# M216: Exercise sheet 10

## Warmup questions

• 1. Show that the following are bilinear maps:

• (a) Matrix multiplication $$M_{m\times n}(\F )\times M_{n\times p}(\F )\to M_{m\times p}(\F )$$.

• (b) Evaluation $$(\phi ,v)\mapsto \phi (v):L(V,W)\times V\to W$$.

• (c) For $$\alpha \in V^{*}$$ and $$w\in W$$, define $$\phi _{\alpha ,w}:V\to W$$ by

\begin{equation*} \phi _{\alpha ,w}(v)=\alpha (v)w. \end{equation*}

• (i) Show that each $$\phi _{\alpha ,w}$$ is linear.

• (ii) Show that the map $$t:V^{*}\times W\to L(V,W)$$ given by $$t(\alpha ,w)=\phi _{\alpha ,w}$$ is bilinear.

• 2. Let $$B:V\times V\to \F$$ be a symmetric bilinear form with diagonalising basis $$\lst {v}1n$$. Suppose that, for some $$v_i$$, $$\bw 1in$$, we have $$B(v_i,v_i)=0$$. Prove that $$v_i\in \rad B$$.

• 3. Let $$B:V\times V\to \F$$ be a real symmetric bilinear form with diagonalising basis $$\lst {v}1n$$. Show that $$B$$ is positive definite if and only if $$B(v_i,v_i)>0$$, for all $$\bw 1{i}n$$.

• 4. Let $$A,B\in M_{n\times n}(\F )$$ be congruent: $$B=P^TAP$$, for some $$P\in \mathrm {GL}(n,\F )$$.

Are the following statements true or false?

• (a) $$\det A=\det B$$.

• (b) $$A$$ is symmetric if and only if $$B$$ is symmetric.

## Rank and signature

• 5. Let $$B=B_A:\R ^4\times \R ^4\to \R$$ where

\begin{equation*} A= \begin{pmatrix} 0&2&1&0\\2&0&0&1\\1&0&0&2\\0&1&2&0 \end {pmatrix}. \end{equation*}

Diagonalise $$B$$ and hence, or otherwise, compute its signature.

• 6. Diagonalise the symmetric bilinear form $$B:\R ^3\times \R ^3\to \R$$ given by $$B(x,y)=x_1y_1+x_1y_2+x_2y_1+2x_2y_{2}+x_2y_3+x_3y_2+x_3y_3$$.

Hence, or otherwise, compute the rank and signature of $$B$$.

• 7. Compute the rank and signature of the quadratic form $$Q(x)=x_1x_2-4x_3x_4$$ on $$\R ^4$$.

December 15, 2023

# M216: Exercise sheet 10—Solutions

• 1.

• (a) The bilinearity amounts to:

\begin{align*} A(C+\lambda D)&=AC+\lambda AD\\ (A+\lambda B)C&=AC+\lambda BC, \end{align*} for all $$A,B\in M_{m\times n}(\F )$$, $$C,D\in M_{n\times p}(\F )$$ and $$\lambda \in \F$$. Both of these are easy to prove. For example,

\begin{multline*} (A(C+\lambda D))_{ij}=\sum _{k=1}^nA_{ik}(C+\lambda D)_{kj}=\sum _{k=1}^nA_{ik}(C_{kj}+\lambda D_{kj})\\ =\sum _{k=1}^n(A_{ik}C_{kj}+\lambda A_{ik}D_{kj}) = (AC)_{ij}+\lambda (AD)_{ij}=(AC+ \lambda AD)_{ij}. \end{multline*}

\begin{align*} (\phi _1+\lambda \phi _{2})(v)&=\phi _1(v)+\lambda \phi _2(v)\\ \phi (u+\lambda v)&=\phi (u)+\lambda \phi (v), \end{align*} for all $$\phi ,\phi _1,\phi _2\in L(V,W)$$, $$u,v\in V$$ and $$\lambda \in \F$$. But the first of these is simply the definition of the pointwise addition and scalar multiplication in $$L(V,W)$$ while the second is simply the assertion that $$\phi$$ is linear!

• (c)

• (i) This comes straight from the linearity of $$\alpha$$: for $$u,v\in V$$ and $$\lambda \in \F$$,

\begin{equation*} \phi _{\alpha ,w}(u+\lambda v)=\alpha (u+\lambda v)w= \alpha (u)w+\lambda \alpha (v)w =\phi _{\alpha ,w}(u)+\lambda \phi _{\alpha ,w}(v). \end{equation*}

• (ii) Bilinearity of $$t$$ amounts to:

\begin{align*} \phi _{\alpha +\lambda \beta ,w}&=\phi _{\alpha ,w}+\lambda \phi _{\beta ,w}\\ \phi _{\alpha ,w_1+\lambda w_2}&=\phi _{\alpha ,w_1}+\lambda \phi _{\alpha ,w_2}, \end{align*} for all $$\alpha ,\beta \in V^{*}$$, $$w,w_1,w_2\in W$$ and $$\lambda \in \F$$. Each is proved by showing that both sides take the same values on each $$v\in V$$. For example:

\begin{multline*} \phi _{\alpha +\lambda \beta ,w}(v)= (\alpha +\lambda \beta )(v)w=\alpha (v)w+\lambda \beta (v)w\\ =\phi _{\alpha ,w}(v)+\lambda \phi _{\beta ,w}(v) =(\phi _{\alpha ,w}+\lambda \phi _{\beta ,w})(v) \end{multline*}

• 2. In this case, we have $$B(v_i,v_j)=0$$, for all $$\bw 1jn$$. So, if $$v\in V$$, write $$v=\sum _j\lambda _jv_j$$ and then

\begin{equation*} B(v_i,v)=\sum _j\lambda _jB(v_i,v_j)=0. \end{equation*}

Otherwise said, $$v_i\in \rad B$$.

• 3. If $$B$$ is positive definite, then $$B(v,v)>0$$ for any non-zero $$v\in V$$ and so, in particular, each $$B(v_i,v_i)>0$$.

Conversely, suppose that each $$B(v_i,v_i)>0$$ and let $$v\in V$$. Write $$v=\lc {\lambda }{v}1n$$ and compute:

\begin{equation*} B(v,v)=B(\sum _i\lambda _{i}v_i,\sum _j\lambda _jv_j)=\sum _{i,j}\lambda _i\lambda _jB(v_i,v_j) =\sum _i\lambda _i^2B(v_i,v_i). \end{equation*}

This last is non-negative and vanishes if and only if each $$\lambda _i^2B(v_i,v_i)=0$$, or, equivalently, $$\lambda _{i}=0$$. Thus $$B$$ is positive definite.

• 4.

• (a) This is false: let $$P=\lambda I_n$$, for $$\lambda \in \F$$. Then $$B=\lambda ^2A$$ so that $$\det B=\lambda ^{2n}\det A$$.

• (b) This is true: if $$A^T=A$$ then

\begin{equation*} B^T=(P^TAP)^T=P^TA^TP=P^TAP=B. \end{equation*}

Conversely, if $$B^T=B$$ we get $$P^TA^TP=P^TAP$$ and multiplying by $$P^{-1}$$ on the right and $$(P^T)^{-1}$$ on the left gives $$A^T=A$$.

• 5. We need to start with $$v_1$$ with $$B(v_1,v_1)\neq 0$$. Those diagonal zeros say that none of the standard basis will do so let us try $$v_1=(1,1,0,0)$$ for which $$B(v_1,v_1)=4$$.

Now seek $$v_2$$ among the $$y$$ with

\begin{equation*} 0=B(v_1,y)= \begin{pmatrix} 1&1&0&0 \end {pmatrix}A\by = \begin{pmatrix} 2&2&1&1 \end {pmatrix}\by =2y_1+2y_2+y_3+y_4. \end{equation*}

We take $$v_2=(0,0,1,-1)$$ with

\begin{equation*} B(v_2,y)= \begin{pmatrix} 0&0&1&-1 \end {pmatrix}A\by = \begin{pmatrix} 1&-1&-2&2 \end {pmatrix}\by =y_1-y_2-2y_3+2y_4. \end{equation*}

Then $$B(v_2,v_2)=-4$$ and we seek $$v_3$$ among the $$y$$ with $$B(v_1,y)=B(v_2,y)=0$$, that is:

\begin{align*} 2y_1+2y_2+y_3+y_4&=0\\ y_1-y_2-2y_3+2y_4&=0. \end{align*} One solution is $$v_3=(-3,5,-4,0)$$ with

\begin{equation*} B(v_3,y)= \begin{pmatrix} -3&5&-4&0 \end {pmatrix}A\by = 3\begin{pmatrix} 2&-2&-1&-1 \end {pmatrix}\by =3(2y_1-2y_2-y_3-y_4). \end{equation*}

Thus $$B(v_3,v_3)=-36$$ and we need to find $$v_4=y$$ with $$B(v_1,y)=B(v_2,y)=B(v_3,y)=0$$:

\begin{align*} 2y_1+2y_2+y_3+y_4&=0\\ y_1-y_2-2y_3+2y_4&=0\\ 2y_1-2y_2-y_3-y_4&=0. \end{align*} A solution is $$v_{4}=(0,4,-5,-3)$$ with $$B(v_4,v_4)=36$$.

We now have a diagonalising basis with $$B(v_i,v_{i})=4,-4,-36,36$$ so $$B$$ has signature $$(2,2)$$ and so has rank $$4$$.

After all this linear equation solving it is probably good to check our answer: let $$P$$ have the $$v_j$$ as columns and check that $$P^TAP$$ is diagonal:

\begin{equation*} \begin{pmatrix} 1&1&0&0\\0&0&1&-1\\-3&5&-4&0\\0&4&-5&-3 \end {pmatrix} \begin{pmatrix} 0&2&1&0\\2&0&0&1\\1&0&0&2\\0&1&2&0 \end {pmatrix} \begin{pmatrix} 1&0&-3&0\\1&0&5&4\\0&1&-4&-5\\0&-1&0&-3 \end {pmatrix}= \begin{pmatrix} 4&0&0&0\\ 0&-4&0&0\\ 0&0&-36&0\\ 0&0&0&36 \end {pmatrix} \end{equation*}

• 6. $$B=B_A$$ where

\begin{equation*} A= \begin{pmatrix} 1&1&0\\1&2&1\\0&1&1 \end {pmatrix}. \end{equation*}

Let us exploit the zero in the $$(1,3)$$ slot: note that

so that we just need to find $$y$$ with

\begin{align*} 0&=B(e_1,y)=y_1+y_2\\ 0&=B(e_3,y)=y_2+y_3. \end{align*} Clearly $$y=(1,-1,1)$$ does the job with $$B(y,y)=0$$. Thus $$e_1,e_3,y$$ are a diagonalising basis with matrix

\begin{equation*} \begin{pmatrix} 1&0&0\\0&1&0\\0&0&0. \end {pmatrix} \end{equation*}

Either way, we see that the signature is $$(2,0)$$ and so the rank is $$2$$.

• 7. The fastest way to do this is to recall that $$xy=\tfrac 14\bigl ((x+y)^2-(x-y)^2\bigr )$$ so that

\begin{equation*} x_1x_2-4x_3x_4=\tfrac 14(x_1+x_2)^2-\tfrac 14(x_1-x_2)^2 -(x_3+x_4)^2+(x_3-x_4)^{2}. \end{equation*}

Moreover, the four linear functionals $$x_1\pm x_2, x_3\pm x_4$$ are linearly independent: one way to see this is that $$x_1\pm x_2=0=x_3\pm x_4$$ forces each $$x_i=0$$ so that Corollary 5.7 applies.

Now two squares appear positively and two negatively giving signature $$(2,2)$$ and so rank $$4$$.