# A Matrix Similar to a Diagonalizable Matrix is Also Diagonalizable

## Problem 213

Let $A, B$ be matrices. Show that if $A$ is diagonalizable and if $B$ is similar to $A$, then $B$ is diagonalizable.

Contents

## Definitions/Hint.

Recall the relevant definitions.

• Two matrices $A$ and $B$ are similar if there exists a nonsingular (invertible) matrix $S$ such that
$S^{-1}BS=A.$
• A matrix $A$ is diagonalizable if $A$ is similar to a diagonal matrix. Namely, $A$ is diagonalizable if there exist a nonsingular matrix $S$ and a diagonal matrix $D$ such that
$S^{-1}AS=D.$

Some useful facts are

• If $S$ and $T$ are invertible matrices, then we have
$(TS)^{-1}=S^{-1}T^{-1}.$ (Note the order of the product.)
• A matrix is nonsingular if and only if its determinant is nonzero.

## Proof.

Since the matrix $A$ is diagonalizable, there exist a nonsingular matrix $S$ and a diagonal matrix $D$ such that
$S^{-1}AS=D. \tag{*}$ Also, since $B$ is similar to $A$, there exist a nonsingular matrix $T$ such that
$T^{-1}BT=A. \tag{**}$

Inserting the expression of $A$ from (**) into the equality (*), we obtain
\begin{align*}
D&=S^{-1}(T^{-1}BT)S\\
&=(S^{-1}T^{-1})B(TS)\\
&=(TS)^{-1}B(TS) \tag{***}.
\end{align*}

Now let us put $U:=TS$. Then the matrix $U$ is nonsingular.
(This is because we have
$\det(U)=\det(TS)=\det(T)\det(S)\neq 0$ since $T$ and $S$ are nonsingular matrices, hence their determinants are not zero.)

Therefore from (***) we have
$D=U^{-1}BU,$ where $D$ is a diagonal matrix and $U$ is a nonsingular matrix.
Thus $B$ is a diagonalizable matrix.

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### 2 Responses

1. 06/13/2017

[…] For a proof, see the post “A matrix similar to a diagonalizable matrix is also diagonalizable“. […]

2. 06/13/2017

[…] For a proof, see the post “A matrix similar to a diagonalizable matrix is also diagonalizable“. […]

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##### How to Diagonalize a Matrix. Step by Step Explanation.

In this post, we explain how to diagonalize a matrix if it is diagonalizable. As an example, we solve the...

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