Eigenvalues of Similarity Transformations

Linear algebra problems and solutions

Problem 452

Let $A$ be an $n\times n$ complex matrix.
Let $S$ be an invertible matrix.

(a) If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix.

(b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.

(c) Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$.

 
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Proof.

Basic Properties of Determinants

We use the following properties of determinants of matrices.
For $n\times n$ matrices $A, B$, we have

  1. $\det(AB)=\det(A)\det(B)$.
  2. $\det(A^{-1})=\det(A)^{-1}$ if $A$ is invertible.
  3. $\det(cA)=c^n\det(A)$ for any complex number $c$.

 

(a) If $SAS^{-1}=\lambda A$, then prove that $\lambda^n=1$ or $A$ is a singular matrix.

Suppose that we have $SAS^{-1}=\lambda A$.
We consider the determinants of both sides and we have
\begin{align*}
\det(SAS^{-1})=\det(\lambda A).
\end{align*}
The left hand side becomes
\begin{align*}
\det(SAS^{-1})&=\det(S)\det(A)\det(S^{-1}) &&\text{by property 1}\\
&=\det(S)\det(A)\det(S)^{-1} &&\text{by property 2}\\
&=\det(A).
\end{align*}
The right hand side is by property 3
\[\det(\lambda A)=\lambda^n \det(A).\]

Hence we obtain
\[\det(A)=\lambda^n \det(A).\] It follows from this that either $\lambda^n=1$ or $\det(A)=0$.
In conclusion, either $\lambda^n=1$ or $A$ is a singular matrix.

(b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.

Suppose that we have $SAS^{-1}=-A$.
Then we have
\begin{align*}
&\det(A)=\det(S)\det(A)\det(S)^{-1} \\
&=\det(S)\det(A)\det(S^{-1}) &&\text{by property 2}\\
&=\det(SAS^{-1}) &&\text{by property 1}\\
&=\det(-A) &&\text{by assumption}\\
&=(-1)^n\det(A) &&\text{by property 3}\\
&=-\det(A) &&\text{since $n$ is odd.}\\
\end{align*}

This yields that $\det(A)=0$.
Note that the product of all eigenvalues of $A$ is $\det(A)$.
(See the post “Determinant/Trace and Eigenvalues of a Matrix” for a proof.)
Thus, $0$ is an eigenvalue of $A$.

(c) Note that as $\det(A) > 0$, the matrix $A$ is invertible.
Suppose that $SAS^{-1}=A^{-1}$.
Then we have
\begin{align*}
&\det(A)=\det(S)\det(A)\det(S)^{-1} \\
&=\det(S)\det(A)\det(S^{-1}) &&\text{by property 2}\\
&=\det(SAS^{-1}) &&\text{by property 1}\\
&=\det(A^{-1}) &&\text{by assumption}\\
&=\det(A)^{-1} &&\text{by property 2}.
\end{align*}
Thus we have $\det(A)^2=1$, hence $\det(A)=1$ as $\det(A) > 0$ by assumption.


Note again that the product of all eigenvalues of $A$ is $\det(A)$.
Since the eigenvalues of $A$ are integers by assumption, the eigenvalues of $A$ are either $1$ or $-1$.

If all of the $n$ eigenvalues of $A$ are $-1$, then the determinant is
\[\det(A)=(-1)^n=-1\] since $n$ is odd, and this is a contradiction.
Thus, at least one of the eigenvalues must be $1$.

This completes the proof.

Similar Transformation (conjugate)

A transformation $A$ to $SAS^{-1}$, for some invertible matrix $S$, is called a similarity transformation or conjugation of the matrix $A$.

We say that matrices $A$ and $B$ are similar if there exists an invertible matrix $S$ such that $B=SAS^{-1}$.
In other words, $A$ is similar to $B$ if there is a similarity transformation from $A$ to $B$.

Check out the following problems about similar matrices.

Problem. Is the matrix $A=\begin{bmatrix}
-1 & 6\\
-2& 6
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
1 & 2\\
-1& 4
\end{bmatrix}$?

For a solution together with similar problems, see the post “Determine whether given matrices are similar“.

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

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


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