Problem trying to find kernel of a linear transformation

Given this linear transformation:

$f\begin{pmatrix}x\\y\\z\\w\end{pmatrix} = \begin{pmatrix}x + 2y − z + 10w\\2x − y + 3z\\y − z + 4w\end{pmatrix}$.

I want to find its kernel and so I try to solve the following equations system:

Equations system

Finding the next solution:

Solution

However, the solution appears to be incorrect, as the answer to the problem is that the kernel of the transformation is:

Real solution

I can't figure out what i'm doing wrong, please help.

• Please use MathJax to type up the question – pureundergrad Apr 30 '18 at 20:57
• There’s no such thing as the “real solution.” Any basis for the kernel will do. – amd Apr 30 '18 at 21:07

2 Answers

So the system could be written as $$f(x,y,z,w) = \begin{bmatrix} 1 & 2 & -1 & 10\\ 2 & -1 & 3 & 0\\ 0 & 1 & -1 & 4\\ \end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ w \end{bmatrix}$$ So let $A$ be the matrix $$A = \begin{bmatrix} 1 & 2 & -1 & 10\\ 2 & -1 & 3 & 0\\ 0 & 1 & -1 & 4\\ \end{bmatrix}$$ To find the null space (Kernel), we should find all vectors such that $$\begin{bmatrix} x\\ y\\ z\\ w \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$ That is

\begin{align} x + 2y − z + 10w &= 0\\ 2x − y + 3z &= 0\\ y − z + 4w &= 0 \end{align} Last equation suggests that $$z = y + 4w$$ Plugging this equation back we get \begin{align} x + 2y − (y + 4w) + 10w &= 0\\ 2x − y + 3(y + 4w) &= 0 \end{align} that is \begin{align} x + 2y − y - 4w + 10w &= 0\\ 2x − y + 3y + 12w &= 0 \end{align} that is \begin{align} x + y +6w &= 0\\ 2x + 2y + 12w &= 0 \end{align} Multiplying equation 1 by a factor of 2: \begin{align} 2x + 2y +12w &= 0\\ 2x + 2y + 12w &= 0 \end{align} We get the same equation. This means that we have the following dependence: \begin{align} z &= y + 4w \\ x &= -y -6w \end{align} So the kernel is all such vectors $(x,y,z,w)$ such that $z = y + 4w$ and $x = -y-6w$ (which is of dimension 2)

Guide:

Both solutions are fine.

Verify that

$$\operatorname{rref} \left(\begin{bmatrix} -1 & 1 & 1 & 0 \\ -2& -4& 0& 1 \end{bmatrix} \right) = \operatorname{rref}\left(\begin{bmatrix} -1 & 1 & 1 & 0 \\ -6& 0&4 & 1 \end{bmatrix} \right)$$