For a real number $a$, consider $2\times 2$ matrices $A, P, Q$ satisfying the following five conditions.

1. $A=aP+(a+1)Q$
2. $P^2=P$
3. $Q^2=Q$
4. $PQ=O$
5. $QP=O$,

where $O$ is the $2\times 2$ zero matrix.
Then do the following problems.

(a) Prove that $(P+Q)A=A$.

(b) Suppose $a$ is a positive real number and let
\[ A=\begin{bmatrix}
a & 0\\
1& a+1
\end{bmatrix}.\] Then find all matrices $P, Q$ satisfying conditions (1)-(5).

(c) Let $n$ be an integer greater than $1$. For any integer $k$, $2\leq k \leq n$, we define the matrix
\[A_k=\begin{bmatrix}
k & 0\\
1& k+1
\end{bmatrix}.\] Then calculate and simplify the matrix product
\[A_nA_{n-1}A_{n-2}\cdots A_2.\]

(Tokyo University Entrance Exam 2007)
 

Solution.

(a) Prove that $(P+Q)A=A$.

We have
\begin{align*}
(P+Q)A&\stackrel{(1)}{=} (P+Q)(aP+(a+1)Q)\\
&=aP^2+(a+1)PQ+aQP+(a+1)Q^2\\
&=aP+(a+1)O+aO+(a+1)Q \\
&\text{[ by (2), (3), (4), (5)]}\\
&=aP+(a+1)Q\stackrel{(1)}{=}A.
\end{align*}
Hence, we obtain
\[(P+Q)A=A\] as required.

(b) Find all matrices $P, Q$

Note that the matrix $A=\begin{bmatrix}
a & 0\\
1& a+1
\end{bmatrix}$ is invertible since the determinant
\[\det(A)=\begin{vmatrix}
a & 0\\
1& a+1
\end{vmatrix}=a(a+1)\neq 0.\] By part (a), we know $(P+Q)A=A$.
Multiplying this by $A^{-1}$ from the right, we have
\[P+Q=I,\] where $I$ is the $2\times 2$ identity matrix.
Substituting $Q=I-P$ into the equality $A=aP+(a+1)Q$ of (1), we have
\begin{align*}
A&=aP+(a+1)(I-P)\\
&=(a+1)I-P.
\end{align*}

Thus, we have
\begin{align*}
P&=(a+1)I-A\\[6pt] &=\begin{bmatrix}
a+1 & 0\\
0& a+1
\end{bmatrix}-\begin{bmatrix}
a & 0\\
1& a+1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
1 & 0\\
-1& 0
\end{bmatrix},
\end{align*}
and thus
\begin{align*}
Q&=I-P\\[6pt] &=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}-\begin{bmatrix}
1 & 0\\
-1& 0
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
0 & 0\\
1& 1
\end{bmatrix}.
\end{align*}
It is straightforward to check that the matrices
\[P=\begin{bmatrix}
1 & 0\\
-1& 0
\end{bmatrix} \text{ and } Q=\begin{bmatrix}
0 & 0\\
1& 1
\end{bmatrix}\] satisfy conditions (1)-(5).
Hence these are the only matrices satisfying conditions (1)-(5).

(c) Calculate and simplify the matrix product $A_nA_{n-1}A_{n-2}\cdots A_2$

In part (2), we showed that for any positive integer $k$
\[ A_k=kP+(k+1)Q,\] where
\[P=\begin{bmatrix}
1 & 0\\
-1& 0
\end{bmatrix} \text{ and } Q=\begin{bmatrix}
0 & 0\\
1& 1
\end{bmatrix}.\] (We just applied the result of (b) with $a=k$.)

Thus we have
\begin{align*}
&A_n A_{n-1} \cdots A_2 \\
&=\left(nP+(n+1)Q\right) \left((n-1)P+nQ\right)\cdots \left (2P+3Q\right)\\[6pt] &=n! P+\frac{(n+1)!}{2} Q
\end{align*}
We used conditions (4) and (5) in the second equality.
Using the explicit matrices for $P$ and $Q$, we have
\begin{align*}
&n! P+\frac{(n+1)!}{2} Q\\[6pt] &=n!\begin{bmatrix}
1 & 0\\
-1& 0
\end{bmatrix}+\frac{(n+1)!}{2}\begin{bmatrix}
0 & 0\\
1& 1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
n! & 0\\
-n!+\frac{(n+1)!}{2}& \frac{(n+1)!}{2}
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
n! & 0\\
n!\frac{n-1}{2}& \frac{(n+1)!}{2}
\end{bmatrix}.
\end{align*}

Note that in the last step, we computed
\begin{align*}
&-n!+\frac{(n+1)!}{2}=-n!+\frac{(n+1)n!}{2}\\[6pt] &=n!(-1+\frac{n+1}{2})\\[6pt] &=n!\frac{n-1}{2}.
\end{align*}

In conclusion, we have obtained
\[A_n A_{n-1} \cdots A_2 =\begin{bmatrix}
n! & 0\\
n!\frac{n-1}{2}& \frac{(n+1)!}{2}
\end{bmatrix}.\]

Comment.

Another way to solve (c) is that one first guesses the formula we obtained and prove it by mathematical induction.

The post Idempotent Matrices. 2007 University of Tokyo Entrance Exam Problem first appeared on Problems in Mathematics.

Let $a$ and $b$ be two distinct positive real numbers. Define matrices
\[A:=\begin{bmatrix}
0 & a\\
a & 0
\end{bmatrix}, \,\,
B:=\begin{bmatrix}
0 & b\\
b& 0
\end{bmatrix}.\]

Find all the pairs $(\lambda, X)$, where $\lambda$ is a real number and $X$ is a non-zero real matrix satisfying the relation
\[AX+XB=\lambda X. \tag{*} \]

(The University of Tokyo Linear Algebra Exam)

Hint.

1. Let $X=\begin{bmatrix}
   x_1 & x_2\\
   x_3& x_4
   \end{bmatrix}$ and compute (*).
2. Compare entries of matrices.
3. Rewrite it as a matrix equation.
4. The problem is now translated to a problem of eigenvalue/eigenvectors.

Solution.

Let $X=\begin{bmatrix}
x_1 & x_2\\
x_3& x_4
\end{bmatrix}$.
Then the relation (*) becomes
\[ a\begin{bmatrix}
x_3 & x_4\\
x_1& x_2
\end{bmatrix}
+b \begin{bmatrix}
x_2 & x_1\\
x_4& x_3
\end{bmatrix}
=
\lambda \begin{bmatrix}
x_1 & x_2\\
x_3& x_4
\end{bmatrix}.\] Comparing the entries of matrices, we obtain four equations
\begin{align*}
ax_3+b x_2 &=\lambda x_1\\
ax_4+ b x_1 &=\lambda x_2\\
ax_1+b x_4 &=\lambda x_3 \\
a x_2+b x_3 &=\lambda x_4.
\end{align*}
We rewrite these equations into the matrix equation
\[ \begin{bmatrix}
-\lambda & b & a & 0 \\
b &-\lambda & 0 & a \\
a & 0 & -\lambda & b \\
0 & a & b & -\lambda
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}=
\begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}.\] Thus the problem is to find all the pairs $(\lambda, X)$ satisfying this matrix equation.
Let
\[C= \begin{bmatrix}
0 & b & a & 0 \\
b & 0 & 0 & a \\
a & 0 & 0 & b \\
0 & a & b & 0
\end{bmatrix}.\]

Then the problem is equivalent to find all eigenvalues and eigenvectors of the matrix $C$.
So we first compute the characteristic polynomial of $C$ to find eigenvalues.
We have
\begin{align*}
\det(C-\lambda I)&= \begin{vmatrix}
-\lambda & b & a & 0 \\
b &-\lambda & 0 & a \\
a & 0 & -\lambda & b \\
0 & a & b & -\lambda
\end{vmatrix}\\
&=\lambda^4-2(a^2+b^2)\lambda^2+(a^2-b^2)^2.
\end{align*}
(Use cofactor expansion and simplify to obtain this.)
We solve
\[\lambda^4-2(a^2+b^2)\lambda^2+(a^2-b^2)^2=0\] for $\lambda$.
By the quadratic formula we have
\begin{align*}
\lambda^2&=a^2+b^2\pm \sqrt{(a^2+b^2)^2-(a^2-b^2)^2}\\
&=a^2+b^2\pm2ab =(a\pm b)^2.
\end{align*}
Hence we obtained four eigenvalues $\lambda=\pm (a\pm b)$.
Note that since we have four distinct eigenvalues, each eigenspace is one dimensional.

Now, let us find eigenvectors. First consider the eigenvalue $\lambda=a+b$.
In this case,
\begin{align*}
C-\lambda I =\begin{bmatrix}
-a-b & b & a & 0 \\
b & -a-b & 0 & a \\
a & 0 & -a-b & b \\
0 & a & b & -a-b
\end{bmatrix}.
\end{align*}
The eigenvector is the solutions of $(C-\lambda I)\mathbf{x}=\mathbf{0}$. The typical method to find eigenvector is to reduce the matrix $C-\lambda I$ into row echelon form.
But as we noted earlier each eigenspace is one dimensional, so if we find one nonzero vector solution, then it is a basis of the eigenspace.

For this specific matrix, we see that $\mathbf{x}=\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}$ satisfies $(C-\lambda I)\mathbf{x}=\mathbf{0}$. Therefore all the eigenvectors corresponding to eigenvalue $\lambda=a+b$ is
\[\mathbf{x}=\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}t\] for any nonzero scalar $t$.

Similarly, we find that
\[\begin{bmatrix}
-1 \\
1 \\
1 \\
-1
\end{bmatrix}t, \quad \begin{bmatrix}
1 \\
-1 \\
1 \\
-1
\end{bmatrix}t, \quad \begin{bmatrix}
1 \\
1 \\
-1 \\
-1
\end{bmatrix}t\] for any nonzero $t$ are all the eigenvector corresponding to eigenvalue $\lambda=-a-b, a-b, -a+b$, respectively.

Therefore the solution to the problem is that the pair $(\lambda, X)$ is one of the following paris.
\begin{align*}
\left(a+b, t\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}\right), \,\,
\left(-a-b, t \begin{bmatrix}
-1 & 1\\
1& -1
\end{bmatrix}\right),\\
\left(a-b, t\begin{bmatrix}
1 & -1\\
1& -1
\end{bmatrix}\right), \,\,
\left(-a+b, t\begin{bmatrix}
1 & 1\\
-1& -1
\end{bmatrix}\right)
\end{align*}
for nonzero $t$ (because $X$ must be a nonzero matrix.)

Comment.

When we find an eigenvector, you could have used elementary row operations to reduce a matrix.
The reduction for this matrix is not impossible, but it takes time and a sheet of paper.

So I think it is better to use the fact that the eigenspace is 1 dimensional (for the current problem)
and guess one nonzero eigenvector like I did above.

The post Find All Matrices Satisfying a Given Relation first appeared on Problems in Mathematics.

Let $A$ be a $4\times 4$ real symmetric matrix. Suppose that $\mathbf{v}_1=\begin{bmatrix}
-1 \\
2 \\
0 \\
-1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $1$ of $A$.
Suppose that the eigenspace for the eigenvalue $2$ is $3$-dimensional.

(a) Find an orthonormal basis for the eigenspace of the eigenvalue $2$ of $A$.

(b) Find $A\mathbf{v}$, where
\[ \mathbf{v}=\begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}.\]

(The University of Tokyo Linear Algebra Exam)

Hint.

1. A symmetric matrix is diagonalizable. Hence the sum of dimensions of eigenspaces is $4$.
2. Show that the eigenspaces for eigenvalues $1$ and $2$ are orthogonal.
3. To obtain an orthonormal basis from any basis, use Gram-Schmidt process and then normalize the lengths.
4. For (b), express the vector $\mathbf{v}$ as a linear combination of a basis consisting of eigenvectors.

Proof.

(a) Find an orthonormal basis for the eigenspace of the eigenvalue $2$ of $A$.

Note that since $A$ is symmetric, it is diagonalizable. Thus $\C^4$ is a direct sum of eigenspaces of $A$.
Since the eigenspace $E_2$ for the eigenvalue $2$ has dimension $3$, the eigenspace $E_1$ for the eigenvalue $1$ must have dimension $1$.

Thus $\mathbf{v}_1$ is a basis for $E_1$. We also see that there is no other eigenvalues. Hence $C^4=E_1\oplus E_2$.

We fist claim that $E_1$ and $E_2$ are orthogonal.

Let $x\in E_1$ and $y \in E_2$.
Then we calculate the inner product
\begin{align*}
(x, y)&=(Ax, y)=\bar{x}^{\trans}\bar{A}^{\trans}y = \bar{x}^{\trans}Ay = (x, Ay)=(x,2y)=2(x,y)
\end{align*}
Thus we have $(x,y)=0$, hence $x$ and $y$ are orthogonal.

Since $\mathbf{v}_1$ is a basis for $E_1$, if $\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in E_2$, we have
\[-x_1+2x_2-x_4=0.\] Solving this we see that
\[x=s\begin{bmatrix}
2 \\
1 \\
0 \\
0
\end{bmatrix}+t \begin{bmatrix}
0 \\
0 \\
1 \\
0
\end{bmatrix}+u\begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix}.\] Thus a basis of $E_3$ is
\[\left\{\, \begin{bmatrix}
2 \\
1 \\
0 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix}\, \right\}. \] Let us call these vectors $b_1, b_2, b_3$.
The note that $b_1\cdot b_2=0$, $b_2\cdot b_3=0$, but $b_1\cdot b_3=-2$.
Hence they are not orthogonal.

We apply Gram-Schmidt orthogonalization process.
Let $b_3’=b_3+ab_1$ be a vector that is orthogonal to $b_1$ for some $a$.
Taking the inner product with $b_1$, we get
$0=b_1\cdot b_3+ab_1\cdot b_1=-2+a5$.
Thus $a=2/5$ and
\[b_3’=b_3+(2/5)b_1=\frac{1}{5}\begin{bmatrix}
-1 \\
2 \\
0 \\
5
\end{bmatrix}.\] Then $b_1, b_2, b_3’$ are orthogonal vectors.

Now we normalize them so that the lengths of them become $1$.
We divide each vector with its length and obtain a orthonormal basis for $E_3$
\[ \left \{\,\frac{1}{5}\begin{bmatrix}
2 \\
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
1 \\
0
\end{bmatrix}, \frac{1}{30}\begin{bmatrix}
-1 \\
2 \\
0 \\
5
\end{bmatrix} \,\right\}.\]

(b) Find $A\mathbf{v}$

We express $\mathbf{v}$ as a linear combination of orthogonal basis $\{\mathbf{v}_1, b_1 b_2, b_3′ \}$ of $C^4=E_1\oplus E_2$.
Let
\[\mathbf{v}=\begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}=c_1\begin{bmatrix}
-1 \\
2 \\
0 \\
-1
\end{bmatrix}+c_2\begin{bmatrix}
2 \\
1 \\
0 \\
0
\end{bmatrix}+c_3\begin{bmatrix}
0 \\
0 \\
1 \\
0
\end{bmatrix}+c_4\begin{bmatrix}
-1 \\
2 \\
0 \\
5
\end{bmatrix}.\] (I scaled $b_3’$ to $5b_3’$. This does not change the orthogonality.)
We solve this for $c_i$. (There are several ways to do this.)
We take the inner product of this linear combination with $\mathbf{v}_1$ and obtain
$ -1=6c_1$, thus $c_1=-1/6$.

Similarly taking inner product with other basis vectors, we obtain $c_2=2/5$, $c_3=0$, and $c_4=-1/30$.
Then we compute
\begin{align*}
A\mathbf{v} &= A\left( \frac{-1}{6}\begin{bmatrix}
-1 \\
2 \\
0 \\
-1
\end{bmatrix}+\frac{2}{5}\begin{bmatrix}
2 \\
1 \\
0 \\
0
\end{bmatrix}+\frac{-1}{30}\begin{bmatrix}
-1 \\
2 \\
0 \\
5
\end{bmatrix} \right)
\\[6pt] &= \frac{-1}{6}A\begin{bmatrix}
-1 \\
2 \\
0 \\
-1
\end{bmatrix}+\frac{2}{5}A\begin{bmatrix}
2 \\
1 \\
0 \\
0
\end{bmatrix}+\frac{-1}{30}A\begin{bmatrix}
-1 \\
2 \\
0 \\
5
\end{bmatrix}
\\[6pt] &= \frac{-1}{6}\begin{bmatrix}
-1 \\
2 \\
0 \\
-1
\end{bmatrix}+\frac{4}{5}\begin{bmatrix}
2 \\
1 \\
0 \\
0
\end{bmatrix}+\frac{-2}{30}\begin{bmatrix}
-1 \\
2 \\
0 \\
5
\end{bmatrix}
\\[6pt] &=\frac{1}{6}\begin{bmatrix}
11 \\
2 \\
0 \\
-1
\end{bmatrix}.
\end{align*}
Therefore the answer is
\[A\mathbf{v}=\frac{1}{6}\begin{bmatrix}
11 \\
2 \\
0 \\
-1
\end{bmatrix}. \]

Comment.

At first glance, there seems insufficient amount of information to answer the question, but everything we need to solve this problem was given.
This is one of the final exam problems of linear algebra course taught at the University of Tokyo.
I like this problem because the statement is concise yet it requires some essential knowledge/skills of linear algebra.

The post Symmetric Matrix and Its Eigenvalues, Eigenspaces, and Eigenspaces first appeared on Problems in Mathematics.