Similar Matrices Have the Same Eigenvalues

Problem 2

Show that if $A$ and $B$ are similar matrices, then they have the same eigenvalues and their algebraic multiplicities are the same.

Contents

Proof.

We prove that $A$ and $B$ have the same characteristic polynomial. Then the result follows immediately since eigenvalues and algebraic multiplicities of a matrix are determined by its characteristic polynomial.

Since $A$ and $B$ are similar, there exists an invertible matrix $S$ such that $S^{-1}AS=B$.
Let $p_A(t)$ and $p_B(t)$ denote the characteristic polynomials of $A$ and $B$, respectively.

We have
\begin{align*}
p_B(t) &= \det(B-tI) =\det(S^{-1}AS-tI)  \\
&=\det(S^{-1}(A-tI)S) = \det(S^{-1}) \det(A-tI) \det(S) \\
&\stackrel{(*)}{=}\det(A-tI)=p_A(t).
\end{align*}

Here the fifth equality (*) follows from the fact that $\det(S^{-1})=\det(S)^{-1}$.
(Also note that even though matrix multiplication is not commutative in general but determinants are just numbers, thus we can change the order of the product of determinants.)

Thus we showed that $p_A(t)=p_B(t)$ and this completes the proof.

Related Facts

We proved that if $A$ and $B$ are similar, then their characteristic polynomials are the same.

Since the determinants and the traces are the coefficients of the characteristic polynomials.
Thus, if $A$ and $B$ are similar, their determinants and traces are the same.

See the problem “If two matrices are similar, then their determinants are the same” for a more direct proof of this fact about determinants.

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4 Responses

1. Owen says:

What property allows us to move the tI term from the outside to matrix A at the start of line 2?

• Yu says:

Dear Owen,

You can confirm the calculation as follows. I think it is easier to think backward.

First $S^{-1}(A-tI)S = S^{-1}AS – S^{-1}(tI)S$ by distributing $S{-1}$ and $S$.
1
Now look at the second term $S^{-1}(tI)S$. As $t$ is just a scalar (number), we have

$S^{-1}(tI)S = t S^{-1}IS = tS^{-1}S= t I$ by the propaty of the identity matrix and inverse matrix.

Combining these we get the desired equality. Let me know if you have further questions.

1. 04/26/2017

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

2. 04/26/2017

[…] We recall that if $A$ and $B$ are similar, then their traces are the same. (See Problem “Similar matrices have the same eigenvalues“.) We compute begin{align*} tr(A)=0+3=3 text{ and } tr(B)=1+3=4, end{align*} and thus […]

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