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If $T$ and $M$ are $3\times 3$ invertible matrices.

Consider the matrix $L=T^tMT$.

I know this matrix is diagonalizable, because it's symmetric. I would like to know why if the field $k$ in the entries of the matrix is algebraically closed, $L$ has just these forms (up to coordinate changes):

$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$

$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$

$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$

I really need help.

Thanks.

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This is false as given, of course. For a simple counter example, just take $$T=I\;\;,\;\;M=2\cdot I\;\implies T^tMT=2\cdot I$$ –  DonAntonio Jan 14 at 21:16
    
You are probably meaning that for each $3\times 3$ matrix $M$, there is an invertible matrix $T$ such that $T^tMT$ has one of those forms. This is the fundamental theorem on the canonical form with respect to matrix congruence, basically due to Sylvester. –  egreg Jan 14 at 21:18
    
@egreg exactly!!! –  user42912 Jan 14 at 21:40
    
@user42912 So, please, reformulate your question. –  egreg Jan 14 at 21:42
    
@egreg do you know where can I find this theorem? –  user42912 Jan 14 at 21:42
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