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.