# 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.

- $A=aP+(a+1)Q$
- $P^2=P$
- $Q^2=Q$
- $PQ=O$
- $QP=O$,

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*)

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## 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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