# Idempotent Matrices. 2007 University of Tokyo Entrance Exam Problem

## Problem 265

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.

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