Let
\[A=\begin{bmatrix}
1-a & a\\
-a& 1+a
\end{bmatrix}\] be a $2\times 2$ matrix, where $a$ is a complex number.
Determine the values of $a$ such that the matrix $A$ is diagonalizable.

(Nagoya University, Linear Algebra Exam Problem)

Proof.

To find eigenvalues of the matrix $A$, we determine the characteristic polynomial $p(t)$ of $A$ as follows.
We have
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
1-a-t & a\\
-a& 1+a-t
\end{vmatrix}\\[6pt] &=(1-a-t)(1+a-t)+a^2\\
&=(1-t)^2-a^2+a^2=(1-t)^2.
\end{align*}

Thus, the eigenvalue of $A$ is $1$ with algebraic multiplicity $2$.

Let us determine the geometric multiplicity (the dimension of the eigenspace $E_1$).
We have
\begin{align*}
A-I=\begin{bmatrix}
-a & a\\
-a& a
\end{bmatrix}
\xrightarrow{R_2-R_1}
\begin{bmatrix}
-a & a\\
0& 0
\end{bmatrix} .
\end{align*}
If $a\neq 0$, then we further reduce it and get
\begin{align*}
\begin{bmatrix}
-a & a\\
0& 0
\end{bmatrix}\xrightarrow{\frac{-1}{a}R_1}
\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}.
\end{align*}

Hence the eigenspace corresponding to the eigenvalue $1$ is
\begin{align*}
E_1=\calN(A-I)=\Span \left\{\, \begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right\},
\end{align*}
and its dimension is $1$, which is less than the algebraic multiplicity.
Thus, when $a\neq 0$, the matrix $A$ is not diagonalizable.

If $a=0$, then the matrix $A=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}$ is already diagonal, hence it is diagonalizable.

In conclusion, the matrix $A$ is diagonalizable if and only if $a=0$.

The post Determine the Values of $a$ such that the 2 by 2 Matrix is Diagonalizable first appeared on Problems in Mathematics.

Suppose that a real symmetric matrix $A$ has two distinct eigenvalues $\alpha$ and $\beta$.
Show that any eigenvector corresponding to $\alpha$ is orthogonal to any eigenvector corresponding to $\beta$.

(Nagoya University, Linear Algebra Final Exam Problem)
 

Hint.

Two vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if their inner (dot) product $\mathbf{u}\cdot \mathbf{v}:=\mathbf{u}^{\trans}\mathbf{v}=0$.

Here $\mathbf{u}^{\trans}$ is the transpose of $\mathbf{u}$.

A fact that we will use below is that for matrices $A$ and $B$, we have $(AB)^{\trans}=B^{\trans}A^{\trans}$.

Proof.

Let $\mathbf{u}, \mathbf{v}$ be eigenvectors corresponding to $\alpha, \beta$, respectively.
Namely we have
\[A\mathbf{u}=\alpha \mathbf{u} \text{ and } A\mathbf{v}=\beta \mathbf{v}. \tag{*}\]

To prove that $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, we show that the inner product $\mathbf{u} \cdot \mathbf{v}=0$.
Keeping this in mind, we compute
\begin{align*}
&\alpha (\mathbf{u} \cdot \mathbf{v}) =(\alpha \mathbf{u}) \cdot \mathbf{v} \\
&\stackrel{(*)}{=} A\mathbf{u}\cdot \mathbf{v} =(A\mathbf{u})^{\trans} \mathbf{v}\\
&=\mathbf{u}^{\trans}A^{\trans}\mathbf{v} \text{ (This follows from the fact mentioned in the hint above)} \\
&=\mathbf{u}^{\trans}A\mathbf{v} \text{ (since $A$ is symmetric.)}\\
& \stackrel{(*)}{=} \mathbf{u}^{\trans}\beta \mathbf{v}=\beta (\mathbf{u}^{\trans} \mathbf{v})=\beta (\mathbf{u}\cdot \mathbf{v}).
\end{align*}

Therefore we obtain
\[\alpha (\mathbf{u} \cdot \mathbf{v})=\beta (\mathbf{u} \cdot \mathbf{v}),\] and thus
\[(\alpha-\beta)(\mathbf{u} \cdot \mathbf{v})=0.\]

Since $\alpha$ and $\beta$ are distinct, $\alpha-\beta \neq 0$.
Hence we must have
\[\mathbf{u} \cdot \mathbf{v}=0,\] and the eigenvectors $\mathbf{u}, \mathbf{v}$ are orthogonal.

The post Orthogonality of Eigenvectors of a Symmetric Matrix Corresponding to Distinct Eigenvalues first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & -1\\
2& 3
\end{bmatrix}.\]

Find the eigenvalues and the eigenvectors of the matrix
\[B=A^4-3A^3+3A^2-2A+8E.\]

(Nagoya University Linear Algebra Exam Problem)
 

Hint.

Apply the Cayley-Hamilton theorem.
That is if $p_A(t)$ is the characteristic polynomial of the matrix $A$, then the matrix $p_A(A)$ is the zero matrix.

Solution.

Let us first find the characteristic polynomial $p_A(t)$ of the matrix $A$.
We have
\begin{align*}
p_A(t)&=\det(A-tI)=\begin{vmatrix}
1-t & -1\\
2& 3-t
\end{vmatrix}\\
&=(1-t)(3-t)-(-1)(2)=t^2-4t+5.
\end{align*}
Solving $t^2-4t+5=0$, we see that the matrix $A$ has the eigenvalues $2\pm i$ but it is not a good idea to use this directly to find the eigenvalues of the matrix $B$.
Instead, note that by the Cayley-Hamilton theorem, we know that
\[p_t(A)=A^2-4A+5I=O,\] where $I$ is the $2\times 2$ identity matrix and $O$ is the $2\times 2$ zero matrix.

Since we have
\[B=A^4-3A^3+3A^2-2A+8E=(A^2-4A+5I)(A^2+A+2I)+A-2I,\] we have
\[B=A-2I=\begin{bmatrix}
-1 & -1\\
2& 1
\end{bmatrix}.\] Since the eigenvalues of $A$ is $2\pm i$, the eigenvalues of $B=A-2I$ are
\[(2\pm i)-2=\pm i.\]

Next, we find eigenvectors.
Let us first find eigenvectors corresponding to the eigenvalue $i$.
We have
\begin{align*}
A-iI&=\begin{bmatrix}
-1-i & -1\\
2& 1-i
\end{bmatrix}
\xrightarrow{(-1+i)R_1}
\begin{bmatrix}
2 & 1-i\\
2 & 1-i
\end{bmatrix}\\
&
\xrightarrow{R_2-R_1}
\begin{bmatrix}
2 & 1-i\\
0 & 0
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_1}
\begin{bmatrix}
1 & (1-i)/2\\
0 & 0
\end{bmatrix}.
\end{align*}
Thus we have
\[x_1=-\frac{1-i}{2}\] and the eigenvectors associated with the eigenvalue $i$ are
\[\mathbf{x}=x_2\begin{bmatrix}
-\frac{1-i}{2} \\
1
\end{bmatrix},\] where $x_2$ is any nonzero complex number.
Or equivalently, scaling the vector by $-1+i$, the eigenvectors corresponding to the eigenvalue $i$ are
\[a\begin{bmatrix}
1 \\
-1-i
\end{bmatrix},\] where $a$ is any nonzero complex number.

Since $B$ is a real matrix and the eigenvalues $i$ and $-i$ are complex conjugate to each other, the eigenvectors of $-i$ are just the conjugates of eigenvectors of $i$. Thus the eigenvectors corresponding to the eigenvalue $-i$ are
\[b\begin{bmatrix}
1 \\
-1+i
\end{bmatrix},\] where $b$ is any nonzero complex number.

The post Find the Eigenvalues and Eigenvectors of the Matrix $A^4-3A^3+3A^2-2A+8E$. first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
\frac{1}{7} & \frac{3}{7} & \frac{3}{7} \\
\frac{3}{7} &\frac{1}{7} &\frac{3}{7} \\
\frac{3}{7} & \frac{3}{7} & \frac{1}{7}
\end{bmatrix}\] be $3 \times 3$ matrix. Find

\[\lim_{n \to \infty} A^n.\]

(Nagoya University Linear Algebra Exam)

Hint.

1. The matrix $A$ is symmetric, hence diagonalizable.
2. Diagonalize $A$.
3. You may want to find eigenvalues
   without computing the characteristic polynomial for $A$.

Solution.

Note that the matrix $A$ is symmetric, hence it is diagonalizable.
We observe several things to simplify the computation.

First, note that the sum of the entries in each row is $1$.
(Such a matrix is call a stochastic matrix. See the comment below.)
Thus the matrix $A$ has eigenvalue $1$ and $\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}$ is an eigenvector.

Also note that if we add $2/7(=-\lambda)$ to diagonal entries, then every entry becomes $3/7$.
That is,
\[A+\frac{2}{7}I=\begin{bmatrix}
\frac{3}{7} & \frac{3}{7} & \frac{3}{7} \\
\frac{3}{7} &\frac{3}{7} &\frac{3}{7} \\
\frac{3}{7} & \frac{3}{7} & \frac{3}{7}
\end{bmatrix},\] which is clearly singular.

From this observation, we notice that $-2/7$ is an eigenvalue of $A$ and eigenvectors are nonzero solution of $x_1+x_2+x_3=0$. Therefore $\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}$ and $\begin{bmatrix}
0 \\
1 \\
-1
\end{bmatrix}$ are linearly independent eigenvectors for the eigenvalue $-2/7$.

Hence the geometric (and hence algebraic) multiplicity of the eigenvalue $-2/7$ is $2$.
(Note that we found all eigenvalues of $A$ without actually computing the characteristic polynomial.)

Now the invertible matix $P=\begin{bmatrix}
1 & 1 & 0 \\
1 &-1 &1 \\
1 & 0 & -1
\end{bmatrix}$ diagonalize $A$. Namely we have
\[P^{-1}AP=\begin{bmatrix}
1 & 0 & 0 \\
0 &-2/7 &0 \\
0 & 0 & -2/7
\end{bmatrix}.\] Hence
\[A=P\begin{bmatrix}
1 & 0 & 0 \\
0 &-2/7 &0 \\
0 & 0 & -2/7
\end{bmatrix}P^{-1}.\] Therefore we have
\[ A^n=P\begin{bmatrix}
1^n & 0 & 0 \\
0 & (-2/7)^n &0 \\
0 & 0 & (-2/7)^n
\end{bmatrix}P^{-1}
\text{ and } \lim_{n \to \infty} A^n=P\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 &0 \\
0 & 0 & 0
\end{bmatrix}P^{-1}. \] Find the inverse $P^{-1}$ by your favorite method
\[P^{-1}=\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt] \frac{2}{3} &\frac{-1}{3} &\frac{-1}{3} \\[6pt] \frac{1}{3} & \frac{1}{3} & \frac{-2}{3}
\end{bmatrix}.\] Then the answer is
\[\lim_{n \to \infty} A^n=\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[6pt] \frac{1}{3} &\frac{1}{3} &\frac{1}{3} \\[6pt] \frac{1}{3} & \frac{1}{3} & \frac{1}{3}
\end{bmatrix}.\]

Comment.

You might come up with the idea to use the diagonalization as it is an exam problem in linear algebra.
But you soon realize that computing the characteristic polynomial for $A$ is a bit messy.
So sometimes it is important to find eigenvalues without finding the roots of the characteristic polynomial.

We observed that

1. the sum of row entries of $A$ is $1$
2. if we add $(2/7)I$ to $A$ then we obtain a matrix whose entries are all identical.

These observation yields the eigenvalues of $A$ without using the characteristic polynomial.

If the sum of entries of each row of a matrix is $1$, then the matrix is called called a stochastic matrix (or Markov matrix, probability matrix).
The matrix $A$ in the problem is an example of a stochastic matrix.

Stochastic matrices have always $1$ as an eigenvalue.
see the post
 Stochastic matrix (Markov matrix) and its eigenvalues and eigenvectors
for an example of a $2\times 2$ stochastic matrix.

The post Find the Limit of a Matrix first appeared on Problems in Mathematics.