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Problem: For $X, Y \in \mathbf{R}^{3}$, and for the matrix $$ A=\left(\begin{array}{lll} 1 & a & 1 \\ a & 0 & 0 \\ 1 & 0 & 1 \end{array}\right) $$ we define the bilinear form $f(X, Y)=X^{t} A Y$. Determine for what values of $a \in \mathbf{R}$, the rank of the form is 3 , and for those values determine the signature of $f$.

Attempt: Clearly $ a\neq 0 $ since otherwise we'd have $ rank(A) \leq 2 $. Hence, applying Algorithm for Congruence Diagonalization of Symmetric matrix on $ A $ to find a non singular matrix $ P $ such that $ D = P^TAP $ is diagonal.
First form the block matrix $ M = [A,I] $; that is, let $M=[A, I]=\left[\begin{array}{rrr:rrr} 1 & a & 1 & 1 & 0 & 0 \\ a & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1\end{array}\right]$.
Apply the row operations "Replace $ R_2 $ by $ -a_{21}R_1+a_{11}R_2 = -aR_1 + R_2 $ " and "Replace $ R_3 $ by $ -a_{31}R_1+a_{11}R_3 = -R_1 + R_3 $ " to $M$, and then apply the corresponding column operations " Replace $ C_2 $ by $ -aC_1 + C_2 $ " and "Replace $ C_3 $ by $-a_{21}C_1+a_{11}C_2 = -C_1+ C_3 $ " to obtain
$ \left[\begin{array}{rrr:rrr} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & -a^2 & -a & -a & 1 & 0 \\ 0 & -a & 0 & -1 & 0 & 1\end{array}\right]$.
( I think so far's correct since $\left(\begin{array}{ccc}1 & 0 & 0 \\ -a & 1 & 0 \\ -1 & 0 & 1\end{array}\right)\left(\begin{array}{lll}1 & a & 1 \\ a & 0 & 0 \\ 1 & 0 & 1\end{array}\right)\left(\begin{array}{ccc}1 & -a & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right) $ $ = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & -a^2 & -a \\ 0 & -a & 0\end{array}\right) $ ). [ Now's the part where I think there's a problem: ]
Next, Apply the row operation " Replace $R_3$ by $ -a_{32}R_2+a_{22}R_3 = aR_2 - a^2R_3 $ " and " Replace $C_3$ by $ -a_{32}C_2+a_{22}C_3 = aC_2 - a^2C_3 $ ", thus we get $ \left[\begin{array}{rrr:rrr} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & -a^2 & 0 & -a & 1 & 0 \\ 0 & 0 & -a^2 & 0 & a & -a^2 \end{array}\right] $

But I got stuck in the last part since $\left(\begin{array}{ccc}1 & 0 & 0 \\ -a & 1 & 0 \\ 0 & a & -a^{2}\end{array}\right)\left(\begin{array}{ccc}1 & a & 1 \\ a & 0 & 0 \\ 1 & 0 & 1\end{array}\right)\left(\begin{array}{ccc}1 & -a & 0 \\ 0 & 1 & a \\ 0 & 0 & -a^{2}\end{array}\right)=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & -a^{2} & 0 \\ 0 & 0 & a^{4}\end{array}\right) $ which is not equal to $ \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & -a^2 & 0 \\ 0 & 0 & -a^2 \end{array}\right) $.

Note: I've been using the following algorithm appearing in the image in the question reference for linear algebra books that teach reverse Hermite method for symmetric matrices

Question: Do you have any idea why in my last step in the algorithm there's a problem? ( If I could proceed any $a\neq 0$ would satisfy $ rank A =3 $), and how can I proceed to find the signature?

Thanks in advance for help!

hazelnut_116
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  • "we define the the quadratic form $f\big(X,Y\big)= X^TAY$ Does your text really say this? A quadratic form has domain $V$ and a bilinear form has domain $V\times V$ yet your 'quadratic form' has domain $V\times V$... – user8675309 Jan 28 '22 at 22:47
  • Thanks, I've edited anyway because that part written in the exam was clearly wrong. – hazelnut_116 Jan 29 '22 at 08:11
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    I'd suggest verifying $\det\big(A\big) = -a^2$ which is a much shorter way of getting to your result of any $a\neq 0$ for full rank. Then you are left with the fact that $A$ has a positive diagonal element (or positive trace if you prefer) so has at least one positive eigenvalue, but the product of the eigenvalues is negative (for $a\neq 0$) so how many positive and negative eigenvalues does $A$ have? And what does this tell you about the signature? – user8675309 Jan 29 '22 at 18:32
  • So now I know that $ det(A) = \lambda_1 \cdot \lambda_2 \cdot \lambda_3 = -a^2 $. Suppose WLOG $ \lambda_1 >0 $. Then $ ( \lambda_2 >0 \land \lambda_3<0 ) $ or $ ( \lambda_2<0 \land \lambda_3>0 ) $. In anycase, $ A $ has 2 positive eigenvalues and 1 negative eigenvalue. ( note that $ \det A \neq 0 $ since that'd mean $ a=0 $ and we have assumed $ a \neq 0 $ ) Thus if one would diagonalize $ A $ there'd be 2 positive values on the diagonal and one negative value, hence the signature is 1 ( $ p_+ = 2 $ , $ p_- = 1 $ hence $ p_+ - p_- = 1 $ ). So $ a$ can be any value besides $0 $. – hazelnut_116 Jan 30 '22 at 11:55

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