A Matrix Similar to a Diagonalizable Matrix is Also Diagonalizable

Diagonalization Problems and Solutions in Linear Algebra

Problem 213

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

 
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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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