Let $A$ and $B$ be $n\times n$ matrices. Then prove that
\[\calN(A)\cap \calN(B) \subset \calN(A+B),\]
where $\calN(A)$ is the null space (kernel) of the matrix $A$.

Let
\[A=\begin{bmatrix}
4 & 1\\
3& 2
\end{bmatrix}\]
and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.
\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]

(a) Prove that the subset $V$ is a subspace of $\R^2$.

(b) Find a basis for $V$ and determine the dimension of $V$.

(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 $W$ 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 $\R^3$.

Suppose that $n\times n$ matrices $A$ and $B$ are similar.

Then show that the nullity of $A$ is equal to the nullity of $B$.
In other words, the dimension of the null space (kernel) $\calN(A)$ of $A$ is the same as the dimension of the null space $\calN(B)$ of $B$.

In this post, we explain how to diagonalize a matrix if it is diagonalizable.

As an example, we solve the following problem.

Diagonalize the matrix
\[A=\begin{bmatrix}
4 & -3 & -3 \\
3 &-2 &-3 \\
-1 & 1 & 2
\end{bmatrix}\]
by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

(Update 10/15/2017. A new example problem was added.) Read solution

Let $A$ be an $m \times n$ real matrix. Then the null space $\calN(A)$ of $A$ is defined by
\[ \calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.\]
That is, the null space is the set of solutions to the homogeneous system $A\mathbf{x}=\mathbf{0}_m$.

Prove that the null space $\calN(A)$ is a subspace of the vector space $\R^n$.
(Note that the null space is also called the kernel of $A$.)

Let $A$ be an $m \times n$ real matrix.
Then the kernel of $A$ is defined as $\ker(A)=\{ x\in \R^n \mid Ax=0 \}$.

The kernel is also called the null space of $A$.
Suppose that $A$ is an $m \times n$ real matrix such that $\ker(A)=0$. Prove that $A^{\trans}A$ is invertible.