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.

Let
\[A=\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\] be a $3\times 3$ matrix. Then find the formula for $A^n$ for any positive integer $n$.

Proof.

We first compute several powers of $A$ and guess the general formula.
We have
\begin{align*}
A^2=\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 3 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix},
\end{align*}
\begin{align*}
A^3=A^2A=\begin{bmatrix}
1 & 1 & 3 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 5 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix},
\end{align*}
\begin{align*}
A^4=A^3A=\begin{bmatrix}
1 & 1 & 5 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 1 & 7 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

From these computations, we guess the general formula of $A^n$ is
\[A^n=\begin{bmatrix}
1 & 1 & 2n-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.\]

We prove this formula by mathematical induction on $n$.
The base case $n=1$ follows from the definition of $A$.

Suppose that the formula is true for $n=k$.
We prove the formula for $n=k+1$.
We have
\begin{align*}
A^{k+1}&=A^{k}A\\
&=\begin{bmatrix}
1 & 1 & 2k-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
&& \text{by the induction hypothesis}\\
&=\begin{bmatrix}
1 & 1 & 2k+1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
1 & 1 & 2(k+1)-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

Thus the formula holds for $n=k+1$.
Hence the formula is true for any positive integer $n$ by induction.

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

Let $A$ be the coefficient matrix of the system of linear equations
\begin{align*}
-x_1-2x_2&=1\\
2x_1+3x_2&=-1.
\end{align*}

(a) Solve the system by finding the inverse matrix $A^{-1}$.

(b) Let $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$ be the solution of the system obtained in part (a).
Calculate and simplify
\[A^{2017}\mathbf{x}.\]

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

Solution.

(a) Solve the system by finding the inverse matrix $A^{-1}$.

The coefficient matrix $A$ of the system is
\[A=\begin{bmatrix}
-1 & -2\\
2& 3
\end{bmatrix}.\] The determinant of the matrix $A$ is $\det(A)=(-1)(3)-(-2)(2)=1\neq 0$, and thus $A$ is invertible and the inverse matrix can be found by the $2\times 2$ inverse matrix formula
\[A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}
d & -b\\
-c& a
\end{bmatrix}\] if $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ and $\det(A)=ad-bc\neq 0$.
Thus
\[A^{-1}=\begin{bmatrix}
3 & 2\\
-2& -1
\end{bmatrix}.\] (Or you could form the augmented matrix $[A\mid I]$ and find $A^{-1}$.)

Now the given system of linear equations can be written as
\[A\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\] Multiplying by $A^{-1}$ on the left, we obtain
\[\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=A^{-1}\begin{bmatrix}
1 \\
-1
\end{bmatrix}=\begin{bmatrix}
3 & 2\\
-2& -1
\end{bmatrix}\begin{bmatrix}
1 \\
-1
\end{bmatrix}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\] Thus the solution of the system is $x_1=1, x_2=-1$, or in the vector form
\[\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\]

(b) Calculate and simplify $A^{2017}\mathbf{x}$

Since $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}$ is the solution of the system $A\mathbf{x}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}$, we see that
\[A\mathbf{x}=\mathbf{x}.\] From this, we have
\begin{align*}
A^2\mathbf{x}&=AA\mathbf{x}=A\mathbf{x}=\mathbf{x}\\
A^3\mathbf{x}&=AA^2\mathbf{x}=A\mathbf{x}=\mathbf{x}\\
A^4\mathbf{x}&=AA^3\mathbf{x}=A\mathbf{x}=\mathbf{x}\\
\vdots &
\end{align*}

Repeating this, we see that
\[A^{2017}\mathbf{x}=\mathbf{x}.\] (To be more precise, you can prove that $A^n\mathbf{x}=\mathbf{x}$ for any positive integer $n$ by mathematical induction on $n$.)
Thus, we have
\[A^{2017}\mathbf{x}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\]

Comment.

This is one of the first midterm exam problems of linear algebra (Math 2568) at the Ohio State University.

Midterm 1 problems and solutions

The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.

1. Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
2. Problem 2 and its solution: The vector form of the general solution of a system
3. Problem 3 and its solution: Matrix operations (transpose and inverse matrices)
4. Problem 4 and its solution: Linear combination
5. Problem 5 and its solution: Inverse matrix
6. Problem 6 and its solution: Nonsingular matrix satisfying a relation
7. Problem 7 and its solution (The current page): Solve a system by the inverse matrix
8. Problem 8 and its solution:A proof problem about nonsingular matrix

The post Solve a System by the Inverse Matrix and Compute $A^{2017}\mathbf{x}$ first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix} \text{ and } \mathbf{u}=\begin{bmatrix}
1\\
0
\end{bmatrix}.\] Compute $A^{2017}\mathbf{u}$.

(The Ohio State University, Linear Algebra Exam)

Solution.

We first compute $A\mathbf{u}$. We have
\[A\mathbf{u} = \begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix} \begin{bmatrix}
1\\
0
\end{bmatrix}=
\begin{bmatrix}
-1\\
0
\end{bmatrix}
=-\begin{bmatrix}
1\\
0
\end{bmatrix}=-\mathbf{u}.\] Thus we have
\[A\mathbf{u}=-\mathbf{u}.\]

Then using this result repeatedly we have
\begin{align*}
A^2\mathbf{u}=A(A\mathbf{u})=A(-\mathbf{u})=-A\mathbf{u}=-(-\mathbf{u})=\mathbf{u}.
\end{align*}
Thus we have
\[A^2\mathbf{u}=\mathbf{u}.\]

Next, we compute
\[A^3\mathbf{u}=A(A^2\mathbf{u})=A(\mathbf{u})=-\mathbf{u}.\] Now you see the pattern and obtain
\begin{align*}
A^n\mathbf{u}=\begin{cases}
-\mathbf{u} & \text{ if } n \text{ is odd}\\
\mathbf{u} & \text{ if } n \text{ is even}.
\end{cases} \tag{*}
\end{align*}
(To prove this pattern, use mathematical induction. See below.)

Thus we conclude that
\[A^{2017}\mathbf{u}=-\mathbf{u}=\begin{bmatrix}
-1\\
0
\end{bmatrix}\] since $2017$ is odd.

The proof of the formula.

For completeness, we prove the pattern (*) we see above.
We use mathematical induction.
When $n=1$, we computed $A\mathbf{u}=-\mathbf{u}$ and the base case is true.
Suppose that(*) is true for $n=k$.

Then we have
\begin{align*}
A^{k+1}\mathbf{u}=A(A^{k}\mathbf{u}) &=
\begin{cases}
A(-\mathbf{u}) &\text{ if } k \text{ is odd}\\
A(\mathbf{u}) & \text{ if } k \text{ is even}
\end{cases} \\
&=\begin{cases}
-A\mathbf{u} &\text{ if } k \text{ is odd}\\
A\mathbf{u} & \text{ if } k \text{ is even}
\end{cases} \\
& =\begin{cases}
\mathbf{u} &\text{ if } k \text{ is odd}\\
-\mathbf{u} & \text{ if } k \text{ is even}
\end{cases}
& =\begin{cases}
\mathbf{u} &\text{ if } k+1 \text{ is even}\\
-\mathbf{u} & \text{ if } k+1 \text{ is odd}
\end{cases}
\end{align*}
Here the second equality follows from the induction hypothesis.

This proves (*) for $n=k+1$ and this completes the induction step and (*) is true for any positive integer $n$.

The post Compute the Product $A^{2017}\mathbf{u}$ of a Matrix Power and a Vector first appeared on Problems in Mathematics.

Let $A=\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}$.
Show that

(1) $A^n=\begin{bmatrix}
a^n & 0\\
0& b^n
\end{bmatrix}$ for any $n \in \N$.

(2) Let $B=S^{-1}AS$, where $S$ be an invertible $2 \times 2$ matrix.
Show that $B^n=S^{-1}A^n S$ for any $n \in \N$

Hint.

Use mathematical induction.

Proof.

(1) We prove $A^n=\begin{bmatrix}
a^n & 0\\
0& b^n
\end{bmatrix}$ by induction on $n$.
The base case $n=1$ is true by definition.

Suppose that $A^k=\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}$. Then we have
\[A^{k+1}=AA^k =\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}
\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}
=\begin{bmatrix}
a^{k+1} & 0\\
0& b^{k+1}
\end{bmatrix}.\] Here we used the induction hypothesis in the second equality.
Hence the inductive step holds. This completes the proof.

(2) We show that $B^n=S^{-1}A^n S$ by induction on $n$.
When $n=1$, this is just the definition of $B$.

For induction step, assume that $B^k=S^{-1} A^k S$.
Then we have
\[B^{k+1}=B B^k=(S^{-1} A S) (S^{-1} A^k S)=S^{-1}A A^k S=S^{-1} A^{k+1} S,\] where we used the induction hypothesis in the second equality and the third equality follows by canceling $S S^{-1}=I_2$ in the middle.

Thus the inductive step holds, and this competes the proof.

The post Powers of a Diagonal Matrix first appeared on Problems in Mathematics.