Let
\[A=\begin{bmatrix}
-5 & 0 & 1 & 2 \\
3 &8 & -3 & 7 \\
0 & 11 & 13 & 28
\end{bmatrix}.\]

(a) What is the size of the matrix $A$?
(b) What is the third column of $A$?
(c) Let $a_{ij}$ be the $(i,j)$-entry of $A$. Calculate $a_{23}-a_{31}$.

Solution.

(a) What is the size of the matrix $A$?

The matrix $A$ has three rows and four columns. Thus, the size of $A$ is $3$ by $4$, or $3\times 4$.

(b) What is the third column of $A$?

The third column of $A$ is $\begin{bmatrix}
1 \\
-3 \\
13
\end{bmatrix}$.

Calculate $a_{23}-a_{31}$.

We see that $a_{23}=-3$ and $a_{31}=0$. Thus, $a_{23}-a_{31}=-3-0=-3$.

The post Elementary Questions about a Matrix first appeared on Problems in Mathematics.

Suppose that an $n \times m$ matrix $M$ is composed of the column vectors $\mathbf{b}_1 , \cdots , \mathbf{b}_m$.

Prove that a vector $\mathbf{v} \in \R^n$ can be written as a linear combination of the column vectors if and only if there is a vector $\mathbf{x}$ which solves the equation $M \mathbf{x} = \mathbf{v}$.

Proof.

Suppose that $\mathbf{v}$ is a linear combination of the vectors $\mathbf{b}_1, \dots, \mathbf{v}_m$. That is, suppose there are coefficients $x_1 , x_2 , \cdots , x_m$ such that
\[x_1 \mathbf{b}_1 + x_2 \mathbf{b}_2 + \cdots + x_m \mathbf{b}_m = \mathbf{v}.\]

Then define the column vector
\[\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_m \end{bmatrix}.\] Then $\mathbf{x}$ satisfies the equation $M \mathbf{x} = \mathbf{v}$.

On the other hand, suppose that there is a vector $\mathbf{x}$ which satisfies the desired equation with components $x_1 , x_2 , \cdots, x_m$.

Then these components can be used to find the linear combination
\[x_1 \mathbf{b}_1 + x_2 \mathbf{b}_2 + \cdots + x_m \mathbf{b}_m = M \mathbf{x} = \mathbf{v}.\] Hence, the vector $\mathbf{v}$ is a linear combination of column vectors $\mathbf{b}_1, \dots, \mathbf{b}_m$.

The post A Condition that a Vector is a Linear Combination of Columns Vectors of a Matrix first appeared on Problems in Mathematics.

Prove the following identity for any positive integer $n$.
\[\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}^n=\begin{bmatrix}
\cos n\theta & -\sin n\theta\\
\sin n\theta& \cos n\theta
\end{bmatrix}.\]

Hint.

Recall the addition theorem of trigonometric functions (sum formula)
\begin{align*}
\sin(x+y)&=\sin x \cos y +\cos x \sin y\\
\cos(x+y)&=\cos x \cos y -\sin x \sin y.
\end{align*}

Use the induction on $n$.

Proof.

We prove the identity by induction on $n$.
The base case $n=1$ is clear.

Suppose that the identity is true for $n=k$.
Then we have
\begin{align*}
&\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta & \cos \theta
\end{bmatrix}^{k+1}\\[6pt] =&
\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta & \cos \theta
\end{bmatrix}
\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta & \cos \theta
\end{bmatrix}^{k}\\[6pt] =&\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta & \cos \theta
\end{bmatrix}
\begin{bmatrix}
\cos k\theta & -\sin k\theta\\
\sin k\theta & \cos k\theta
\end{bmatrix}
\\
&\text{(by the induction hypothesis)}\\[6pt] =&\begin{bmatrix}
\cos \theta \cos k\theta – \sin \theta \sin k \theta & -\cos \theta \sin k\theta-\sin \theta \cos k \theta\\
\sin \theta \cos k \theta + \cos \theta \sin k \theta & -\sin \theta \sin k\theta +\cos \theta \cos k \theta
\end{bmatrix}
\\
&\text{(by matrix multiplication)}\\[6pt] =&\begin{bmatrix}
\cos (\theta+k\theta) & -\sin (\theta+ k\theta)\\
\sin (\theta+k\theta) & \cos (\theta+k\theta)
\end{bmatrix}\\
&\text{(by the addition theorem of trigonometric functions)}\\[6pt] =&\begin{bmatrix}
\cos (k+1)\theta & -\sin (k+1)\theta\\
\sin (k+1)\theta & \cos (k+1)\theta
\end{bmatrix}.
\end{align*}

This proves that the identity holds for $n=k+1$.
Thus by induction, the identity is true for all positive integers $n$.

The post The Powers of the Matrix with Cosine and Sine Functions 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 $n\times n$ matrix. Assume that every vector $\mathbf{x}$ in $\R^n$ is an eigenvector for some eigenvalue of $A$.
Prove that there exists $\lambda\in \R$ such that $A=\lambda I$, where $I$ is the $n\times n$ identity matrix.

Proof.

Let us write $A=(a_{ij})$.
Let $\mathbf{e}_i$ be the unit vector in $\R^n$ whose $i$-th entry is $1$, and $0$ elsewhere.
Then the vector $\mathbf{e}_i$ is an eigenvector corresponding to some eigenvalue $\lambda_{i}$. That is, we have
\[A\mathbf{e}_i =\lambda_{i} \mathbf{e}_i.\]

More explicitly, this shows that
\[\begin{bmatrix}
a_{1i} \\
a_{2i} \\
\vdots \\
a_{n-1 i}\\
a_{ni}
\end{bmatrix}=
\begin{bmatrix}
0 \\
\vdots \\
\lambda_{i} \\
\vdots \\
0
\end{bmatrix}.\] Therefore, it follows that $a_{ki}=0$ if $k\neq i$ and $a_{i i}=\lambda_{i}$.
So we have
\[A=\begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \\
0 &\lambda_2 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n
\end{bmatrix}.\]

Now we consider the vector $\mathbf{v}=\begin{bmatrix}
1 \\
1 \\
\vdots \\
1
\end{bmatrix} \in \R^n$ with all entries $1$. Then $\mathbf{v}$ is an eigenvector of an eigenvalue $\lambda$, that is, $A\mathbf{v}=\lambda \mathbf{v}$.
It follows that we have
\begin{align*}
\begin{bmatrix}
\lambda \\
\lambda \\
\vdots \\
\lambda
\end{bmatrix}
=
\lambda \mathbf{v}=A\mathbf{v}=
\begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \\
0 &\lambda_2 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n
\end{bmatrix}
\begin{bmatrix}
1 \\
1 \\
\vdots \\
1
\end{bmatrix}
=\begin{bmatrix}
\lambda_1 \\
\lambda_2 \\
\vdots \\
\lambda_n
\end{bmatrix}
\end{align*}

As a result, we must have $\lambda=\lambda_1=\cdots =\lambda_n$.
Thus, we obtain
\begin{align*}
A=\begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \\
0 &\lambda_2 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n
\end{bmatrix}
=\begin{bmatrix}
\lambda & 0 & \dots & 0 \\
0 &\lambda & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda
\end{bmatrix}
=\lambda I
\end{align*}
as required.

The post If Every Vector is Eigenvector, then Matrix is a Multiple of Identity Matrix first appeared on Problems in Mathematics.

Each of the following sets are not a subspace of the specified vector space. For each set, give a reason why it is not a subspace.
(1) \[S_1=\left \{\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \R^3 \quad \middle | \quad x_1\geq 0 \,\right \}\] in the vector space $\R^3$.

(2) \[S_2=\left \{\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \R^3 \quad \middle | \quad x_1-4x_2+5x_3=2 \,\right \}\] in the vector space $\R^3$.

(3) \[S_3=\left \{\, \begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2 \quad \middle | \quad y=x^2 \quad \,\right \}\] in the vector space $\R^2$.

(4) Let $P_4$ be the vector space of all polynomials of degree $4$ or less with real coefficients.
\[S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}\] in the vector space $P_4$.

(5) \[S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}\] in the vector space $P_4$.

(6) Let $M_{2 \times 2}$ be the vector space of all $2\times 2$ real matrices.
\[S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\} \] in the vector space $M_{2\times 2}$.

(7) \[S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\} \] in the vector space $M_{2\times 2}$.

(Linear Algebra Exam Problem, the Ohio State University)

(8) Let $C[-1, 1]$ be the vector space of all real continuous functions defined on the interval $[a, b]$.
\[S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\} \] in the vector space $C[-2, 2]$.

(9) \[S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$.

(10) Let $C^2[a, b]$ be the vector space of all real-valued functions $f(x)$ defined on $[a, b]$, where $f(x), f'(x)$, and $f^{\prime\prime}(x)$ are continuous on $[a, b]$. Here $f'(x), f^{\prime\prime}(x)$ are the first and second derivative of $f(x)$.
\[S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$.

(11) Let $S_{11}$ be the set of real polynomials of degree exactly $k$, where $k \geq 1$ is an integer, in the vector space $P_k$.

(12) Let $V$ be a vector space and $W \subset V$ a vector subspace. Define the subset $S_{12}$ to be the complement of $W$,
\[ V \setminus W = \{ \mathbf{v} \in V \mid \mathbf{v} \not\in W \}.\]

Solution.

Recall the following subspace criteria.
A subset $W$ of a vector space $V$ over the scalar field $K$ is a subspace of $V$ if and only if the following three criteria are met.

Thus, to prove a subset $W$ is not a subspace, we just need to find a counterexample of any of the three criteria.
  

Solution (1). $S_1=\{ \mathbf{x} \in \R^3 \mid x_1\geq 0 \}$

The subset $S_1$ does not satisfy condition 3. For example, consider the vector
\[\mathbf{x}=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}.\] Then since $x_1=1\geq 0$, the vector $\mathbf{x}\in S_1$. Then consider the scalar product of $\mathbf{x}$ and the scalar $-1$. Then we have
\[(-1)\cdot\mathbf{x}=\begin{bmatrix}
-1 \\
0 \\
0
\end{bmatrix},\] and the first entry is $-1$, hence $-\mathbf{x}$ is not in $S_1$. Thus $S_1$ does not satisfy condition 3 and it is not a subspace of $\R^3$.
(You can check that conditions 1, 2 are met.)
  

Solution (2). $S_2= \{ \mathbf{x}\in \R^3\mid x_1-4x_2+5x_3=2 \}$

The zero vector of the vector space $\R^3$ is
\[\mathbf{0}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.\] Since the zero vector $\mathbf{0}$ does not satisfy the defining relation $x_1-4x_2+5x_3=2$, it is not in $S_2$. Hence condition 1 is not met, hence $S_2$ is not a subspace of $\R^3$.
(You can check that conditions 2, 3 are not met as well.)
  

Solution (3). $S_3=\{\mathbf{x}\in \R^2 \mid y=x^2 \quad \}$

Consider vectors
\[\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ and } \begin{bmatrix}
-1 \\
1
\end{bmatrix}.\] These are vectors in $S_3$ since both vectors satisfy the defining relation $y=x^2$.

However, their sum
\[\begin{bmatrix}
1 \\
1
\end{bmatrix} + \begin{bmatrix}
-1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
0 \\
1
\end{bmatrix}\] is not in $S_3$ since $1\neq 0^2$.
Hence condition 2 is not met, and thus $S_3$ is not a subspace of $\R^2$.
(You can check that condition 1 is fulfilled yet condition 3 is not met.)
  

Solution (4). $S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}$

Consider the polynomial $f(x)=x$. Since the degree of $f(x)$ is $1$ and $f(1)=1$ is an integer, it is in $S_4$. Consider the scalar product of $f(x)$ and the scalar $1/2\in \R$.
Then we evaluate the scalar product at $x=1$ and we have
\begin{align*}
\frac{1}{2}f(1)=\frac{1}{2},
\end{align*}
which is not an integer.
Thus $(1/2)f(x)$ is not in $S_4$, hence condition 3 is not met. Thus $S_4$ is not a subspace of $P_4$.
(You can check that conditions 1, 2 are met.)
  

Solution (5). $S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}$

Let $f(x)=x$. Then $f(x)$ is a degree $1$ polynomial and $f(1)=1$ is a rational number.
However, the scalar product $\sqrt{2} f(x)$ of $f(x)$ and the scalar $\sqrt{2} \in \R$ is not in $S_5$ since
\[\sqrt{2}f(1)=\sqrt{2},\] which is not a rational number. Hence condition 3 is not met and $S_5$ is not a subspace of $P_4$.
(You can check that conditions 1, 2 are met.)
  

Solution (6). $S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\}$

The zero vector of the vector space $M_{2 \times 2}$ is the $2\times 2$ zero matrix $O$.
Since the determinant of the zero matrix $O$ is $0$, it is not in $S_6$. Thus, condition 1 is not met and $S_6$ is not a subspace of $M_{2 \times 2}$.
(You can check that conditions 2, 3 are not met as well.)
  

Solution (7). $S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\}$

Consider the matrices
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
0& 1
\end{bmatrix}.\] The determinants of $A$ and $B$ are both $0$, hence they belong to $S_7$.
However, their sum
\[A+B=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}\] has the determinant $1$, hence the sum $A+B$ is not in $S_7$.
So condition 2 is not met and $S_7$ is not a subspace of $M_{2 \times 2}$.
(You can check that conditions 1, 3 are met.)
  

Solution (8). $S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\}$

Consider the continuous functions
\[f(x)=x-1 \text{ and } g(x)=x+1.\] (These are polynomials, hence they are continuous.)
We have
\begin{align*}
&f(-1)f(1)=(-2)\cdot(0)=0 \text{ and }\\
&g(-1)g(1)=(0)\cdot 2=0.
\end{align*}
So these functions are in $S_8$.

However, their sum $h(x):=f(x)+g(x)$ does not belong to $S_8$ since we have
\begin{align*}
h(-1)h(1)&=\big(f(-1)+g(-1)\big) \big(f(1)+g(1) \big)\\
&=(-2+0)(0+2)=-4\neq 0.
\end{align*}
Therefore, condition 2 is not met and $S_8$ is not a subspace of $C[-1, 1]$.
(You can check that conditions 1, 3 are met.)
  

Solution (9). $S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}$

Let $f(x)=x^2$, an open-up parabola.
Then $f(x)$ is continuous and non-negative for $-1 \leq x \leq 1$. Hence $f(x)=x^2$ is in $S_9$.
However, the scalar product $(-1)f(x)$ of $f(x)$ and the scalar $-1$ is not in $S_9$ since, say,
\[(-1)f(1)=-1\] is negative.
So condition 3 is not met and $S_9$ is not a subspace of $C[-1, 1]$.
(You can check that conditions 1, 2 are met.)
  

Solution (10). $S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}$

The zero vector of the vector space $C^2[-1, 1]$ is the zero function $\theta(x)=0$.
The second derivative of the zero function is still the zero function.
Thus,
\[\theta^{\prime\prime}(x)+\theta(x)=0\] and since $\sin(x)$ is not the zero function, $\theta(x)$ is not in $S_{10}$.
Hence $S_{10}$ is not a subspace of $C^2[-1, 1]$.

(You can check that conditions 2, 3 are not met as well.
For example, consider the function $f(x)=-\frac{1}{2}x\cos(x)\in S_{10}$.)

Solution (11). Let $S_{11}$ be the set of real polynomials of degree exactly $k$.

The set $S_{11}$ is not a vector subspace of $\mathbf{P}_k$. One reason is that the zero function $\mathbf{0}$ has degree $0$, and so does not lie in $S_{11}$. The set $S_{11}$ is also not closed under addition. Consider the two polynomials $f(x) = x^k + 1$ and $g(x) = – x^k + 1$. Both of these polynomials lie in $S_{11}$, however $f(x) + g(x) = 2$ has degree $0$ and so does not lie in $S_{11}$.

Solution (12). The complement

The complement $S_{12}= V \setminus W$ is not a vector subspace. Specifically, if $\mathbf{0} \in V$ is the zero vector, then we know $\mathbf{0} \in W$ because $W$ is a subspace. But then $\mathbf{0} \not\in V \setminus W$, and so $V \setminus W$ cannot be a vector subspace.

Linear Algebra Midterm Exam 2 Problems and Solutions

• True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
• Problem 1 and its solution (current problem): See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
• Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
• Problem 3 and its solution: Orthonormal basis of null space and row space
• Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
• Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
• Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
• Problem 7 and its solution: Find matrix representation of linear transformation from $R^2$ to $R^2$
• Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

The post 12 Examples of Subsets that Are Not Subspaces of Vector Spaces first appeared on Problems in Mathematics.

Let $G=\GL(n, \R)$ be the general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices.
Consider the subset of $G$ defined by
\[\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.\] Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$.
The subgroup $\SL(n,\R)$ is called special linear group

Hint.

We are going to use the following facts from linear algebra about the determinant of a matrix.
For any $n\times n$ matrices $A, B$, we have
\begin{align*}
\det(AB)&=\det(A)\det(B) \text{ and }\\
\det(A^{-1})&=\det(A)^{-1}
\end{align*}
if $A$ is invertible.

We give two proofs.
The first one proves that $\SL(n,\R)$ is a normal subgroup of $\GL(n,\R)$ by directly verifying the defining property.

The second proof uses a fact about group homomorphism. If you are familiar with group homomorphism, the second proof is concise and nice.

Proof 1.

The special linear group $\SL(n,\R)$ is a subgroup.

Let $X, Y\in \SL(n,\R)$ be arbitrary elements. We have
\[\det(X)=\det(Y)=1\] by definition of $\SL(n,\R)$.

Then we obtain
\begin{align*}
\det(XY)=\det(X)\det(Y)=1,
\end{align*}
and hence $XY$ is in $\SL(n,\R)$.

Also, we have
\[\det(X^{-1})=\det(X)^{-1}=1,\] and it follows that $X^{-1}$ is in $\SL(n,\R)$.
Thus, $\SL(n,\R)$ is a subgroup of $G$.

The special linear group $\SL(n,\R)$ is normal.

To prove that $\SL(n,\R)$ is a normal subgroup of $G$, let $X\in \SL(n,\R)$ and let $P\in G$.
Then we have
\begin{align*}
\det(PXP^{-1})=\det(P)\det(X)\det(P)^{-1}=\det(X)=1,
\end{align*}
and hence the conjugate $PXP^{-1}$ is in $\SL(n,\R)$.
Therefore, $\SL(n,\R)$ is a normal subgroup of $G$.

Proof 2.

Let $\R^*$ be the multiplicative group of nonzero real numbers.

Let $\phi:\GL(n,\R) \to \R^*$ be the map given by
\[\phi(X)=\det(X),\] for each $X\in \GL(n,\R)$.

Note that this is well-defined since $\det(X)\neq 0$ for $X\in \GL(n,\R)$.

By the property of the determinant, we know that
\[\det(XY)=\det(X)\det(Y)\] for any $X, Y \in \GL(n,\R)$.

This implies that the map $\phi$ is a group homomorphism.

Then the kernel of $\phi$ is given by
\begin{align*}
\ker(\phi)=\{X \in \GL(n,\R) \mid \phi(X)=\det(X)=1\}=\SL(n, \R).
\end{align*}

As the kernel of the group homomorphism $\phi:\GL(n,\R) \to \R^*$ is alway a normal subgroup of $\GL(n,\R)$, we conclude that $\SL(n, \R)$ is a normal subgroup of $\GL(n,\R)$.

The post Special Linear Group is a Normal Subgroup of General Linear Group 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.

Suppose that $A$ is $2\times 2$ matrix that has eigenvalues $-1$ and $3$.
Then for each positive integer $n$ find $a_n$ and $b_n$ such that
\[A^{n+1}=a_nA+b_nI,\] where $I$ is the $2\times 2$ identity matrix.

Solution.

Since $-1, 3$ are eigenvalues of the matrix $A$, the characteristic polynomial of $A$ is
\[(t+1)(t-3)=t^2-2t-3.\] By Cayley-Hamilton theorem, we have
\[A^2-2A-3I=O,\] where $O$ is the $2\times 2$ zero matrix.
Thus we have
\[A^2=2A+3I, \tag{*}\] and hence $a_1=2, b_1=3$.

Multiplying (*) by $A$, we have
\begin{align*}
A^3&=AA^2=2A^2+3A\\
&=2(2A+3I)+3A\\
&=7A+6I.
\end{align*}
Thus, $a_2=7, b_2=6$.

In general, we have
\begin{align*}
A^{n+2}&=AA^{n+1}\\
&=A(a_nA+b_nI)\\
&=a_nA^2+b_nA\\
&=a_n(2A+3I)+b_n A\\
&=(2a_n+b_n)A+(3a_n)I.
\end{align*}
Thus we obtain
\begin{align*}
a_{n+1}&=2a_n+b_n\\
b_{n+1}&=3a_n.
\end{align*}

This yields the linear recurrence relation
\[a_{n+2}=2a_{n+1}+3a_{n}\] with initial values $a_1=2, a_2=7$.

The general term $a_n$ of the sequence $(a_n)_{i=n}^{\infty}$ is obtained in the previous post Solve a linear recurrence relation using vector space technique:
\[a_n=\frac{1}{4} \big( (-1)^n+3^{n+1} \big).\] (You may alternatively find this using an elementary technique.)

Then we also have
\[b_n=3a_{n-1}=\frac{3}{4} \big( (-1)^{n-1}+3^{n} \big).\] (This formula is true for $n=1$ as well.)

In conclusion, we have obtained
\begin{align*}
A^n=\frac{1}{4} \big( (-1)^n+3^{n+1} \big)A+\frac{3}{4} \big( (-1)^{n-1}+3^{n} \big)I
\end{align*}
for any positive integer $n$.

The post Find the Formula for the Power of a Matrix Using Linear Recurrence Relation first appeared on Problems in Mathematics.

Let $V$ be a real vector space of all real sequences
\[(a_i)_{i=1}^{\infty}=(a_1, a_2, \dots).\] Let $U$ be a subspace of $V$ defined by
\[U=\{(a_i)_{i=1}^{\infty}\in V \mid a_{n+2}=2a_{n+1}+3a_{n} \text{ for } n=1, 2,\dots \}.\] Let $T$ be the linear transformation from $U$ to $U$ defined by
\[T\big((a_1, a_2, \dots)\big)=(a_2, a_3, \dots). \]

(a) Find the eigenvalues and eigenvectors of the linear transformation $T$.

(b) Use the result of (a), find a sequence $(a_i)_{i=1}^{\infty}$ satisfying $a_1=2, a_2=7$.

Solution.

(a) Find the eigenvalues and eigenvectors of the linear transformation $T$.

Note that each sequence $(a_i)_{i=1}^{\infty}$ in $U$ is determined by the first two numbers $a_1, a_2$ since the rests can be obtained from the linear recurrence relation $a_{n+2}=2a_{n+1}+3a_{n}$.
Thus, for example, we take $(a_1, a_2)=(1,0), (0, 1)$ and obtain a basis $B=\{\mathbf{u}_1, \mathbf{u}_2\}$ of $U$, where
\begin{align*}
\mathbf{u}_1 &=(1, 0, 3, 6, 21, \cdots )\\
\mathbf{u}_2 &=(0, 1, 2, 7, 20, \dots).
\end{align*}

We find the matrix $A$ of $T$ with respect to the basis $B$.
We have
\begin{align*}
T(\mathbf{u}_1)&=(0, 3, 6, 21, \cdots )=0 \mathbf{u}_1+3\mathbf{u}_2\\
T(\mathbf{u}_2) &=(1, 2, 7, 20, \dots)=\mathbf{u}_1+2\mathbf{u}_2.
\end{align*}

Thus, the matrix $A$ for $T$ is
\[A=\begin{bmatrix}
0 & 1\\
3 & 2
\end{bmatrix}.\]

Then the eigenvalues of $T$ are eigenvalues of $A$.
The characteristic polynomial for $A$ is
\begin{align*}
p_A(t)=\det(A-tI)=t^2-2t-3=(t+1)(t-3).
\end{align*}
Thus the eigenvalues are $t=-1, 3$.

Let us find eigenvectors corresponding to $t=-1$.
We solve
\[T\big((a_1, a_2, \dots)\big)=-(a_1, a_2, \dots),\] or equivalently,
\[(a_2, a_3, \dots)=-(a_1, a_2, \dots).\] This yields the relation
\[a_{n+1}=-a_{n}\] for $n=1, 2, \dots$.

Hence the sequence $(a_i)_{i=1}^{\infty}$ is a geometric progression with initial value $a_1$ and ratio $-1$.
Therefore, the sequence $(a_i)_{i=1}^{\infty}$ such that
\[a_i=(-1)^{i-1}a_1\] are eigenvectors corresponding to the eigenvalue $-1$.

Similarly, eigenvectors corresponding to the eigenvalue $3$ are geometric progression with initial value $a_1$ and ratio $3$, thus
\[a_i=3^{i-1}a_1.\]  

(b) Find a sequence $(a_i)_{i=1}^{\infty}$ satisfying $a_1=2, a_2=7$.

From part (a), we see that sequences
\[\mathbf{v}_1=\big( (-1)^{i-1} \big)_{i=1}^{\infty} \text{ and } \mathbf{v}_2=\big( 3^{i-1})_{i=1}^{\infty}\] form a basis of $U$ consisting of eigenvectors of $T$. (We chose $a_1=1$.)
Hence any sequence $(a_i)_{i=1}^{\infty}$ can be written as a linear combination of these two vectors:
\begin{align*}
(a_i)_{i=1}^{\infty} &=c_1\mathbf{v}_1+c_2\mathbf{v}_2\\
&=c_1\big( (-1)^{i-1} \big)_{i=1}^{\infty}+c_2\big( 3^{i-1})_{i=1}^{\infty}
\end{align*}
for some $c_1, c_2$.

If the sequence satisfies the initial condition $a_1=2, a_2=7$, then we have
\begin{align*}
2&=a_1=c_1+c_2\\
7&=a_2=-c_1+3c_2.
\end{align*}

Solving this, we obtain $c_1=-1/4$ and $c_2=9/4$.
Hence we have
\begin{align*}
(a_i)_{i=1}^{\infty}&=-\frac{1}{4}\big( (-1)^{i-1} \big)_{i=1}^{\infty}+\frac{9}{4}\big( 3^{i-1})_{i=1}^{\infty}\\[6pt] &=\frac{1}{4}\big( (-1)^{i} \big)_{i=1}^{\infty}+\frac{1}{4}\big( 3^{i+1})_{i=1}^{\infty}\\[6pt] &=\frac{1}{4}\big( (-1)^{i}+ 3^{i+1} \big)_{i=1}^{\infty}.
\end{align*}
In summary, the sequence $(a_i)_{i=1}^{\infty}$ satisfying the linear recurrence relation $a_{n+2}=2a_{n+1}+3a_{n}$ with initial conditions $a_1=2, a_2=7$ is
\[(a_i)_{i=1}^{\infty}=\frac{1}{4}\big( (-1)^{i}+ 3^{i+1} \big)_{i=1}^{\infty}.\]

Related Question.

Check out the following problems about how to solve a linear recurrence relation using the vector space techniques, like the current problem.

1. Sequences satisfying linear recurrence relation form a subspace
2. Matrix representation of a linear transformation of subspace of sequences satisfying recurrence relation
3. Solve linear recurrence relation using linear algebra (eigenvalues and eigenvectors)

The post Solve a Linear Recurrence Relation Using Vector Space Technique first appeared on Problems in Mathematics.