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I am asked to show that $A ^3$ is congruent to $A$ for all symmetric real matrices $A$.

If $A$ is invertible, then -
$A^3 = A * A * A = A ^t * (A) * A$
and they are congruent by definition, because $A$ is non-singular and symmetric as stated.

I am stuck on the case where $A$ is singular, and I believe I shouldn't make that distinction of cases from the first place.
Thanks for any suggestions.

Eliyahu Abadi
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  • Perhaps this question wants you to use the fact that a real symmetric matrix has an orthonormal basis of eigenvectors for real eigenvalues. So particularly $A$ is congruent to $\Lambda$ for some real diagonal matrix $\Lambda$, and $A^3$ is congruent to $\Lambda^3$. Since congruence is an equivalence relation, you will need to show that $\Lambda$ is congruent to $\Lambda^3$. – Izaak van Dongen Oct 18 '20 at 13:45

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Since $A$ is a symmetric matrix, there exists an orthogonal matrix $P$ such that $P^TAP$- say $D$- is a diagonal matrix. We can assume WLOG that $$D=\begin{pmatrix} D_* &0\\0&0\end{pmatrix}$$ where $D_*$ is a diagonal invertible square matrix of order $k=\mathrm{rank}(A)$. We have$$A^3=(PDP^T)^3=PD^3P^T=P\begin{pmatrix} D_*^3 &0\\0&0\end{pmatrix}P^T=P\begin{pmatrix} D_* &0\\0&I\end{pmatrix}\begin{pmatrix} D_* &0\\0&0\end{pmatrix}\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^T=P\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^TPDP^TP\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^T=P\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^TAP\begin{pmatrix} D_* &0\\0&I\end{pmatrix}P^T=QAQ$$ Since $Q$ is a symmetric invertible matrix, we are done!

framago
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