If Two Matrices are Similar, then their Determinants are the Same

Linear algebra problems and solutions

Problem 390

Prove that if $A$ and $B$ are similar matrices, then their determinants are the same.

 
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Proof.

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.

Related Question.

More generally, we can prove that if $A$ and $B$ are similar, then their characteristic polynomials are the same.
From this, we also can deduce that the determinants of $A$ and $B$ are the same as well as their traces are the same.

For a proof, see the post “Similar matrices have the same eigenvalues“.


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1 Response

  1. 04/26/2017

    […] that if $A$ and $B$ are similar, then their determinants are the same. We compute begin{align*} det(A)=(1)(3)-(2)(0)=3 text{ and } det(B)=(3)(2)-(0)(1)=6. […]

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