MA22020: Exercise sheet 6

Warmup questions

• 1. Let $B:V\times V\to \mathbb{F}$ be a symmetric bilinear form with diagonalising basis $v_1,\dots ,v_n$. Suppose that, for some $v_i$, $1\le i\le n$, we have $B(v_i,v_i)=0$. Prove that $v_i\in \mathrm{rad}B$.

• 2. Let $B:V\times V\to \mathbb{F}$ be a real symmetric bilinear form with diagonalising basis $v_1,\dots ,v_n$. Show that $B$ is positive definite if and only if $B(v_i,v_i)>0$, for all $1\le i\le n$.

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

  Are the following statements true or false?

  • (a) $detA=detB$.

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

  Rank and signature

• 4. Let $B=B_A:{\mathbb{R}}^4\times {\mathbb{R}}^4\to \mathbb{R}$ where

  $$

  $$A=\left(\begin{array}{cccc}0& 2& 1& 0\\ 2& 0& 0& 1\\ 1& 0& 0& 2\\ 0& 1& 2& 0\end{array}\right).$$

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

• 5. Diagonalise the symmetric bilinear form $B:{\mathbb{R}}^3\times {\mathbb{R}}^3\to \mathbb{R}$ given by $B(x,y)=x_1{y}_1+x_1{y}_2+x_2{y}_1+2x_2{y}_2+x_2{y}_3+x_3{y}_2+x_3{y}_3$.

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

• 6. Compute the rank and signature of the quadratic form $Q(x)=x_1{x}_2-4x_3{x}_4$ on ${\mathbb{R}}^4$.

MA22020: Exercise sheet 6—Solutions

• 1. In this case, we have $B(v_i,v_j)=0$, for all $1\le j\le n$. So, if $v\in V$, write $v=\sum _j{\lambda }_j{v}_j$ and then

  $$

  $$B(v_i,v)=\sum _j{\lambda }_{j}B(v_i,v_j)=0.$$

  Otherwise said, $v_i\in \mathrm{rad}B$.

• 2. 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 $v={\lambda }_1{v}_1+\cdots +{\lambda }_n{v}_n$ and compute:

  $$

  $$B(v,v)=B(\sum _i{\lambda }_i{v}_i,\sum _j{\lambda }_j{v}_j)=\sum _{i,j}{\lambda }_i{\lambda }_{j}B(v_i,v_j)=\sum _i{\lambda }_i^{2}B(v_i,v_i).$$

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

• 3.

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

  • (b) This is true: if $A^T=A$ then

    $$

    $$B^T=(P^{T}AP{)}^T=P^T{A}^{T}P=P^{T}AP=B.$$

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

• 4. We need to start with $v_1$ with $B(v_1,v_1)\ne 0$. 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

  $$

  $$0=B(v_1,y)=\left(\begin{array}{cccc}1& 1& 0& 0\end{array}\right)A\mathbf{y}=\left(\begin{array}{cccc}2& 2& 1& 1\end{array}\right)\mathbf{y}=2y_1+2y_2+y_3+y_4.$$

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

  $$

  $$B(v_2,y)=\left(\begin{array}{cccc}0& 0& 1& -1\end{array}\right)A\mathbf{y}=\left(\begin{array}{cccc}1& -1& -2& 2\end{array}\right)\mathbf{y}=y_1-y_2-2y_3+2y_4.$$

  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:

  $$

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

  $$

  $$B(v_3,y)=\left(\begin{array}{cccc}-3& 5& -4& 0\end{array}\right)A\mathbf{y}=3\left(\begin{array}{cccc}2& -2& -1& -1\end{array}\right)\mathbf{y}=3(2y_1-2y_2-y_3-y_4).$$

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

  $$

  $$\begin{array}{rl}2y_1+2y_2+y_3+y_4& =0\\ y_1-y_2-2y_3+2y_4& =0\\ 2y_1-2y_2-y_3-y_4& =0.\end{array}$$ 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 $P^{T}AP$ is diagonal:

  $$

  $$\left(\begin{array}{cccc}1& 1& 0& 0\\ 0& 0& 1& -1\\ -3& 5& -4& 0\\ 0& 4& -5& -3\end{array}\right)\left(\begin{array}{cccc}0& 2& 1& 0\\ 2& 0& 0& 1\\ 1& 0& 0& 2\\ 0& 1& 2& 0\end{array}\right)\left(\begin{array}{cccc}1& 0& -3& 0\\ 1& 0& 5& 4\\ 0& 1& -4& -5\\ 0& -1& 0& -3\end{array}\right)=\left(\begin{array}{cccc}4& 0& 0& 0\\ 0& -4& 0& 0\\ 0& 0& -36& 0\\ 0& 0& 0& 36\end{array}\right)$$

• 5. $B=B_A$ where

  $$

  $$A=\left(\begin{array}{ccc}1& 1& 0\\ 1& 2& 1\\ 0& 1& 1\end{array}\right).$$

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

  $$

  $$B(e_1,e_1)=B(e_3,e_3)=1,B(e_1,e_3)=0$$

  so that we just need to find $y$ with

  $$

  $$\begin{array}{rl}0& =B(e_1,y)=y_1+y_2\\ 0& =B(e_3,y)=y_2+y_3.\end{array}$$ 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

  $$

  $$\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0.\end{array}\right)$$

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

• 6. The fastest way to do this is to recall that $xy=\frac{1}{4}((x+y{)}^2-(x-y{)}^2)$ so that

  $$

  $$x_1{x}_2-4x_3{x}_4=\frac{1}{4}(x_1+x_2{)}^2-\frac{1}{4}(x_1-x_2{)}^2-(x_3+x_4{)}^2+(x_3-x_4{)}^2.$$

  Moreover, the four linear functions $x_1\pm x_2,x_3\pm x_4$ have linearly independent coefficients: $(1,\pm 1,0,0)$ and $(0,0,1,\pm 1)$.

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