Let $A$ and $B$ be $n\times n$ matrices and assume that they commute: $AB=BA$.
Then prove that the matrices $A$ and $B$ share at least one common eigenvector.

Proof.

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be an eigenvector corresponding to $\lambda$. That is, we have $A\mathbf{x}=\lambda \mathbf{x}$.
Then we claim that the vector $\mathbf{v}:=B\mathbf{x}$ belongs to the eigenspace $E_{\lambda}$ of $\lambda$.
In fact, as $AB=BA$ we have
\begin{align*}
A\mathbf{v}=AB\mathbf{x}=BA\mathbf{x} =\lambda B\mathbf{x}=\lambda \mathbf{v}.
\end{align*}
Hence $\mathbf{v}\in E_{\lambda}$.

Now, let $\{\mathbf{x}_1, \dots, \mathbf{x}_k\}$ be an eigenbasis of the eigenspace $E_{\lambda}$.
Set $\mathbf{v}_i=B\mathbf{x}_i$ for $i=1, \dots, k$.
The above claim yields that $\mathbf{v}_i \in E_{\lambda}$, and hence we can write
\[\mathbf{v}_i=c_{1i} \mathbf{x}_1+c_{2i}\mathbf{x}_2+\cdots+c_{ki}\mathbf{x}_k \tag{*}\] for some scalars $c_{1i}, c_{2i}, \dots, c_{ki}$.

Extend the basis $\{\mathbf{x}_1, \dots, \mathbf{x}_k\}$ of $E_{\lambda}$ to a basis
\[\{\mathbf{x}_1, \dots, \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n\}\] of $\R^n$ by adjoining vectors $\mathbf{x}_{k+1}, \dots, \mathbf{x}_n$.

Then we obtain using (*)
\begin{align*}
&B [\mathbf{x}_1,\dots , \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n]\\
&=[B\mathbf{x}_1,\dots , B\mathbf{x}_k, B\mathbf{x}_{k+1}, \dots, B\mathbf{x}_n]\\
&=[\mathbf{v}_1,\dots , \mathbf{v}_k, B\mathbf{x}_{k+1}, \dots, B\mathbf{x}_n]\\[6pt] &=[\mathbf{x}_1,\dots , \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n] \left[\begin{array}{c|c}
C & D\\
\hline
O & F
\end{array}
\right],\tag{**}
\end{align*}

where $C=(c_{ij})$ is the $k\times k$ matrix whose entries are the coefficients $c_{ij}$ of the linear combination (*), $O$ is the $(n-k) \times k$ zero matrix, $D$ is a $k \times (n-k)$ matrix, and $F$ is an $(n-k) \times (n-k)$ matrix.

Let $P=[\mathbf{x}_1,\dots , \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n]$.
As the column vectors of $P$ are linearly independent, $P$ is invertible.

From (**), we obtain
\[P^{-1}BP=\left[\begin{array}{c|c}
C & D\\
\hline
O & F
\end{array}
\right].\] It follows that
\begin{align*}
\det(B-tI)&=\det(P^{-1}BP-tI)=\left|\begin{array}{c|c}
C-tI & D\\
\hline
O & F-tI
\end{array}
\right|=\det(C-tI)\det(F-tI).
\end{align*}

Let $\mu$ be an eigenvalue of the matrix $C$ and let $\mathbf{a}$ be an eigenvector corresponding to $\mu$.
Then as $\det(C-\mu I)=0$, we see that $\det(B-\mu I)=0$ and $\mu$ is an eigenvalue of $B$.

Write
\[\mathbf{a}=\begin{bmatrix}
a_1 \\
a_2 \\
\vdots \\
a_k
\end{bmatrix}\neq \mathbf{0}\] and define a new vector by
\[\mathbf{y}=a_1\mathbf{x}_1+\cdots +a_k \mathbf{x}_k\in E_{\lambda}.\] Then $\mathbf{y}$ is an eigenvector in $E_{\lambda}$ since it is a nonzero (as $\mathbf{a}\neq \mathbf{0}$) linear combination of the basis $E_{\lambda}$.

Multiplying $BP=P\left[\begin{array}{c|c}
C & D\\
\hline
O & F
\end{array}
\right]$ from (**) by the $n$-dimensional vector
\[\begin{bmatrix}
\mathbf{a} \\
0 \\
\vdots\\
0
\end{bmatrix}=\begin{bmatrix}
a_1 \\
\vdots \\
a_k \\
0 \\
\vdots\\
0
\end{bmatrix}\] on the right, we have
\[BP\begin{bmatrix}
\mathbf{a} \\
0 \\
\vdots\\
0
\end{bmatrix}=P\left[\begin{array}{c|c}
C & D\\
\hline
O & F
\end{array}
\right] \begin{bmatrix}
\mathbf{a} \\
0 \\
\vdots\\
0
\end{bmatrix}.\]

The left hand side is equal to
\[B[a_1\mathbf{x}_1+\cdots+a_k \mathbf{x}_k]=B\mathbf{y}.\]

On the other hand, the right hand side is equal to
\begin{align*}
P\begin{bmatrix}
C \mathbf{a}\\
\mathbf{0}
\end{bmatrix}
=P\begin{bmatrix}
\mu \mathbf{a} \\
\mathbf{0}
\end{bmatrix}=\mu P\begin{bmatrix}
\mathbf{a} \\
\mathbf{0}
\end{bmatrix}
=\mu [a_1\mathbf{x}_1+\cdots+a_k \mathbf{x}_k]=\mu \mathbf{y}.
\end{align*}

Therefore, we obtain
\[B\mathbf{y}=\mu \mathbf{y}.\] This proves that the vector $\mathbf{y}$ is eigenvector of both $A$ and $B$.
Hence, this completes the proof.

Let $\lambda$ be an eigenvalue of $n\times n$ matrices $A$ and $B$ corresponding to the same eigenvector $\mathbf{x}$.

(a) Show that $2\lambda$ is an eigenvalue of $A+B$ corresponding to $\mathbf{x}$.

(b) Show that $\lambda^2$ is an eigenvalue of $AB$ corresponding to $\mathbf{x}$.

(The Ohio State University, Linear Algebra Final Exam Problem)
 

Proof.

(a) Show that $2\lambda$ is an eigenvalue of $A+B$ corresponding to $\mathbf{x}$.

Since $\lambda$ is an eigenvalue of $A$ and $B$, and $\mathbf{x}$ is a corresponding eigenvector, we have
\[A\mathbf{x}=\lambda \mathbf{x} \text{ and } B\mathbf{x}=\lambda \mathbf{x} \tag{*}.\] Then we compute
\begin{align*}
(A+B)\mathbf{x}&=A\mathbf{x}+B\mathbf{x}\\
&=\lambda \mathbf{x}+ \lambda \mathbf{x} && \text {by (*)}\\
&=2\lambda \mathbf{x}.
\end{align*}

Since $\mathbf{x}$ is an eigenvector, it is a nonzero vector by definition.
Hence from the equality
\[(A+B)\mathbf{x}=2\lambda \mathbf{x},\] we see that $2\lambda$ is an eigenvalue of the matrix $A+B$ and $\mathbf{x}$ is an associated eigenvector.

(b) Show that $\lambda^2$ is an eigenvalue of $AB$ corresponding to $\mathbf{x}$.

We have
\begin{align*}
(AB)\mathbf{x}&=A(B\mathbf{x})\\
&=A(\lambda \mathbf{x}) && \text{by (*)}\\
&=\lambda (A\mathbf{x})\\
&=\lambda (\lambda \mathbf{x}) && \text{by (*)}\\
&=\lambda^2 \mathbf{x}.
\end{align*}

Since $\mathbf{x}$ is a nonzero vector as it is an eigenvector, it follows from the equality
\[(AB)\mathbf{x}=\lambda^2 \mathbf{x}\] that $\lambda^2$ is an eigenvalue of the matrix $AB$ and $\mathbf{x}$ is a corresponding eigenvector.

Let $A$ and $B$ be an $n \times n$ matrices.
Suppose that all the eigenvalues of $A$ are distinct and the matrices $A$ and $B$ commute, that is $AB=BA$.

Then prove that each eigenvector of $A$ is an eigenvector of $B$.

(It could be that each eigenvector is an eigenvector for distinct eigenvalues.)

Hint.

Each eigenspace for $A$ is one dimensional.

Proof.

Since $A$ has $n$ distinct eigenvalues, the characteristic polynomial for $A$ factors into the product of degree $1$ polynomials.

Thus, the algebraic multiplicity of each eigenvalue is, $1$ and hence the geometric multiplicity is also $1$.
(The geometric multiplicity is always less than or equal to the algebraic multiplicity and greater than 0 by definition.)

Thus the dimension of each eigenspace, which is the geometric multiplicity, is $1$.

Let $\lambda$ be an eigenvalue of the matrix $A$ and let $\mathbf{x}$ be the eigenvector corresponding to $\lambda$.
Since the eigenspace $E_{\lambda}$ for $\lambda$ is one dimensional and $\mathbf{x}\in E_{\lambda}$ is a nonzero vector in it, the vector $\mathbf{x}$ is a basis.
That is, we have $E_{\lambda}=\{t\mathbf{x} \mid t\in \C \}$.

Now we multiply $A\mathbf{x}=\lambda \mathbf{x}$ by the matrix $B$ on the left and obtain
\begin{align*}
BA\mathbf{x}&=\lambda B\mathbf{x}\\
\iff \,\,\,\, AB\mathbf{x}&=\lambda B\mathbf{x} \text{ since } AB=BA.
\end{align*}

This implies that $B\mathbf{x} \in E_{\lambda}=\{t\mathbf{x} \mid t\in \C \}$. Therefore there exists $t\in \C$ such that $B \mathbf{x}= t\mathbf{x}$.

Hence the vector $\mathbf{x}$ is also an eigenvector corresponding to the eigenvalue $t$ of the matrix $B$.
This completes the proof.

Let $A$ and $B$ be $n\times n$ matrices.
Suppose that these matrices have a common eigenvector $\mathbf{x}$.

Show that $\det(AB-BA)=0$.

Steps.

1. Write down eigenequations of $A$ and $B$ with the eigenvector $\mathbf{x}$.
2. Show that AB-BA is singular.
3. A matrix is singular if and only if the determinant of the matrix is zero.

Proof.

Let $\alpha$ and $\beta$ be eigenvalues of $A$ and $B$ such that the vector $\mathbf{x}$ is a corresponding eigenvector.
Namely we have $A \mathbf{x}=\alpha \mathbf{x}$ and $B\mathbf{x}=\beta \mathbf{x}$.

Then we have
\begin{align*}
(AB-BA)\mathbf{x}&=AB\mathbf{x}-BA\mathbf{x}=A(\beta \mathbf{x}) -B( \alpha \mathbf{x}) \\
& = \beta A \mathbf{x}- \alpha B\mathbf{x} =\beta \alpha -\alpha \beta=0.
\end{align*}

By the definition of eigenvector,  $\mathbf{x}$ is a non-zero vector. Thus the matrix $AB-BA$ is singular.
Equivalently the determinant of $AB-BA$ is zero.

Comment.

This is a simple necessary condition that $A$ and $B$ have a common eigenvector.

Here are few derived questions.

• Is this a sufficient condition?
• If so prove it.
• If not give a counterexample,
• and find a necessary and sufficient condition.