I have the following matrix:
\begin{bmatrix}1&2\\3&4\end{bmatrix}
and I'd like to find the vectors that span the kernel.
The book I'm reading isn't helping me understand this concept at all.
$\endgroup$ 34 Answers
$\begingroup$The kernel (or null space) of a matrix is the set of all vectors that are mapped to the zero vector by the matrix. So you are looking for all the solutions $(x,y)$ to the system: $$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix} \iff \left\{ \begin{array}{rcl} x+2y & = & 0 \\ 3x+4y & = & 0 \end{array}\right. $$ The only solution to this system is the zero vector: $(0,0)$.
The empty set spans the zero vector, the dimension of this kernel is $0$. A basis is thus an empty set.
Since this is a rather trivial case and your book isn't really helping you, I'll add an example. Consider the matrix: $$\begin{bmatrix} 2 & -1 \\ -4 & 2 \end{bmatrix}$$ The kernel consists of the solutions to: $$\begin{bmatrix} 2 & -1 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix} \iff \left\{ \begin{array}{rcl} 2x-y & = & 0 \\ -4x+2y & = & 0 \end{array}\right. \iff 2x = y $$ Here, there are an infinite number of solutions to this system, any vector of the form $(x,2x) \;,\; x \in \mathbb{R}$ is in the kernel. This means the kernel is spanned by, for example, the vector $(1,2)$ (or any non-zero multiply of this vector). The dimension of the kernel is now 1 because it can be spanned by 1 vector (any basis will consist of 1 vector).
$\endgroup$ $\begingroup$Generally, we know that the vector $X$ is in kernel of matrix $A$ when $AX=0$. So, we need to find the solutions of the system $AX=0$ for your given matrix. To do so, assuming $$X=\begin{bmatrix}x\\y\end{bmatrix}$$ we have
$$\begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$
and so, $ \ \ \ x+2y=0\ \ \ \ $ and $\ \ \ \ \ 3x+4y=0$. The answer is $x=y=0$.
Simple solution for your case:
The matrix is non-singular, and so its kernel is $\left\{\begin{bmatrix}0\\0\end{bmatrix}\right\}$
$\endgroup$ $\begingroup$I assume that the field is $\mathbb{Q}$. The determinant of this matrix is 4-6=-2. Since the determinant is nonzero this matrix induces an isomorphism and has therefore kernel $\{0\}$
$\endgroup$ 2 $\begingroup$The columns are linearly independent - they span $\mathbb{R}^2$.
∴ by rank-nullity theorem, kernel = $\{\vec{0}\}$
$\endgroup$ 2
