The following problems are True or False.

Let $A$ and $B$ be $n\times n$ matrices.

(a) If $AB=B$, then $B$ is the identity matrix.
(b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions.
(c) If $A$ is invertible, then $ABA^{-1}=B$.
(d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix.
(e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

Solution.

(a) True or False: if $AB=B$, then $B$ is the identity matrix.

False. For example, if $B$ is the zero matrix, then of course we have $AB=B$ as both sides are the zero matrix.

(b) True or False: if the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions.

False. If the coefficient matrix $A$ is invertible, the system has a unique solution $\mathbf{x}=A^{-1}\mathbf{b}$.

(c) True or False: if $A$ is invertible, then $ABA^{-1}=B$.

False. The given equality is equivalent to $AB=BA$. Even $A$ is invertible, matrix multiplication is not commutative. As a counterexample, consider
\[A=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0\\
-1& 1
\end{bmatrix}.\] Note that the determinant of $A$ is $\det(A)=1\neq 0$. Hence $A$ is invertible.
Yet, we have
\[AB=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
-1& 1
\end{bmatrix}=\begin{bmatrix}
0 & 1\\
-1& 1
\end{bmatrix}\] and
\[BA=\begin{bmatrix}
1 & 0\\
-1& 1
\end{bmatrix}\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}=\begin{bmatrix}
1 & 1\\
-1& 0
\end{bmatrix},\] and hence $AB\neq BA$.

(d) True or False: if $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix.

True. Since $A$ is idempotent, we have $A^2=A$. As $A$ is nonsingular, it is invertible. Thus, the inverse matrix $A^{-1}$ exists. Then we have
\[I=A^{-1}A=A^{-1}A^2=IA=A.\] Hence, such a matrix $A$ must be the identity matrix $I$.

(e) True or False: if $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

Note that any homogeneous system has the zero solution. In addition to the zero solution, this system has a solution $x_1=0, x_2=0, x_3=1$. So, it has at least two solutions. The only possibilities for the number of solutions of a system of linear equations are zero, one, or infinitely many.
So, we conclude that the system must have infinitely many solutions.

The post True or False Problems on Midterm Exam 1 at OSU Spring 2018 first appeared on Problems in Mathematics.

A square matrix $A$ is called idempotent if $A^2=A$.

(a) Let $\mathbf{u}$ be a vector in $\R^n$ with length $1$.
Define the matrix $P$ to be $P=\mathbf{u}\mathbf{u}^{\trans}$.

Prove that $P$ is an idempotent matrix.

(b) Suppose that $\mathbf{u}$ and $\mathbf{v}$ be unit vectors in $\R^n$ such that $\mathbf{u}$ and $\mathbf{v}$ are orthogonal.
Let $Q=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}$.

Prove that $Q$ is an idempotent matrix.

(c) Prove that each nonzero vector of the form $a\mathbf{u}+b\mathbf{v}$ for some $a, b\in \R$ is an eigenvector corresponding to the eigenvalue $1$ for the matrix $Q$ in part (b).

Proof.

(a) Prove that $P=\mathbf{u}\mathbf{u}^{\trans}$ is an idempotent matrix.

The length of the vector $\mathbf{u}$ is given by
\[\|\mathbf{u}\|=\sqrt{\mathbf{u}^{\trans}\mathbf{u}}.\] Since $\|\mathbf{u}\|=1$ by assumption, it yields that
\[\mathbf{u}^{\trans}\mathbf{u}=1.\]

Let us compute $P^2$ using the associative properties of matrix multiplication.
We have
\begin{align*}
P^2&=(\mathbf{u}\mathbf{u}^{\trans})(\mathbf{u}\mathbf{u}^{\trans})\\
&=\mathbf{u}(\mathbf{u}^{\trans}\mathbf{u})\mathbf{u}^{\trans}\\
&=\mathbf{u}( 1 ) \mathbf{u}^{\trans}=\mathbf{u}\mathbf{u}^{\trans}=P.
\end{align*}

Thus, we have obtained $P^2=P$, and hence $P$ is an idempotent matrix.

(b) Prove that $Q=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}$ is an idempotent matrix

Since $\mathbf{u}$ and $\mathbf{v}$ are unit vectors, we have as in part (a)
\[\mathbf{u}^{\trans}\mathbf{u}=1 \text{ and } \mathbf{v}^{\trans}\mathbf{v}=1.\] Since $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, their dot (inner) product is $0$.
Thus we have
\[\mathbf{u}^{\trans}\mathbf{v}=\mathbf{v}^{\trans}\mathbf{u}=0.\]

Using these identities, we compute $Q^2$ as follows.
We have
\begin{align*}
Q^2&=(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})\\
&=\mathbf{u}\mathbf{u}^{\trans}(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})
+\mathbf{v}\mathbf{v}^{\trans}(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})\\
&=\mathbf{u}\underbrace{\mathbf{u}^{\trans}\mathbf{u}}_{1}\mathbf{u}^{\trans}+\mathbf{u}\underbrace{\mathbf{u}^{\trans}\mathbf{v}}_{0}\mathbf{v}^{\trans}
+\mathbf{v}\underbrace{\mathbf{v}^{\trans}\mathbf{u}}_{0}\mathbf{u}^{\trans}+\mathbf{v}\underbrace{\mathbf{v}^{\trans}\mathbf{v}}_{1}\mathbf{v}^{\trans}\\[6pt] &=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}=Q.
\end{align*}
It follows that $Q$ is an idempotent matrix.

(c) Prove that $a\mathbf{u}+b\mathbf{v}$ is an eigenvector

Let us first compute $Q\mathbf{u}$. We have
\begin{align*}
Q\mathbf{u}&=(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})\mathbf{u}\\
&=\mathbf{u}\underbrace{\mathbf{u}^{\trans}\mathbf{u}}_{1}+\mathbf{v}\underbrace{\mathbf{v}^{\trans}\mathbf{u}}_{0}=\mathbf{u}.
\end{align*}

Note that $\mathbf{u}$ is a nonzero vector because it is a unit vector.
Thus, the equality $Q\mathbf{u}=\mathbf{u}$ implies that $1$ is an eigenvalue of $Q$ and $\mathbf{v}$ is a corresponding eigenvector.

Similarly, we can check that $\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1$.

Now let $a\mathbf{u}+b\mathbf{v}$ be a nonzero vector.
Then we have
\begin{align*}
Q(a\mathbf{u}+b\mathbf{v})&=aQ\mathbf{u}+bQ\mathbf{v}=a\mathbf{u}+b\mathbf{v}.
\end{align*}
It follows that $a\mathbf{u}+b\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1$.

Another way to prove (c)

Another way to see this is as follows.
As we saw above, the vectors $\mathbf{u}$ and $\mathbf{v}$ are eigenvectors corresponding to the eigenvalue $1$.
Hence $\mathbf{u}, \mathbf{v} \in E_{1}$, where $E_1$ is an eigenspace of the eigenvalue $1$

Note that $E_1$ is a vector space, hence $a\mathbf{u}+b\mathbf{v}$ is a nonzero vector in $E_{1}$.
Thus, $a\mathbf{u}+b\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1$ as well.

Related Question.

See the post ↴
Idempotent Matrix and its Eigenvalues
for solutions of this problem.

The post Unit Vectors and Idempotent Matrices first appeared on Problems in Mathematics.

Let $A$ be a square matrix such that
\[A^{\trans}A=A,\] where $A^{\trans}$ is the transpose matrix of $A$.
Prove that $A$ is idempotent, that is, $A^2=A$. Also, prove that $A$ is a symmetric matrix.

Hint.

Recall the basic properties of transpose matrices.
For matrices $A, B$, we have

1. $(AB)^{\trans}=B^{\trans}A^{\trans}$.
2. $A^{\trans \trans}=A$.

Proof.

We first prove that $A$ is a symmetric matrix.
We have
\begin{align*}
A^{\trans}&=(A^{\trans}A)^{\trans}\\
&=A^{\trans}A^{\trans\trans} && \text{by property 1}\\
&=A^{\trans}A && \text{by property 2}\\
&=A.
\end{align*}
Hence we obtained $A^{\trans}=A$, and thus $A$ is a symmetric matrix.

Now we prove that $A$ is idempotent.
We compute
\begin{align*}
A^2&=AA\\
&=A^{\trans}A && \text{since $A$ is symmetric}\\
&=A && \text{by assumption}.
\end{align*}
Therefore, the matrix $A$ satisfies $A^2=A$, and hence it is idempotent.

The post If $A^{\trans}A=A$, then $A$ is a Symmetric Idempotent Matrix first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.

 

We give three proofs of this problem. The first one proves that $\R^n$ is a direct sum of eigenspaces of $A$, hence $A$ is diagonalizable.

The second proof proves the direct sum expression as in proof 1 but we use a linear transformation.

The third proof discusses the minimal polynomial of $A$.

Range=Image, Null space=Kernel

In the following proofs, we use the terminologies range and null space of a linear transformation. These are also called image and kernel of a linear transformation, respectively.

Proof 1.

Recall that only possible eigenvalues of an idempotent matrix are $0$ or $1$.
(For a proof, see the post “Idempotent matrix and its eigenvalues“.)

Let
\[E_0=\{\mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}\} \text{ and } E_{1}\{\mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{x}\}\] be subspaces of $\R^n$.
(Thus, if $0$ and $1$ are eigenvalues, then $E_0$ and $E_1$ are eigenspaces.)

Let $r$ be the rank of $A$. Then by the rank-nullity theorem, the nullity of $A$
\[\dim(E_0)=n-r.\]

The rank of $A$ is the dimension of the range
\[\calR(A)=\{\mathbf{y} \in \R^n \mid \mathbf{y}=A\mathbf{x} \text{ for some } \mathbf{x}\in \R^n\}.\]

Let $\mathbf{y}_1, \dots, \mathbf{y}_r$ be basis vectors of $\calR(A)$.
Then there exists $\mathbf{x}_i\in \R^n$ such that
\[\mathbf{y}_i=A\mathbf{x}_i,\] for $i=1, \dots, r$.

Then we have
\begin{align*}
A\mathbf{y}_i&=A^2\mathbf{x}_i\\
&=A\mathbf{x}_i && \text{since $A$ is idempotent}\\
&=\mathbf{y}_i.
\end{align*}

It follows that $y_i\in E_1$.
Since $y_i, i=1,\dots, r$ form a basis of $\calR(A)$, they are linearly independent and thus we have
\[r\leq \dim(E_1).\]

We have
\begin{align*}
&n=\dim(\R^n)\\
&\geq \dim(E_0)+\dim(E_1) && \text{since } E_0\cap E_1=\{\mathbf{0}\}\\
&\geq (n-r)+r=n.
\end{align*}

So in fact all inequalities are equalities, and hence
\[\dim(\R^n)=\dim(E_0)+\dim(E_1).\]

This implies
\[\R^n=E_0 \oplus E_1.\] Thus $\R^n$ is a direct sum of eigenspaces of $A$, and hence $A$ is diagonalizable.

Proof 2.

Let $E_0$ and $E_1$ be as in proof 1.
Consider the linear transformation $T:\R^n \to \R^n$ represented by the idempotent matrix $A$, that is, $T(\mathbf{x})=A\mathbf{x}$.

Then the null space $\calN(T)$ of the linear transformation $T$ is $E_0$ by definition.

We claim that the range $\calR(T)$ is $E_1$.
If $\mathbf{x}\in \calR(T)$, then we have $\mathbf{y}\in \R^n$ such that $\mathbf{x}=T(\mathbf{y})=A\mathbf{y}$.

Then we have
\begin{align*}
\mathbf{x}&=A\mathbf{y}=A^2\mathbf{y} =A(A\mathbf{y})
=A\mathbf{x}.
\end{align*}
(The second equality follows since $A$ is idempotent.)

This implies that $\mathbf{x}\in E_1$, and hence $\calR(T) \subset E_1$.

On the other hand, if $\mathbf{x}\in E_1$, then we have
\[\mathbf{x}=A\mathbf{x}=T(\mathbf{x})\in \calR(T).\] Thus, we have $E_1\subset \calR(T)$. Putting these two inclusions together gives $E_1=\calR(T)$.

By the isomorphism theorem of vector spaces, we have
\[\R^n=\calN(A)\oplus \calR(T)=E_0\oplus E_1.\] Thus, $\R^n$ is a direct sum of eigenspaces of $A$ and hence $A$ is diagonalizable.

Proof 3.

Since $A$ is idempotent we have $A^2=A$.
Thus we have $A^2-A=O$, the zero matrix, and so $A$ satisfies the polynomial $x^2-x$.

If $x^2-x=x(x-1)$ is not the minimal polynomial of $A$, then $A$ must be either the identity matrix or the zero matrix.
Since these matrices are diagonalizable (as they are already diagonal matrices), we consider the case when $x^2-x$ is the minimal polynomial of $A$.

Since the minimal polynomial has two distinct simple roots $0, 1$, the matrix $A$ is diagonalizable.

Another Proof

A slightly different proof is given in the post “Idempotent (Projective) Matrices are Diagonalizable“.

The proof there is a variation of Proof 2.

The post Idempotent Matrices are Diagonalizable first appeared on Problems in Mathematics.

A square matrix $A$ is called idempotent if $A^2=A$.

(a) Suppose $A$ is an $n \times n$ idempotent matrix and let $I$ be the $n\times n$ identity matrix. Prove that the matrix $I-A$ is an idempotent matrix.

(b) Assume that $A$ is an $n\times n$ nonzero idempotent matrix. Then determine all integers $k$ such that the matrix $I-kA$ is idempotent.

(c) Let $A$ and $B$ be $n\times n$ matrices satisfying
\[AB=A \text{ and } BA=B.\] Then prove that $A$ is an idempotent matrix.

Proof.

(a) Prove that the matrix $I-A$ is an idempotent matrix.

To prove that $I-A$ is an idempotent matrix, we show that $(I-A)^2=I-A$.
We compute
\begin{align*}
(I-A)^2&=(I-A)(I-A)\\
&=I(I-A)-A(I-A)\\
&=I-A-A+A^2\\
&=I-2A+A^2\\
&=I-2A+A && \text{since $A$ is idempotent}\\
&=I-A.
\end{align*}
Thus, we have $(I-A)^2=I-A$ and $I-A$ is an idempotent matrix.

(b) Determine all integers $k$ such that the matrix $I-kA$ is idempotent.

Let us find out the condition on $k$ so that $I-kA$ is an idempotent matrix.
We have
\begin{align*}
&(I-kA)^2=(I-kA)(I-kA)\\
&=I(I-kA)-kA(I-kA)\\
&=I-kA-kA+k^2A^2\\
&=I-2kA+k^2A && \text{ since $A$ is idempotent}\\
&=I-(2k-k^2)A.
\end{align*}

It follows that $I-kA$ is idempotent if and only if $I-kA=I-(2k-k^2)A$, or equivalently $(k^2-k)A=O$, the zero matrix.
Since $A$ is not the zero matrix, we see that $I-kI$ is idempotent if and only if $k^2-k=0$.

Since $k^2-k=k(k-1)$, we conclude that $I-kA$ is an idempotent matrix if and only if $k=0, 1$.

(c) Prove that $A$ is an idempotent matrix.

Let $A$ and $B$ be $n\times n$ matrices satisfying
\[AB=A \tag{*}\] and
\[ BA=B. \tag{**}\] Our goal is to show that $A^2=A$.
We compute
\begin{align*}
&A^2=AA\\
&=(AB)A && \text{by (*)}\\
&=A(BA)\\
&=AB && \text{by (**)}\\
&=A && \text{by (*)}.
\end{align*}
Therefore, we have obtained the identity $A^2=A$, and we conclude that $A$ is an idempotent matrix.

Related Question.

The proofs are given in the post ↴
Unit Vectors and Idempotent Matrices

The post If $A$ is an Idempotent Matrix, then When $I-kA$ is an Idempotent Matrix? first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ idempotent complex matrix.
Then prove that $A$ is diagonalizable.

Definition.

An $n\times n$ matrix $A$ is said to be idempotent if $A^2=A$.
It is also called projective matrix.

Proof.

In general, an $n \times n$ matrix $B$ is diagonalizable if there are $n$ linearly independent eigenvectors. So if eigenvectors of $B$ span $\R^n$, then $B$ is diagonalizable.

We prove that $\R^n$ is spanned by eigenspaces. Every vector $\mathbf{v}\in \R^n$ can be expresses as
\[\mathbf{v}=(\mathbf{v}-A\mathbf{v})+A\mathbf{v}=\mathbf{v}_0+\mathbf{v}_1,\] where we put $\mathbf{v}_0=\mathbf{v}-A\mathbf{v}$ and $\mathbf{v}_1=A\mathbf{v}$.

We claim that $\mathbf{v}_0$ and $\mathbf{v}_1$ are elements in the eigenspaces corresponding to (possible) eigenvalues $0$ and $1$, respectively.
To see this, we compute
\begin{align*}
A\mathbf{v}_0&=A(\mathbf{v}-A\mathbf{v})\\
&=A\mathbf{v}-A^2\mathbf{v}\\
&=A\mathbf{v}-A\mathbf{v} && \text{since $A$ is idempotent}\\
&=O=0\mathbf{v}_0.
\end{align*}

Thus, we have $A\mathbf{v}_0=0\mathbf{v}_0$, and this means that $\mathbf{v}_0$ is a vector in the eigenspace corresponding to the eigenvalue $0$.
(If $0$ is not an eigenvalue of $A$, then $\mathbf{v}_0=\mathbf{0}$.)

We also have
\begin{align*}
A\mathbf{v}_1=A(A\mathbf{v})=A^2\mathbf{v}=A\mathbf{v}=\mathbf{v}_1,
\end{align*}
where the third equality holds as $A$ is idempotent.
This implies that $\mathbf{v}_1$ is a vector in the eigenspace corresponding to eigenvalue $1$. (If $1$ is not an eigenvalue of $A$, then $\mathbf{v}_1=\mathbf{0}$.)

It follows that every vector $\mathbf{v}\in \R^n$ is a sum of eigenvectors (or the zero vector).
That is, $\R^n$ is spanned by eigenvectors.

By the general fact mentioned at the beginning of the proof, we conclude that the idempotent matrix $A$ is diagonalizable.

Different Proofs

Three other different proofs of the fact that every idempotent matrix is diagonalizable are given in the post “Idempotent Matrices are Diagonalizable“.

The post Idempotent (Projective) Matrices are Diagonalizable first appeared on Problems in Mathematics.

Let $A$ be the matrix for a linear transformation $T:\R^n \to \R^n$ with respect to the standard basis of $\R^n$.
We assume that $A$ is idempotent, that is, $A^2=A$.
Then prove that
\[\R^n=\im(T) \oplus \ker(T).\]

Proof.

To prove the equality $\R^n=\im(T) \oplus \ker(T)$, we need to prove
(a) $\R^n=\im(T) + \ker(T)$, and
(b) $\im(T) \cap \ker(T)=\{0\}$.

By definition, the image $\im(T)$ and $\ker(T)$ are subspaces of $\R^n$, hence $\im(T) + \ker(T) \subset \R^n$.
To prove the reverse inclusion, for any $x\in \R^n$, we write
\begin{align*}
x=Ax+(x-Ax).
\end{align*}
Then the first term is in $\im(T)$ since
\[Ax=T(x)\in \im(T).\] The second term $x-Ax$ is in $\ker(T)$ since
\begin{align*}
T(x-Ax)&=A(x-Ax)\\
&=Ax-A^2x\\
&=Ax-Ax && (\text{since $A$ is idempotent})\\
&=0.
\end{align*}

Thus we have
\[x=\underbrace{Ax}_{\in \im(T)}+\underbrace{(x-Ax)}_{\in \ker(T)}\in \im(T) + \ker(T) .\] Since $x$ is arbitrary element in $\R^n$, we have
\[\R^n\subset \im(T) + \ker(T),\] and putting the two inclusions together yields
\[\R^n= \im(T) + \ker(T),\] and we proved (a).

To prove (b), let $x\in \im(T) \cap \ker(T)$. Thus $x\in \im(T)$ and $x\in \ker(T)$.
Since $x\in \im(T)$, there exists $x’\in \R^n$ such that $T(x’)=x$, or equivalently, $Ax’=x$.

Then, we have
\begin{align*}
&0=T(x)=Ax\\
&=A(Ax’)=A^2x’\\
&=Ax’ && (\text{since $A$ is idempotent})\\
&=x.
\end{align*}

Hence we have proved an arbitrary element $x$ in the intersection is $x=0$, and thus we have
\[\im(T) \cap \ker(T)=\{0\}.\] So (b) is proved.

The facts (a), (b) implies that we have
\[\R^n=\im(T) \oplus \ker(T),\] as required.

The post Idempotent Linear Transformation and Direct Sum of Image and Kernel first appeared on Problems in Mathematics.

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.

Suppose the following information is known about a $3\times 3$ matrix $A$.
\[A\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}=6\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix},
\quad
A\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}=3\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}, \quad
A\begin{bmatrix}
2 \\
-1 \\
0
\end{bmatrix}=3\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}.\]

(a) Find the eigenvalues of $A$.

(b) Find the corresponding eigenspaces.

(c) In each of the following questions, you must give a correct reason (based on the theory of eigenvalues and eigenvectors) to get full credit.
Is $A$ a diagonalizable matrix?
Is $A$ an invertible matrix?
Is $A$ an idempotent matrix?

(Johns Hopkins University Linear Algebra Exam)
 

Solution.

(a) Find the eigenvalues of $A$.

From the first and the second equations, we see that $6$ and $3$ are eigenvalues of $A$.
From the second and the third equations, we have
\[A\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}=3\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}=A\begin{bmatrix}
2 \\
-1 \\
0
\end{bmatrix}\] and thus we have
\[\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}=A\left( \begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}- \begin{bmatrix}
2 \\
-1 \\
0
\end{bmatrix}\right)=A \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}.\] This yields that
\[A\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}=0\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \tag{*}\] and we conclude that $0$ is an eigenvalue.
Since the size of the matrix $A$ is $3\times 3$, it has at most three eigenvalues. Thus we found them all. Eigenvalues of $A$ are $0, 3, 6$.

(b) Find the corresponding eigenspaces.

From the first equation, the vector $\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $6$.
We also see from the second equation, the vector $\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $3$.
The equation (*) in the solution (a) also tells us that the vector $\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $0$.

Note that we have three distinct eigenvalues for $3\times 3$ matrix $A$. So all the algebraic multiplicities for eigenvalues are $1$, hence the geometric multiplicities must be $1$ since the geometric multiplicity is always less than or equal to the algebraic multiplicity.
Therefore, each eigenspace has dimension $1$. We already found a nonzero vector in each eigenspace, and thus the vector is a basis vector for each eigenspace.
We denote $E_{\lambda}$ for the eigenspace for eigenvector $\lambda$. Then we have
\[E_0=\Span\left(\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\right), \quad
E_3=\Span\left(\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}\right), \quad
E_6=\Span\left(\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}\right).\]

(c) Diagonalizable matrix? Invertible matrix? Idempotent matrix?

Since $A$ has three distinct eigenvalues, $A$ is diagonalizable. (Or, algebraic multiplicities are the same as geometric multiplicities.)
Since the matrix $A$ has $0$ as an eigenvalue. Thus $A$ is not invertible.

If $A$ is idempotent matrix, then the eigenvalues of $A$ is either $0$ or $1$. Thus $A$ is not an idempotent matrix.
(For a proof of this fact, see the post Eigenvalues of an idempotent matrix.)

The post Determine Eigenvalues, Eigenvectors, Diagonalizable From a Partial Information of a Matrix first appeared on Problems in Mathematics.

Let $A$ be an $n \times n$ matrix. We say that $A$ is idempotent if $A^2=A$.

(a) Find a nonzero, nonidentity idempotent matrix.

(b) Show that eigenvalues of an idempotent matrix $A$ is either $0$ or $1$.

(The Ohio State University, Linear Algebra Final Exam Problem)
 

Proof.

(a) Nonzero, nonidentity idempotent matrix

Let $A=\begin{bmatrix}
0 & 1\\
0 & 1
\end{bmatrix}$. Then $A$ is a nonzero, nonidentity matrix and $A$ is idempotent since we have
\[A^2=\begin{bmatrix}
0 & 1\\
0 & 1
\end{bmatrix}\begin{bmatrix}
0 & 1\\
0 & 1
\end{bmatrix}=\begin{bmatrix}
0 & 1\\
0 & 1
\end{bmatrix}=A.\]

(b) Eigenvalues of an idempotent matrix $A$ is either $0$ or $1$

Let $\lambda$ be an eigenvalue of the idempotent matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Namely we have
\[A\mathbf{x}=\lambda \mathbf{x}, \mathbf{x}\neq \mathbf{0}. \tag{*} \] Then we compute $A^2\mathbf{x}$ in two ways.
First, since $A$ is idempotent we have $A^2=A$ and thus we have
\[A^2\mathbf{x}=A\mathbf{x}\stackrel{(*)}{=} \lambda \mathbf{x}.\]

Next, we compute as follows.
\begin{align*}
A^2\mathbf{x}=A(A\mathbf{x})\stackrel{(*)}{=}A(\lambda \mathbf{x})=\lambda (A\mathbf{x})\stackrel{(*)}{=}\lambda (\lambda \mathbf{x})=\lambda^2\mathbf{x}.
\end{align*}

Comparing these two computations, we obtain
\[\lambda \mathbf{x}=\lambda^2 \mathbf{x}.\] Since $\mathbf{x}$ is a nonzero vector (because $\mathbf{x}$ is an eigenvector), we must have
\[\lambda=\lambda^2.\] Hence solving $\lambda(\lambda-1)=0$, the possible values for $\lambda$ is either $0$ or $1$.
Thus, the idempotent matrix $A$ only have eigenvalues $0$ or $1$.

Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

1. Find All the Eigenvalues of 4 by 4 Matrix
2. Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue
3. Diagonalize a 2 by 2 Matrix if Diagonalizable
4. Find an Orthonormal Basis of the Range of a Linear Transformation
5. The Product of Two Nonsingular Matrices is Nonsingular
6. Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not
7. Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials
8. Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable
9. Idempotent Matrix and its Eigenvalues
10. Diagonalize the 3 by 3 Matrix Whose Entries are All One
11. Given the Characteristic Polynomial, Find the Rank of the Matrix
12. Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$
13. Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$

The post Idempotent Matrix and its Eigenvalues first appeared on Problems in Mathematics.