MA22020: Exercise sheet 6

    Warmup questions

  • 1. 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 \(v_i\in \rad B\).

  • 2. 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 \(\bw 1{i}n\).

  • 3. 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

  • 4. 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.

  • 5. 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 \(B\).

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

December 10, 2024

MA22020: Exercise sheet 6—Solutions

  • 1. 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, \(v_i\in \rad B\).

  • 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 \(v=\lc {\lambda }{v}1n\) 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.

  • 3.

    • (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 \(A^T=A\).

  • 4. 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*}

  • 5. \(B=B_A\) 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

    \begin{equation*} B(e_1,e_1)=B(e_3,e_3)=1,\qquad B(e_1,e_3)=0 \end{equation*}

    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 \(2\).

  • 6. 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 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 \(4\).