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If $A$ and $B$ are symmetric matrices related by $A = P^tBP$, where P is invertible, does it follow that the ranks of $A$ and $B$ are equal?

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In fact multiplying by an invertible matrix does not change rank, hence $\operatorname{rank}(B)=\operatorname{rank}(BP)=\operatorname{rank}(P^TBP)$. – Martin Sleziak Sep 11 '12 at 18:53
up vote 3 down vote accepted

Hint: $x\in \operatorname{Ker}A$ then $Px\in \operatorname{Ker}B$. $x\in \operatorname{Ker}B$ then $P^{-1}x\in \operatorname{Ker}A$. Conclude that $\operatorname{dim} (\operatorname{Ker}A)= \operatorname{dim}(\operatorname{Ker}B)$

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