Matrix Similarity proofs Say that $A$ and $B$ are similar i.e. there exists an invertible matrix $P$ such that $A = P^{−1}BP$. Prove that: (a) $tr(A) = tr(B)$ (b) $|A| = |B|$ (Notation: for an $n × n$ matrix $M$, $|M| = \det M$). (c) $|tIn − A| = |tIn − B|$ (hence $A$ and $B$ have the same eigenvalues with the same algebraic multiplicities). (d) If $x$ is an eigenvector for $A$ corresponding to eigenvalue $λ$, then $Px$ is an eigenvector for $B$ corresponding to $λ$. I have a special problem solving proofs. I would greatly appreciate it if someone can show me. I just seem to not know where to start from.

Hints: For a) use the fact that tr($XY$)= tr($YX$).

So, tr$(P^{-1}BP)$= tr($PP^{-1}B$)= tr($B$)

For b, use $|XY|=|YX|$ and use similar methods as above.

For c again use $|XY|=|YX|$ and note that $P^{-1}P=I$

For d, by definition $Ax=\lambda x$. So, $P^{-1}BPx=\lambda x$ i.e., $BPx=P(\lambda x)=\lambda P(x)$, and hence you are done.