Let $A$, $P$ and $D$ be $n \times n$ matrices, with $P$ being invertible. If $A = PDP^{-1}$, then $\det A = \det D$
I used an example to show its false, but just looking for verification to see if it is indeed false.
Let $A$, $P$ and $D$ be $n \times n$ matrices, with $P$ being invertible. If $A = PDP^{-1}$, then $\det A = \det D$
I used an example to show its false, but just looking for verification to see if it is indeed false.
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Suppose that $A$ and $B$ are similar. Then there exists a nonsingular matrix $S$ such that $[S^{-1}AS=B]$ by definition. Then we have \begin{align*} &\det(B)\\ &=\det(S^{-1}AS)\\ &=\det(S)^{-1}\det(A)\det(S) &\text{(by multiplicative properties of determinants)}\\ &=\det(A) &\text{(since determinants are just numbers, hence commutative)} \end{align*}
Thus, we obtain $\det(A)=\det(B)$ as required.