Two questions regarding real symmetric matrices.

1) $A$, $B$, $C$, $D$ are $n\times n$ real matrices, $A \sim B$ and $C \sim D$. Then $AC\sim BD$ ($\sim$ = similar).

  • I know if $A \sim B$ then $A = PBP^{-1}$. But I believe $AC\sim BD$ is not true.

2) If $A$ and $B$ are $n\times n$ real symmetric matrices and congruent then they are also similar.

  • I know if they are congruent then there exists an invertible matrix $P$ such that $P^TBP = A$ and they have the same rank. I see that $P^T = P^{-1}$ then it is orthogonal also. But not sure if "similar".

1 Answer 1



  1. Consider $A=B=C=\pmatrix{0&1\\ 0&0}$. Find a matrix $D$ to disprove the statement.

  2. Similar matrices must have identical determinants. Now, consider the $1\times1$ case. The determinant of a $1\times1$ matrix (i.e. a scalar) is simply the value of the scalar. Is it possible to find two real scalars $A, B$ such that $A=P^TBP$ for some $P\ne0$ but $A\ne B$?

  • $\begingroup$ I got #1. But for #2, I know that two real symmetric nxn matrices A and B are congruent if there is a non-singular real symmetric matrix R such that B = R^TAR. I think this is true as similarity and congruence only applies to real symmetric matrices which A and B are. Is that correct? Not sure on a proof though. $\endgroup$
    – Tessa
    Nov 17, 2014 at 3:58

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