Let $G$ be a group with the identity element $e$ and suppose that we have a group homomorphism $\phi$ from the direct product $G \times G$ to $G$ satisfying
\[\phi(e, g)=g \text{ and } \phi(g, e)=g, \tag{*}\]
for any $g\in G$.

Let $\mu: G\times G \to G$ be a map defined by
\[\mu(g, h)=gh.\]
(That is, $\mu$ is the group operation on $G$.)

Then prove that $\phi=\mu$.
Also prove that the group $G$ is abelian.

Let $A$ be an $n \times n$ matrix satisfying
\[A^2+c_1A+c_0I=O,\]
where $c_0, c_1$ are scalars, $I$ is the $n\times n$ identity matrix, and $O$ is the $n\times n$ zero matrix.

Prove that if $c_0\neq 0$, then the matrix $A$ is invertible (nonsingular).
How about the converse? Namely, is it true that if $c_0=0$, then the matrix $A$ is not invertible?

For a real number $a$, consider $2\times 2$ matrices $A, P, Q$ satisfying the following five conditions.

$A=aP+(a+1)Q$

$P^2=P$

$Q^2=Q$

$PQ=O$

$QP=O$,

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
\[A_nA_{n-1}A_{n-2}\cdots A_2.\]

Let $A$ and $B$ are matrices such that the matrix product $AB$ is defined and $AB$ is a square matrix.
Is it true that the matrix product $BA$ is also defined and $BA$ is a square matrix? If it is true, then prove it. If not, find a counterexample.

(a) Solve the following system by transforming the augmented matrix to reduced echelon form (Gauss-Jordan elimination). Indicate the elementary row operations you performed.
\begin{align*}
x_1+x_2-x_5&=1\\
x_2+2x_3+x_4+3x_5&=1\\
x_1-x_3+x_4+x_5&=0
\end{align*}

(b) Determine all possibilities for the solution set of a homogeneous system of $2$ equations in $2$ unknowns that has a solution $x_1=1, x_2=5$.

Let $I$ be the $n\times n$ identity matrix, where $n$ is a positive integer. Prove that there are no $n\times n$ matrices $X$ and $Y$ such that
\[XY-YX=I.\]
Read solution

(a) Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$.

(b) Find a basis for the null space of $A$.

(c) Find a basis for the range of $A$ that consists of columns of $A$. For each columns, $A_j$ of $A$ that does not appear in the basis, express $A_j$ as a linear combination of the basis vectors.

Let
\[A=\begin{bmatrix}
a & -1\\
1& 4
\end{bmatrix}\]
be a $2\times 2$ matrix, where $a$ is some real number.
Suppose that the matrix $A$ has an eigenvalue $3$.

(a) Determine the value of $a$.

(b) Does the matrix $A$ have eigenvalues other than $3$?

Suppose that $\lambda$ and $\mu$ are two distinct eigenvalues of a square matrix $A$ and let $\mathbf{x}$ and $\mathbf{y}$ be eigenvectors corresponding to $\lambda$ and $\mu$, respectively.
If $a$ and $b$ are nonzero numbers, then prove that $a \mathbf{x}+b\mathbf{y}$ is not an eigenvector of $A$ (corresponding to any eigenvalue of $A$).

Let $P_4$ be the vector space consisting of all polynomials of degree $4$ or less with real number coefficients.
Let $W$ be the subspace of $P_2$ by
\[W=\{ p(x)\in P_4 \mid p(1)+p(-1)=0 \text{ and } p(2)+p(-2)=0 \}.\]
Find a basis of the subspace $W$ and determine the dimension of $W$.

Let $B=\{\mathbf{v}_1, \mathbf{v}_2 \}$ be a basis for the vector space $\R^2$, and let $T:\R^2 \to \R^2$ be a linear transformation such that
\[T(\mathbf{v}_1)=\begin{bmatrix}
1 \\
-2
\end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
3 \\
1
\end{bmatrix}.\]

If $\mathbf{e}_1=\mathbf{v}_1+2\mathbf{v}_2 \text{ and } \mathbf{e}_2=2\mathbf{v}_1-\mathbf{u}_2$, where $\mathbf{e}_1, \mathbf{e}_2$ are the standard unit vectors in $\R^2$, then find the matrix of $T$ with respect to the basis $\{\mathbf{e}_1, \mathbf{e}_2\}$.

Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are
\[\|\mathbf{a}\|=\|\mathbf{b}\|=1\]
and the inner product
\[\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.\]

Then determine the length $\|\mathbf{a}-\mathbf{b}\|$.
(Note that this length is the distance between $\mathbf{a}$ and $\mathbf{b}$.)

Determine whether the following is true or false. If it is true, then give a proof. If it is false, then give a counterexample.

Let $W_1$ and $W_2$ be subspaces of the vector space $\R^n$.
If $B_1$ and $B_2$ are bases for $W_1$ and $W_2$, respectively, then $B_1\cap B_2$ is a basis of the subspace $W_1\cap W_2$.

Let $W$ be the subset of $\R^3$ defined by
\[W=\left \{ \mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}\in \R^3 \quad \middle| \quad 5x_1-2x_2+x_3=0 \right \}.\]
Exhibit a $1\times 3$ matrix $A$ such that $W=\calN(A)$, the null space of $A$.
Conclude that the subset $W$ is a subspace of $\R^3$.

Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n$, and let $I$ be the $n \times n$ identity matrix. Suppose that the inner product of $\mathbf{u}$ and $\mathbf{v}$ satisfies
\[\mathbf{v}^{\trans}\mathbf{u}\neq -1.\]
Define the matrix
\[A=I+\mathbf{u}\mathbf{v}^{\trans}.\]

Prove that $A$ is invertible and the inverse matrix is given by the formula
\[A^{-1}=I-a\mathbf{u}\mathbf{v}^{\trans},\]
where
\[a=\frac{1}{1+\mathbf{v}^{\trans}\mathbf{u}}.\]
This formula is called the Sherman-Woodberry formula.