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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.

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    $\begingroup$ Can you find the kernel? $\endgroup$ – StackTD Jul 15 '16 at 13:59
  • $\begingroup$ Which book are you refering to? @Kevin R. $\endgroup$ – Nebo Alex Jul 15 '16 at 14:05
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    $\begingroup$ You should specify the field in which the entries of this matrix live. $\endgroup$ – ThorbenK Jul 15 '16 at 14:05
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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).

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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 $\left\{\begin{bmatrix}0\\0\end{bmatrix}\right\}$

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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 $\{0\}$

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  • $\begingroup$ Why would you "assume that the field is Q"? Why not R? Of course, the answer is the same either way. $\endgroup$ – user247327 Jul 15 '16 at 14:08
  • $\begingroup$ That is exactly, why I'm assuming its $\mathbb{Q}$ this gives an answer for $\mathbb{R}$ and $\mathbb{C}$ aswell. If it would be a finite field the answer would change. $\endgroup$ – ThorbenK Jul 15 '16 at 14:45
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The columns are linearly independent - they span $\mathbb{R}^2$.

∴ by rank-nullity theorem, kernel = $\{\vec{0}\}$

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  • $\begingroup$ I'm not sure that this 2-years old question, with already 3 answers, needs a 4th answer. $\endgroup$ – Watson May 30 '18 at 21:27
  • $\begingroup$ I wanted to give a different perspective (4th answer) for people who are still reading it 2 years later (like you and me). $\endgroup$ – user566182 May 30 '18 at 22:02

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