# M216: Exercise sheet 8

## Warmup questions

• 1. Let $$\lst \alpha 1k$$ span $$E\leq V^{*}$$. Show that

\begin{equation*} \sol E=\bigcap _{i=1}^k\ker \alpha _i. \end{equation*}

• 2. Define $$\alpha ,\beta \in (\R ^3)^{*}$$ be given by

\begin{align*} \alpha (x)&=2x_{1}+x_2-x_3\\ \beta (x)&=x_1-x_2+x_3, \end{align*} for $$x\in \R ^3$$.

Let $$E=\Span {\alpha ,\beta }$$ and compute $$\sol E$$.

## Homework

• 3. Let $$A,B\in M_4(\C )$$ be given by

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

Compute the Jordan normal forms of $$A$$ and $$B$$.

Are $$A$$ and $$B$$ similar?

• 4. Let $$U\leq V$$ and $$v\in V$$ with $$v\notin U$$. Show that there is $$\alpha \in V^{*}$$ such that $$\alpha$$ is zero on $$U$$ but $$\alpha (v)\neq 0$$.

Hint: Apply theorem 5.3 to $$V/U$$.

## Extra questions

• 5. Let $$V$$ be a vector space over a field $$\F$$ and let $$\alpha ,\beta \in V^{*}$$ be non-zero linear functionals.

Prove that $$\ker \alpha =\ker \beta$$ if and only there is non-zero $$\lambda \in \F$$ such that $$\alpha =\lambda \beta$$.

Hint: If $$v_0\notin \ker \alpha$$, show that $$V=\Span {v_0}+\ker \alpha$$.

• 6. Let $$V$$ be a vector space over $$\F$$. For $$v\in V$$, define $$\ev (v):V^{*}\to \F$$ by

\begin{equation*} \ev (v)(\alpha )=\alpha (v). \end{equation*}

• (a) Show that $$\ev (v)$$ is linear so that $$\ev (v)\in V^{**}$$.

• (b) We therefore have a map $$\ev :V\to V^{**}$$. Show that $$\ev$$ is linear.

• (c) Show that $$\ev$$ is injective.

• (d) Deduce that if $$V$$ is finite-dimensional then $$\ev :V\to V^{**}$$ is an isomorphism.

Please hand in at 4W level 1 by NOON on Friday December 1st

# M216: Exercise sheet 8—Solutions

• 1. Let $$v\in \sol E$$ so that $$\alpha (v)=0$$, for all $$\alpha \in E$$. Then, in particular, each $$\alpha _i(v)=0$$ so that $$v\in \ker \alpha _i$$, for $$\bw 1ik$$. That is, $$v\in \bigcap _{i=1}^k\ker \alpha _i$$ and $$\sol E\leq \bigcap _{i=1}^k\ker \alpha _i$$.

Conversely, let $$v\in \bigcap _{i=1}^k\ker \alpha _i$$ so that $$\alpha _i(v)=0$$, for $$\bw 1ik$$. Let $$\alpha \in E$$. Then $$\alpha =\sum _{i=1}^k\lambda _i\alpha _i$$, for some $$\lst \lambda 1k\in \F$$, since the $$\alpha _i$$ span $$E$$, and

\begin{equation*} \alpha (v)=\sum _{i=1}^k\lambda _i\alpha _i(v)=0 \end{equation*}

so that $$v\in \sol E$$. Thus $$\bigcap _{i=1}^k\ker \alpha _i\leq \sol E$$ and we are done.

• 2. According to question 1, $$\sol E$$ consists of those $$x\in \R ^3$$ such that $$\alpha (x)=\beta (x)=0$$, that is, such that

\begin{align*} 2x_{1}+x_2-x_3&=0\\x_1-x_2+x_3&=0. \end{align*} Adding these gives $$3x_1=0$$ and then the first gives $$x_2=x_3$$ so that $$\sol E=\Span {(0,1,1)}$$.

• 3. Both being upper triangular, we see that $$\Delta _A=\Delta _B=x^4$$ so that the only eigenvalue of $$A$$ or $$B$$ is $$0$$. Moreover, we compute to see that $$A^2=B^2=0$$ so that $$m_A=x^2$$. Thus both $$A$$ and $$B$$ have at least one $$2\times 2$$ Jordan block $$J_{2}$$. Thus the possibilities for the Jordan normal form of either are $$J_2\oplus J_2$$ or $$J_{2}\oplus J_1\oplus J_1$$. To distinguish these, recall that the number of Jordan blocks with eigenvalue 0 is the dimension of the kernel. Now $$A$$ has clearly has row rank $$1$$ and so $$3$$-dimensional kernel. Thus $$A$$ has Jordan normal form $$J_2\oplus J_1\oplus J_1$$.

Meanwhile $$B$$ has row rank $$2$$, thus nullity $$2$$ so that it has JNF $$J_2\oplus J_2$$.

Since they have different JNF, $$A$$ and $$B$$ are not similar.

• 4. Let $$q:V\to V/U$$ be the quotient map so that $$q$$ is a linear surjection with kernel $$U$$ (this is all we need to know about the quotient construction). Since $$v\notin U$$, $$q(v)\neq 0$$ so that, by the Sufficiency Principle (Theorem 5.3), there is $$\beta \in (V/U)^{*}$$ such that $$\beta (q(v))\neq 0$$.

Let $$\alpha =\beta \circ q:V\to \F$$. This is linear, being a composition of linear maps, so $$\alpha \in V^{*}$$. Moreover, $$\alpha (v)=\beta (q(v))\neq 0$$ while, if $$u\in U$$, $$q(u)=0$$ so that $$\alpha (u)=\beta (0)=0$$.

• 5. The reverse implication is clear: if $$\lambda \neq 0$$ and $$\alpha =\lambda \beta$$ then $$\alpha (v)=0$$ if and only if $$\lambda \alpha (v)=\beta (v)=0$$.

Now suppose that $$\ker \alpha =\ker \beta$$ with $$\alpha \neq 0$$. Thus there is $$v_0\in V$$ such that $$\alpha (v_0)\neq 0$$. Following the hint, let $$v\in V$$ and observe that $$v-(\alpha (v)/\alpha (v_0))v_0\in \ker \alpha$$ so that $$V=\Span {v_0}+\ker \alpha$$.

Now, since $$v_0\notin \ker \alpha =\ker \beta$$, $$\beta (v_0)\neq 0$$ also. Set $$\lambda =\alpha (v_0)/\beta (v_0)$$ so that

\begin{equation*} \alpha (v_0)=\lambda \beta (v_0). \end{equation*}

Further $$\alpha (v)=\lambda \beta (v)$$, for all $$v\in \ker \alpha$$, since both sides are zero. It follows that $$\alpha =\lambda \beta$$ on $$\Span {v_0}+\ker \alpha =V$$.

• 6. This is a case of thinking carefully what each statement means after which it will be very easy to prove.

• (a) To see that $$\ev (v):V^{*}\to \F$$ is linear, we must show that

\begin{equation*} \ev (v)(\alpha +\lambda \beta )=\ev (v)(\alpha )+\lambda \ev (v)(\beta ), \end{equation*}

for all $$\alpha ,\beta \in V^{*}$$ and $$\lambda \in \F$$. Using the definition of $$\ev (v)$$, this reads

\begin{equation*} (\alpha +\lambda \beta )(v)=\alpha (v)+\lambda \beta (v) \end{equation*}

which is exactly the definition of the (pointwise) addition and scalar multiplication in $$V^{*}$$.

• (b) Linearity of $$\ev :V\to V^{**}$$ means that for $$v,w\in V$$ and $$\lambda \in \F$$, we have

\begin{equation*} \ev (v+\lambda w)=\ev (v)+\lambda \ev (w). \end{equation*}

This is supposed to be equality of elements of $$V^{**}$$, that is to say, equality of two functions on $$V^{*}$$. This holds when the two functions give the same answers on any $$\alpha \in V^{*}$$ so we need

\begin{equation*} \ev (v+\lambda w)(\alpha )=\ev (v)(\alpha )+\lambda \ev (w)(\alpha ). \end{equation*}

However, using the definition of $$\ev$$, this reads

\begin{equation*} \alpha (v+\lambda w)=\alpha (v)+\lambda \alpha (w) \end{equation*}

which is true since $$\alpha$$ is linear!

• (c) $$\ev$$ is injective if and only if $$\ker \ev =\set 0$$. Let $$v\in \ker \ev$$. Thus $$\ev (v)=0\in V^{**}$$, the zero functional on $$V^{*}$$. Otherwise said, $$\ev (v)(\alpha )=0$$, for all $$\alpha \in V^{*}$$, or equivalently, $$\alpha (v)=0$$, for all $$\alpha \in V^{*}$$. But the Sufficiency Principle now forces $$v=0$$ so that $$\ev$$ injects.

• (d) If $$v$$ is finite-dimensional, $$\dim V=\dim V^{*}=\dim V^{**}$$ so that $$\ev$$ is an isomorphism by rank-nullity since we have just seen that it injects.