Find vectors that span the kernel of $\begin{bmatrix}1&2\\3&4\end{bmatrix}$ 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. 
 A: 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).
A: 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 $\left\{\begin{bmatrix}0\\0\end{bmatrix}\right\}$
A: 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 $\{0\}$
A: The columns are linearly independent - they span $\mathbb{R}^2$.
∴ by rank-nullity theorem, kernel =  $\{\vec{0}\}$
