# M216: Exercise sheet 4

## Warmup questions

• 1. Let $$p,q\in \R [x]$$ be given by $$p=x^2-2x-3$$, $$q=x^3-2x^2+2x-5$$.

Let $$A\in M_2(\R )$$ and $$B\in M_3(\R )$$ be given by

\begin{equation*} A= \begin{pmatrix*}[r] 1&2\\2&1 \end {pmatrix*}\qquad B= \begin{pmatrix*}[r] 1&2&1\\-2&0&1\\2&1&1 \end {pmatrix*}. \end{equation*}

Compute $$p(A),p(B),q(A),q(B)$$.

• 2. Compute the characteristic polynomials of $$A$$ and $$B$$, from question 1.

What do you notice?

• 3. Let $$\F =\Z _2$$, the field of two elements and let $$p=x^2+x\in \F [x]$$.

Show that $$p(t)=0$$, for all $$t\in \F$$.

## Homework questions

• 4. Compute the minimum polynomial of $$A\in M_5(\R )$$ given by

\begin{equation*} \begin{pmatrix*}[r] 0&0&0&0&-3\\1&0&0&0&6\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0 \end {pmatrix*}. \end{equation*}

• 5. Let $$\phi \in L(V)$$ be an operator on a finite-dimensional vector space over $$\F$$ and let $$p=m_{\phi }\in \F [x]$$.

Let $$\lambda$$ be a root of $$p$$.

• (a) Show there is $$q\in \F [x]$$ with $$\deg q<\deg p$$ such that

\begin{equation*} p=(x-\lambda )q. \end{equation*}

• (b) Prove that $$q(\phi )$$ is non-zero.

• (c) Deduce that $$\lambda$$ is an eigenvalue of $$\phi$$.

This shows that the roots of $$p$$ are exactly the eigenvalues of $$\phi$$ without recourse to the Cayley–Hamilton theorem.

• (d) Deduce that $$\phi$$ is invertible if and only if $$p$$ has non-zero constant term.

## Extra questions

• 6. Let $$\phi \in L(V)$$ have minimal polynomial $$p=4+5x+6x^{2}-7x^3-8x^{4}+x^{5}$$, so that $$\phi$$ is invertible by question 5(d).

Compute the minimal polynomial of $$\phi ^{-1}$$.

Hint: Think about multiplying $$a_0\id _V+\dots +\phi ^n$$ by $$\phi ^{-n}$$.

Please hand in at 4W level 1 by NOON on Friday 3rd November

# M216: Exercise sheet 4—Solutions

• 1. We just compute:

\begin{equation*} A^{2}= \begin{pmatrix*}[r] 5&4\\4&5 \end {pmatrix*},\qquad A^3= \begin{pmatrix*}[r] 13&14\\14&13 \end {pmatrix*} \end{equation*}

so that

\begin{align*} p(A)&=A^2-2A-3I_2=\begin{pmatrix*}[r] 5&4\\4&5 \end {pmatrix*}-2\begin{pmatrix*}[r] 1&2\\2&1 \end {pmatrix*}-3 \begin{pmatrix*}[r] 1&0\\0&1 \end {pmatrix*}= \begin{pmatrix*}[r] 0&0\\0&0 \end {pmatrix*}\\ q(A)&=A^{3}-2A^2+2A-5I_{3}=\begin{pmatrix*}[r] 13&14\\14&13 \end {pmatrix*}-2\begin{pmatrix*}[r] 5&4\\4&5 \end {pmatrix*}+2\begin{pmatrix*}[r] 1&2\\2&1 \end {pmatrix*}-5\begin{pmatrix*}[r] 1&0\\0&1 \end {pmatrix*}= \begin{pmatrix*}[r] 0&10\\10&0 \end {pmatrix*}. \end{align*} Similarly,

\begin{align*} p(B)&= \begin{pmatrix*}[r] -6&-1&2\\4&-6&-3\\-2&3&-1 \end {pmatrix*},\\ q(B)&= \begin{pmatrix*}[r] 0&0&0\\ 0&0&0\\ 0&0&0 \end {pmatrix*}. \end{align*}

• 2. Again, we just compute:

\begin{equation*} \Delta _A= \begin{vmatrix} 1-x&2\\2&1-x \end {vmatrix}=(1-x)^2-4=x^2-2x-3. \end{equation*}

Similarly,

\begin{equation*} \begin{split} \Delta _B= \begin{vmatrix} 1-x&2&1\\-2&-x&1\\2&1&1-x \end {vmatrix}=(1-x)\bigl (x(x-1)-1\bigr ) -2\bigl (2(x-1)-2\bigr ) +(-2+2x)\\ =(-x^3+2x^2-1)-4x+8+2x-2=-x^3+2x^2-2x+5. \end {split} \end{equation*}

We notice that, with $$p,q$$ as in question 1, $$p=\Delta _A$$ and $$q=-\Delta _B$$ and so, again from question 1,

\begin{equation*} \Delta _A(A)=\Delta _B(B)=0. \end{equation*}

As we shall soon see, this is the Cayley–Hamilton theorem in action.

• 3. We recall that $$\Z _2=\set {\bz ,\bo }$$ with addition and multiplication given by

\begin{align*} \bz &=\bz +\bz =\bo +\bo &\bo &=\bz +\bo =\bo +\bz \\ \bz &=\bz \bz =\bz \bo =\bo \bz &\bo &=\bo \bo . \end{align*} We immediately conclude that $$\bo ^{2}+\bo =\bz =\bz ^2+\bz$$ so that $$p(t)=\bo$$, for both $$t\in \F$$.

• 4. Let us compute the first few powers of $$A$$:

\begin{gather*} A^2= \begin{pmatrix*}[r] 0&0&0&-3&0\\0&0&0&6&-3\\1&0&0&0&6\\0&1&0&0&0\\0&0&1&0&0 \end {pmatrix*}\quad A^3= \begin{pmatrix*}[r] 0&0&-3&0&0\\0&0&6&-3&0\\0&0&0&6&-3\\1&0&0&0&6\\0&1&0&0&0 \end {pmatrix*}\quad A^4= \begin{pmatrix*}[r] 0&-3&0&0&0\\0&6&-3&0&0\\0&0&6&-3&0\\0&0&0&6&-3\\1&0&0&0&6 \end {pmatrix*}\\ A^{5}= \begin{pmatrix*}[r] -3&0&0&0&-18\\6&-3&0&0&36\\0&6&-3&0&0\\0&0&6&-3&0\\0&0&0&6&-3 \end {pmatrix*} \end{gather*} Stare at the top row to see that there can be no monic polynomial $$p=a_0\plus {}{x^{k}}$$ with $$k\leq 4$$ with $$p(A)=0$$: the $$-3$$ on the top row of the leading term would give $$a_0\plus 0{a_{k-1}}0-3=0$$. On the other hand, we readily see that $$A^{5}-6A+3I_{5}=0$$ so that $$m_A=x^5-6x+3$$.

• 5.

• (a) The remainder theorem says we can write $$p=(x-\lambda )q+r$$ with $$\deg r<\deg (x-\lambda )=1$$ so that $$r$$ is degree zero and so constant. Evaluating at $$\lambda$$ gives $$0=p(\lambda )=0q+r=r$$ and we are done.

• (b) $$q(\phi )$$ cannot be zero unless $$q=0$$ since $$\deg q<\deg p$$ and $$p$$ is the minimal polynomial of $$\phi$$. But $$q$$ cannot be zero since $$p$$ is non-zero.

• (c) Since $$q(\phi )$$ is non-zero, there is $$v\in V$$ such that $$q(\phi )v\neq 0$$. Now

\begin{equation*} 0=p(\phi )(v)=(\phi -\lambda \id _V)(q(\phi )(v)) \end{equation*}

so that $$q(\phi )v$$ is an eigenvector with eigenvalue $$\lambda$$.

• (d) $$\phi$$ is invertible if and only if $$\phi$$ is injective if and only if zero is not an eigenvalue if and only if (thanks to the previous part) zero is not a root of $$p$$ if and only if $$p$$ has non-zero constant term.

• 6. If $$a_0\id _V+a_1\plus \phi {\phi ^n}=0$$ then, multiplying by $$\phi ^{-n}$$ gives $$a_0\phi ^{-n}+a_1\plus {\phi ^{n-1}}{a_n} \id _V=0$$. In the case at hand, this means that

\begin{equation*} 4\phi ^{-5}+5\phi ^{-4}+6\phi ^{-3}-7\phi ^{-2}-8\phi ^{-1}+\id _V=0. \end{equation*}

If there was a non-zero polynomial $$q=\sum _{k=1}^{4}b_kx^k$$ of lower degree with $$q(\phi ^{-1})=0$$ gives

\begin{equation*} b_4\plus {\id _V}{b_0}\phi ^4=0, \end{equation*}

contradicting the minimality of $$p$$. Thus, dividing by $$4$$ to get a monic polynomial, the minimum polynomial of $$\phi ^{-1}$$ is $$1/4-2x-7/4x^2+3/2x^{3}+5/4x^4+x^5$$.

More generally, the same argument says that if $$\sum _{k=0}^na_kx^k$$ is the minimal polynomial of invertible $$\phi$$ with degree $$n$$ then $$1/a_{0}\sum _{k=0}^{n}a_{n-k}x^k$$ is the minimal polynomial of $$\phi ^{-1}$$.