Prove that the matrix
\[A=\begin{bmatrix}
0 & 1\\
-1& 0
\end{bmatrix}\]
is diagonalizable.
Prove, however, that $A$ cannot be diagonalized by a real nonsingular matrix.
That is, there is no real nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix.

A square matrix $A$ is called nilpotent if some power of $A$ is the zero matrix.
Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.

Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample.

Let $V$ be a vector space over a scalar field $K$.
Let $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ be vectors in $V$ and consider the subset
\[W=\{a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \mid a_1, a_2, \dots, a_k \in K \text{ and } a_1+a_2+\cdots+a_k=0\}.\]
So each element of $W$ is a linear combination of vectors $\mathbf{v}_1, \dots, \mathbf{v}_k$ such that the sum of the coefficients is zero.

Let $V$ be a subset of $\R^4$ consisting of vectors that are perpendicular to vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$, where
\[\mathbf{a}=\begin{bmatrix}
1 \\
0 \\
1 \\
0
\end{bmatrix}, \quad \mathbf{b}=\begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix}, \quad \mathbf{c}=\begin{bmatrix}
0 \\
1 \\
-1 \\
0
\end{bmatrix}.\]

Namely,
\[V=\{\mathbf{x}\in \R^4 \mid \mathbf{a}^{\trans}\mathbf{x}=0, \mathbf{b}^{\trans}\mathbf{x}=0, \text{ and } \mathbf{c}^{\trans}\mathbf{x}=0\}.\]

Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup.
Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.

Then prove that $H$ is an abelian normal subgroup of odd order.

The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).
The time limit was 55 minutes.

This post is Part 3 and contains Problem 7, 8, and 9.
Check out Part 1 and Part 2 for the rest of the exam problems.

Problem 7. Let $A=\begin{bmatrix}
-3 & -4\\
8& 9
\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}
-1 \\
2
\end{bmatrix}$.

(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.

(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.

Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.

Problem 9.
Determine whether each of the following sentences is true or false.

(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.

(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.

(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.

(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.

The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).
The time limit was 55 minutes.

This post is Part 2 and contains Problem 4, 5, and 6.
Check out Part 1 and Part 3 for the rest of the exam problems.

Problem 4. Let
\[\mathbf{a}_1=\begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}, \mathbf{a}_2=\begin{bmatrix}
2 \\
-1 \\
4
\end{bmatrix}, \mathbf{b}=\begin{bmatrix}
0 \\
a \\
2
\end{bmatrix}.\]

Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.

Problem 5.
Find the inverse matrix of
\[A=\begin{bmatrix}
0 & 0 & 2 & 0 \\
0 &1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
1 & 0 & 0 & 1
\end{bmatrix}\]
if it exists. If you think there is no inverse matrix of $A$, then give a reason.

Problem 6.
Consider the system of linear equations
\begin{align*}
3x_1+2x_2&=1\\
5x_1+3x_2&=2.
\end{align*}

(a) Find the coefficient matrix $A$ of the system.

(b) Find the inverse matrix of the coefficient matrix $A$.

(c) Using the inverse matrix of $A$, find the solution of the system.

(Linear Algebra Midterm Exam 1, the Ohio State University)

The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).
The time limit was 55 minutes.

This post is Part 1 and contains the first three problems.
Check out Part 2 and Part 3 for the rest of the exam problems.

Problem 1. Determine all possibilities for the number of solutions of each of the systems of linear equations described below.

(a) A consistent system of $5$ equations in $3$ unknowns and the rank of the system is $1$.

(b) A homogeneous system of $5$ equations in $4$ unknowns and it has a solution $x_1=1$, $x_2=2$, $x_3=3$, $x_4=4$.

Problem 2. Consider the homogeneous system of linear equations whose coefficient matrix is given by the following matrix $A$. Find the vector form for the general solution of the system.
\[A=\begin{bmatrix}
1 & 0 & -1 & -2 \\
2 &1 & -2 & -7 \\
3 & 0 & -3 & -6 \\
0 & 1 & 0 & -3
\end{bmatrix}.\]

Problem 3. Let $A$ be the following invertible matrix.
\[A=\begin{bmatrix}
-1 & 2 & 3 & 4 & 5\\
6 & -7 & 8& 9& 10\\
11 & 12 & -13 & 14 & 15\\
16 & 17 & 18& -19 & 20\\
21 & 22 & 23 & 24 & -25
\end{bmatrix}
\]
Let $I$ be the $5\times 5$ identity matrix and let $B$ be a $5\times 5$ matrix.
Suppose that $ABA^{-1}=I$.
Then determine the matrix $B$.

(Linear Algebra Midterm Exam 1, the Ohio State University)

For an $m\times n$ matrix $A$, we denote by $\mathrm{rref}(A)$ the matrix in reduced row echelon form that is row equivalent to $A$.
For example, consider the matrix $A=\begin{bmatrix}
1 & 1 & 1 \\
0 &2 &2
\end{bmatrix}$
Then we have
\[A=\begin{bmatrix}
1 & 1 & 1 \\
0 &2 &2
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_2}
\begin{bmatrix}
1 & 1 & 1 \\
0 &1 & 1
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &1
\end{bmatrix}\]
and the last matrix is in reduced row echelon form.
Hence $\mathrm{rref}(A)=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &1
\end{bmatrix}$.

Find an example of matrices $A$ and $B$ such that
\[\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B).\]

Let $I$ be the $2\times 2$ identity matrix.
Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\times 2$ matrices $A$ and $B$ with determinant $1$.