# M216: Exercise sheet 5

## Warmup questions

• 1. Write down matrices $$A\in M_n(\R )$$ of the following forms:

• (a) $$A_1\oplus A_2\oplus A_3$$ with each $$A_i\in M_2(\R )$$.

• (b) $$\oplst {A}15$$ with each $$A_i\in M_1(\R )$$.

• (c) $$A\in M_3(\R )$$ such that $$A$$ is not of the form $$\oplst {A}1k$$ with $$k>1$$.

• 2. Let $$\lst {V}1k\leq V$$ and $$\phi _i\in L(V_i)$$, $$\bw 1ik$$. Suppose that $$V=\oplst {V}1k$$ and set $$\phi =\oplst \phi 1k$$.

• (a) If $$U_i\leq V_i$$, $$\bw 1ik$$, show that the sum $$\plst {U}1k$$ is direct.

• (b) Prove that $$\im \phi =\oplst {\im \phi }1k$$.

## Homework

• 3. Let $$\phi \in L(V)$$ be a linear operator on a vector space $$V$$.

Prove that $$\im \phi ^{k}\geq \im \phi ^{k+1}$$, for all $$k\in \N$$. Moreover, if $$\im \phi ^{k}=\im \phi ^{k+1}$$ then $$\im \phi ^k=\im \phi ^{k+n}$$, for all $$n\in \N$$.

• 4. Compute the characteristic and minimum polynomials of

\begin{equation*} A= \begin{pmatrix*}[r] 1&-5&-7\\1&4&2\\0&1&4 \end {pmatrix*}. \end{equation*}

• 5. Let $$\phi \in L(V)$$ be a linear operator on a vector space $$V$$ and $$v\in V$$, $$k\in \N$$ such that $$\phi ^{k+1}(v)=0$$ but $$\phi ^{k}(v)\neq 0$$.

Show that $$v,\phi (v),\dots ,\phi ^k(v)$$ are linearly independent.

Hint: Induct on $$k$$.

• 6. In the situation of Question 2, prove:

• (a) $$m_{\phi _i}$$ divides $$m_{\phi }$$, for each $$\bw 1ik$$.

• (b) If each $$m_{\phi _i}$$ divides $$p\in \F [x]$$, then $$p(\phi )=0$$.

Thus $$m_{\phi }$$ is the monic polynomial of smallest degree divided by each $$m_{\phi _i}$$. Otherwise said, $$m_{\phi }$$ is the least common multiple of $$m_{\phi _1},\dots ,m_{\phi _k}$$.

Please hand in at 4W level 1 by NOON on Friday 10th November

# M216: Exercise sheet 5—Solutions

• 1. There are a gazillion possibilities.

• (a)

\begin{equation*} \begin{pmatrix} 1&2&0&0&0&0\\ 2&4&0&0&0&0\\ 0&0&5&6&0&0\\ 0&0&7&8&0&0\\ 0&0&0&0&9&0\\ 0&0&0&0&1&2 \end {pmatrix}= \begin{pmatrix} 1&2\\3&4 \end {pmatrix}\oplus \begin{pmatrix} 5&6\\7&8 \end {pmatrix}\oplus \begin{pmatrix} 9&0\\1&2 \end {pmatrix}. \end{equation*}

• (b) Any $$5\times 5$$ diagonal matrix will do:

\begin{equation*} \begin{pmatrix} \lambda _1&0&0&0&0\\ 0&\lambda _2&0&0&0\\ 0&0&\lambda _3&0&0\\ 0&0&0&\lambda _4&0\\ 0&0&0&0&\lambda _5 \end {pmatrix}= (\lambda _1)\oplus \dots \oplus (\lambda _5). \end{equation*}

• (c) Any block matrix with more than one block will have zeros so

\begin{equation*} A= \begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&1 \end {pmatrix} \end{equation*}

cannot be written $$\oplst {A}1k$$ with $$k>1$$.

• 2.

• (a) Let $$u\in \plst {U}1k$$ so we can write $$u=\plst {u}1k$$, with $$u_i\in U_i\leq V_i$$. However, since the sum of the $$V_i$$ is direct, there is only one way to write $$u$$ as a sum of elements of the $$V_i$$ and so, in particular, as a sum of elements of the $$U_i$$. Thus the sum of the $$U_i$$ is direct.

• (b) Let $$v\in \im \phi$$ so that $$v=\phi (w)$$, for some $$w\in V$$. Then, writing $$w=\plst {w}1k$$ with each $$w_i\in V_i$$, we have

\begin{equation*} v=\phi (w)=\phi _1(w_1)+\dots +\phi _k(w_k)\in \oplst {\im \phi }1k. \end{equation*}

Thus $$\im \phi \leq \oplst {\im \phi }1k$$.

For the converse, let $$v\in \oplst {\im \phi }1k$$ so that $$v=\phi _1(w_1)+\dots +\phi _k(w_k)$$ with $$w_i\in V_i$$, $$\bw 1ik$$. Since each $$\phi _i=\phi \restr {V_i}$$, this reads

\begin{equation*} v=\phi (w_1)+\dots +\phi (w_k)=\phi (\plst {w}1k)\in \im \phi . \end{equation*}

Thus $$\oplst {\im \phi }1k\leq \im \phi$$ and we are done.

• 3. Let $$v\in \im \phi ^{k+1}$$ so that $$v=\phi ^{k+1}(w)$$, for some $$w\in V$$. Then $$v=\phi ^k(\phi (w))\in \im \phi ^k$$. Thus $$\im \phi ^{k}\geq \im \phi ^{k+1}$$.

Suppose now that $$\im \phi ^k=\im \phi ^{k+1}$$. We prove that $$\im \phi ^k=\im \phi ^{k+n}$$ by induction on $$n$$. We are given that this holds for $$n=1$$ so we suppose this holds for some $$n$$ ($$\im \phi ^k=\im \phi ^{k+n}$$) and prove it then holds for $$n+1$$. Thus, let $$v\in \im \phi ^{k}=\im \phi ^{k+1}$$ so that $$v=\phi (\phi ^k(w))$$, for some $$w\in V$$. Then $$\phi ^k(w)\in \im \phi ^k=\im \phi ^{k+n}$$, by the induction hypothesis, so that $$\phi ^k(w)=\phi ^{k+n}(u)$$, some $$u\in V$$, whence $$v=\phi (\phi ^{k+n}(u))=\phi ^{k+n+1}(u)\in \im \phi ^{k+n+1}$$. We conclude that $$\im \phi ^k\leq \im \phi ^{k+n+1}$$. The converse inclusion always holds so we have equality. Induction now bakes the cake.

• 4. We compute the characteristic polynomial of $$A$$ to be

\begin{equation*} \Delta _A=-x^3+9x^2-27x+27=-(x-3)^3. \end{equation*}

We learn from the Cayley–Hamilton theorem that $$m_A=(x-3)^k$$, for some $$k$$ with $$\bw {k}13$$. Clearly $$k=1$$ is out, since $$A$$ is not diagonal, so we try $$k=2$$:

\begin{equation*} (A-3I)^2= \begin{pmatrix*}[r] -2&-5&-7\\1&1&2\\0&1&1 \end {pmatrix*} \begin{pmatrix*}[r] -2&-5&-7\\1&1&2\\0&1&1 \end {pmatrix*} = \begin{pmatrix*}[r] -1&-2&-3\\ -1&-2&-3\\ 1&2&3 \end {pmatrix*}, \end{equation*}

which is non-zero. This means we must have $$m_A=(x-3)^3$$.

• 5. We follow the hint and induct on $$k$$. For $$k=0$$, the assertion is that if $$v\neq 0$$ and $$\phi (v)=0$$ then the one element list $$v$$ is linearly independent which is certainly true (whether or not $$\phi (v)=0$$).

Suppose now that the result holds for $$\ell <k$$ so that $$\phi ^{\ell +1}(w)=0$$ and $$\phi ^{\ell }(w)\neq 0$$ forces $$w,\dots ,\phi ^{\ell }(w)$$ to be linearly independent. Now suppose that $$\phi ^{k+1}(v)=0$$ and $$\phi ^k(v)\neq 0$$. If there are $$\lambda _i\in \F$$ with

$$\label {eq:1} \lambda _1v+\dots +\lambda _k\phi ^k(v)=0,$$

then taking $$\phi$$ of this gives

\begin{equation*} \lambda _1\phi (v)+\dots +\lambda _{k-1}\phi ^{k-1}(\phi (v))=0. \end{equation*}

But with $$w=\phi (v)$$, we have $$\phi ^k(w)=0$$ and $$\phi ^{k-1}(w)\neq 0$$ so the induction hypothesis applies with $$\ell =k-1<k$$ to give that $$w,\dots ,\phi ^{k-1}(w)$$ are linearly independent. Therefore $$\lst \lambda 1{k-1}$$ all vanish. Plugging this back into (1) gives $$\lambda _k\phi ^k(v)=0$$ whence $$\lambda _k=0$$ also. We conclude that $$v,\dots ,\phi ^k(v)$$ are linearly independent and so the result holds for all $$k\in \N$$ by induction.

• 6.

• (a) We have that $$m_{\phi }(\phi )=0$$ so that $$0=m_{\phi }(\phi )\restr {V_i}=m_{\phi }(\phi _i)$$. It follows that $$m_{\phi _i}$$ divides $$m_{\phi }$$.

• (b) Since $$m_{\phi _i}$$ divides $$p$$, $$p(\phi _i)=0$$ for each $$i$$. But then

\begin{equation*} p(\phi )=p(\phi _1)\oplus \dots \oplus p(\phi _k)=0 \end{equation*}

so that $$m_{\phi }$$ divides $$p$$.