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 $A$.

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.

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

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

The post Eigenvalues of Similarity Transformations first appeared on Problems in Mathematics.

(a) Is the matrix $A=\begin{bmatrix}
1 & 2\\
0& 3
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
3 & 0\\
1& 2
\end{bmatrix}$?  

(b) Is the matrix $A=\begin{bmatrix}
0 & 1\\
5& 3
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
1 & 2\\
4& 3
\end{bmatrix}$? 

(c) Is the matrix $A=\begin{bmatrix}
-1 & 6\\
-2& 6
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
3 & 0\\
0& 2
\end{bmatrix}$? 

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

Solution.

(a) Is the matrix $A=\begin{bmatrix}
1 & 2\\
0& 3
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
3 & 0\\
1& 2
\end{bmatrix}$?

Recall 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.
\end{align*}
Thus, $\det(A)\neq \det(B)$, and hence $A$ and $B$ are not similar.

(b) Is the matrix $A=\begin{bmatrix}
0 & 1\\
5& 3
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
1 & 2\\
4& 3
\end{bmatrix}$?

It is straightforward to check that $\det(A)=-5=\det(B)$. Thus determinants does not help here.
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 $\tr(A)\neq\tr(B)$.
Hence $A$ and $B$ are not similar.

(c) Is the matrix $A=\begin{bmatrix}
-1 & 6\\
-2& 6
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
3 & 0\\
0& 2
\end{bmatrix}$?

We see that
\[\det(A)=6=\det(B) \text{ and } \tr(A)=5=\tr(B).\] Thus, the determinants and traces do not give any information about similarity.
The characteristic polynomial of $A$ is given by
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
-1-t & 6\\
-2& 6-t
\end{vmatrix}\\
&=(-1-t)(6-t)-(6)(-2)\\
&=t^2-5t+6.
\end{align*}
(Note that since we found the determinant and trace of $A$, we could have found the characteristic polynomial from the formula $p(t)=t^2-\tr(A)t+\det(A)$.)

Since $p(t)=(t-2)(t-3)$, the eigenvalue of $A$ are $2$ and $3$.
Since $A$ has two distinct eigenvalues, it is diagonalizable.
That is, there exists a nonsingular matrix $S$ such that
\[S^{-1}AS=\begin{bmatrix}
2 & 0\\
0& 3
\end{bmatrix}=B.\] Thus, $A$ and $B$ are similar.

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

We see that
\[\det(A)=6=\det(B) \text{ and } \tr(A)=5=\tr(B).\] It follows from the formula $p(t)=t^2-\tr(A)t+\det(A)$ (or just computing directly) that the characteristic polynomials of $A$ and $B$ are both
\[t^2-5t+6=(t-2)(t-3).\] Thus, the eigenvalues of $A$ and $B$ are $2, 3$. Hence both $A$ and $B$ are diagonalizable.
There exist nonsingular matrices $S$ and $P$ such that
\[S^{-1}AS=\begin{bmatrix}
2 & 0\\
0& 3
\end{bmatrix} \text{ and } P^{-1}BP=\begin{bmatrix} 2 & 0\\
0& 3
\end{bmatrix}. \] So we have $S^{-1}AS=P^{-1}BP$, and hence
\[PS^{-1}ASP^{-1}=B.\] Putting $U=SP^{-1}$, we have
\[U^{-1}AU=B.\] (Since the product of invertible matrices is invertible, the matrix $U$ is invertible.)
Therefore $A$ and $B$ are similar.

Related Question.

For more problems about similar matrices, check out the following posts:

• Problems and solutions about similar matrices
• A matrix similar to a diagonalizable matrix is also diagonalizable

The post Determine Whether Given Matrices are Similar first appeared on Problems in Mathematics.

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

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“.

The post If Two Matrices are Similar, then their Determinants are the Same first appeared on Problems in Mathematics.

Let $A, B$, and $C$ be $n \times n$ matrices and $I$ be the $n\times n$ identity matrix.
Prove the following statements.

(a) If $A$ is similar to $B$, then $B$ is similar to $A$.

(b) $A$ is similar to itself.

(c) If $A$ is similar to $B$ and $B$ is similar to $C$, then $A$ is similar to $C$.

(d) If $A$ is similar to the identity matrix $I$, then $A=I$.

(e) If $A$ or $B$ is nonsingular, then $AB$ is similar to $BA$.

(f) If $A$ is similar to $B$, then $A^k$ is similar to $B^k$ for any positive integer $k$.

Definition.

We say that a matrix $A$ is similar to a matrix $B$ is there exists a nonsingular (invertible) matrix $P$ such that
\[B=P^{-1}AP.\]

Proof.

(a) If $A$ is similar to $B$, then $B$ is similar to $A$.

If $A$ is similar to $B$, then there exists a nonsingular matrix $P$ such that $B=P^{-1}AP$.
Let $Q=P^{-1}$. Since $P$ is nonsingular, so is $Q$.
Then we have
\begin{align*}
Q^{-1}BQ&=(P^{-1})^{-1}BP^{-1}=PBP^{-1}\\
&=P(P^{-1}AP)P^{-1}=IAI=A.
\end{align*}
Hence $B$ is similar to $A$.

(b) $A$ is similar to iteself.

Since the identity matrix $I$ is nonsingular and we have
\[A=I^{-1}AI,\] the matrix $A$ is similar to $A$ itself.

(c) If $A$ is similar to $B$ and $B$ is similar to $C$, then $A$ is similar to $C$.

If $A$ is similar to $B$, we have
\[B=P^{-1}AP,\] for some nonsingular matrix $P$.
Also, if $B$ is similar to $C$, we have
\[C=Q^{-1}BQ,\] for some nonsingular matrix $Q$.
Then we have
\begin{align*}
C&=Q^{-1}BQ\\
&=Q^{-1}(P^{-1}AP)Q\\
&=(PQ)^{-1}A(PQ).
\end{align*}
Let $R=PQ$. Since both $P$ and $Q$ are nonsingular, $R=PQ$ is also nonsingular.
The above computation yields that we have
\[C=R^{-1}AR,\] hence $A$ is similar to $C$.

(d) If $A$ is similar to the identity matrix $I$, then $A=I$.

Since $A$ is similar to $I$, there exists a nonsingular matrix $P$ such that
\[A=P^{-1}IP.\] Since $P^{-1}IP=I$, we have $A=I$.

(e) If $A$ or $B$ is nonsingular, then $AB$ is similar to $BA$.

Suppose first that $A$ is nonsingular. Then $A$ is invertible, hence the inverse matrix $A^{-1}$ exists.
Then we have
\begin{align*}
A^{-1}(AB)A=A^{-1}ABA=IBA=BA,
\end{align*}
hence $AB$ and $BA$ are similar.

Analogously, if $B$ is nonsingular, then the inverse matrix $B^{-1}$ exists.
We have
\begin{align*}
B^{-1}(BA)B=B^{-1}BAB=IAB=AB,
\end{align*}
hence $AB$ and $BA$ are similar.

(f) If $A$ is similar to $B$, then $A^k$ is similar to $B^k$.

If $A$ is similar to $B$, then we have
\[B=P^{-1}AP\] for some nonsingular matrix $P$.
Then we have for a positive integer $k$
\begin{align*}
B^{k}&=(P^{-1}AP)^k\\
&=\underbrace{(P^{-1}AP)(P^{-1}AP)\cdots (P^{-1}AP)}_{k \text{ times}} \\
&=P^{-1}A^kP
\end{align*}
since we can cancel $P$ and $P^{-1}$ in between.
Hence $A^k$ and $B^k$ are similar.

Comment.

Part (a), (b), (c) show that similarity is an equivalence relation.

The post Problems and Solutions About Similar Matrices first appeared on Problems in Mathematics.

Suppose that $n\times n$ matrices $A$ and $B$ are similar.

Then show that the nullity of $A$ is equal to the nullity of $B$.
In other words, the dimension of the null space (kernel) $\calN(A)$ of $A$ is the same as the dimension of the null space $\calN(B)$ of $B$.

Definitions/Hint.

The null space (kernel) of an $m \times n$ matrix $A$ is the subspace of $\R^m$ defined by
\[\calN(A)=\{\mathbf{x} \in \R^m \mid A\mathbf{x}=\mathbf{0}\}.\]

Two matrices $A$ and $B$ are similar if there exists an invertible matrix $S$ such that
\[A=S^{-1}BS.\]

To show that the dimensions of $\calN(A)$ and $\calN(B)$ are equal, find an isomorphism between these vector spaces using the fact that matrices $A$ and $B$ are similar.

Proof.

Since $A$ and $B$ are similar, there exists an invertible matrix $S$ such that
\[A=S^{-1}BS.\] Observe that if $\mathbf{x}\in \calN(A)$, then we have
\begin{align*}
A\mathbf{x}=\mathbf{0}\\
\Leftrightarrow
(S^{-1}BS)\mathbf{x}=\mathbf{0}\\
\Leftrightarrow
B(S\mathbf{x})=\mathbf{0}.
\end{align*}

Therefore we have
\[S\mathbf{x} \in \calN(B).\] From this observation, we define the map
\[\Psi: \calN(A) \to \calN(B)\] by sending $\mathbf{x} \in \calN(A)$ to $S\mathbf{x}\in \calN(B)$.

We claim that the map $\Psi$ is an isomorphism of vector spaces.

To see that $\Psi$ is a linear transformation, let $\mathbf{x}, \mathbf{y} \in \calN(A)$, and $c$ be a scalar.
Then we have
\begin{align*}
\Psi(\mathbf{x}+\mathbf{y})=S(\mathbf{x}+\mathbf{y})=S\mathbf{x}+S\mathbf{y}=\Psi(\mathbf{x})+\Psi(\mathbf{y})
\end{align*}
and
\begin{align*}
\Psi(c\mathbf{x})=S(c\mathbf{x})=cS\mathbf{x}=c\Psi(\mathbf{x}).
\end{align*}
Thus $\Psi$ is a linear transformation.

To show that $\Psi$ is an isomorphism, we give the inverse linear transformation of $\Psi$.

We define $\Phi:\calN(B) \to \calN(A)$ to be a map sending $\mathbf{x} \in \calN(B)$ to $S^{-1}\mathbf{x} \in \calN(A)$.
By a similar argument as above, we can show that the element $S^{-1}\mathbf{x}$ is indeed in $\calN(A)$ and $\Phi$ is a linear transformation and it is straightforward to see that $\Psi\circ \Phi=\id_{\calN(B)}$ and $\Phi \circ \Psi=\id_{\calN(A)}$.
Hence $\Phi$ is the inverse of $\Psi$, and $\Psi$ is an isomorphism.

Therefore the vector spaces $\calN(A)$ and $\calN(B)$ are isomorphic, and hence their dimensions are the same.

Comment.

Instead of finding the inverse linear transformation, you may directly show that the map $\Psi$ is bijective (injective and surjective).

The post Dimension of Null Spaces of Similar Matrices are the Same first appeared on Problems in Mathematics.

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

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.

The post A Matrix Similar to a Diagonalizable Matrix is Also Diagonalizable first appeared on Problems in Mathematics.

Let $A=\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}$.
Show that

(1) $A^n=\begin{bmatrix}
a^n & 0\\
0& b^n
\end{bmatrix}$ for any $n \in \N$.

(2) Let $B=S^{-1}AS$, where $S$ be an invertible $2 \times 2$ matrix.
Show that $B^n=S^{-1}A^n S$ for any $n \in \N$

Hint.

Use mathematical induction.

Proof.

(1) We prove $A^n=\begin{bmatrix}
a^n & 0\\
0& b^n
\end{bmatrix}$ by induction on $n$.
The base case $n=1$ is true by definition.

Suppose that $A^k=\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}$. Then we have
\[A^{k+1}=AA^k =\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}
\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}
=\begin{bmatrix}
a^{k+1} & 0\\
0& b^{k+1}
\end{bmatrix}.\] Here we used the induction hypothesis in the second equality.
Hence the inductive step holds. This completes the proof.

(2) We show that $B^n=S^{-1}A^n S$ by induction on $n$.
When $n=1$, this is just the definition of $B$.

For induction step, assume that $B^k=S^{-1} A^k S$.
Then we have
\[B^{k+1}=B B^k=(S^{-1} A S) (S^{-1} A^k S)=S^{-1}A A^k S=S^{-1} A^{k+1} S,\] where we used the induction hypothesis in the second equality and the third equality follows by canceling $S S^{-1}=I_2$ in the middle.

Thus the inductive step holds, and this competes the proof.

The post Powers of a Diagonal Matrix first appeared on Problems in Mathematics.