Show that $trace P^{-1}MP = trace M$

Suppose that $$M =\begin{bmatrix}a & b \\c & d\end{bmatrix},$$ $$P =\begin{bmatrix}\alpha & \beta \\\gamma & \delta\end{bmatrix}$$ with P invertible. show that $trace P^{-1} M P = trace M$ by calculating the matrix $P^{-1}MP$ explicitly and then reading off its trace.

my professor gave this question word by word:

So I started to do this question as he requested: $P^{-1}MP$ = $1 \over detP$ $P^{-1} M P$ and so on...

But my question is that why do we have to solve this question by calculating the matrix explicitly??

why can't we just say there is $trace P^{-1} M P = trace M$ and $P^{-1} and P$ gets cancelled out therefore its $trace M?$

For instance $det P^{-1} M P = det M$ can we use the same logic here?

• @GeoffRobinson can i say 'Similar matrices have the same trace' if $B = P^{-1} A P, then, tr(B) = tr{(P^{-1} A) P} = tr {P(P^{-1}A)} = tr(A)?$ assuming A = B Dec 19 '15 at 8:09
• yes, that is the "Commutative property", one of the trace properties. Please check: math.tutorvista.com/algebra/trace-matrix.html Dec 19 '15 at 8:13
• Yes, that more or less follows from the comment I wrote ( except I am not sure what you mean by "assuming A=B", which is an unnecessary restriction). Dec 19 '15 at 8:24

$\det(P^{-1}MP)=\det M$ because $\det(AB)=\det A\det B$ and $\det (A^{-1})=(\det A)^{-1}$. How do you know that any of these things are true about the trace?
You just need to proof that $tr(A*B)=tr(B*A)\;$ which you can easily do using the definition of matrix multiplication.
$$tr(A*B)=\sum_{k=1}^n{(A*B)_{k,k}}=\sum_{k=1}^n{\sum_{p=1}^n{A_{k,p}*B_{p,k}}}=\sum_{k=1}^n{\sum_{p=1}^n{B_{p,k}*A_{k,p}}}=\sum_{p=1}^n{\sum_{k=1}^n{B_{p,k}*A_{k,p}}}=tr(A*B)$$
Then for your problem you get $tr(P^{-1}MP)=tr((P^{-1}M)P)=tr(P(P^{-1}M))=tr(M)$