Idempotent Matrices. 2007 University of Tokyo Entrance Exam Problem

Tokyo University Linear Algebra Exam Problems and Solutions

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)
 
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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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