0,1,0)$$. Our Jordan basis is therefore $$(0,1,1),(1,0,2),(0,1,0)$$.

• 5. We follow the hint which tells us that when $$A=\oplst {A}1k$$, $$m_{A}$$ is the smallest monic polynomial divided by each $$m_{A_i}$$.

In the case at hand, with $$A=J_{n_1}\oplus \dots \oplus J_{n_k}$$, we know from sheet 6 that $$m_{J_{n_i}}=x^{n_i}$$. Thus $$m_A$$ is the monic polynomial of smallest degree divided by each $$x^{n_i}$$ which is $$x^s$$, for $$s=\max \set {\lst {n}1k}$$.

• 6. From sheet 6, we have that $$m_{\phi }=x^2(x-5)$$ so that the JNF is $$J(0,2)\oplus J(5,1)$$. A Jordan basis for $$G_{\phi }(5)$$ is any non-zero eigenvector with eigenvalue $$5$$. We know from sheet 6 that $$(0,0,1)$$ is such a vector.

For $$G_{\phi }(0)$$, a Jordan basis is $$\phi (v),v$$ with $$\phi ^2(v)=0$$. As usual, work backwards from $$w\in \ker \phi$$: take $$w=(0,1,0)$$ and solve

\begin{equation*} A\mathbf {v}=\begin{pmatrix*} 0&0&0\\4&0&0\\0&0&5 \end {pmatrix*}\mathbf {v}= \begin{pmatrix} 0\\1\\0 \end {pmatrix} \end{equation*}

to get $$v=(1/4,0,0)$$, for example.

Thus, a Jordan basis is given by $$(0,1,0),(1/4,0,0),(0,0,1)$$.