I was trying to write this linear transformation as a matrix,
the exercise is the following:
let "f: $R^3 –> R^4$" be a linear transformation and Kerf = (x, y , z) in $R^3$: $x-2y+z = 0$ is its null space. Determine the canonical matrix.
I know how to solve this type of exercises (generally), as follow:
- I use the basis of the first vector space (R3 in this case),
- I plug x, y, z, in the equations of the linear transformation,
- then, I'm ready to use each component of the vectors (of R4, in this case) to create the matrix. (the canonical matrix has basis of the entire space, for example, 1 0 0, 0 1 0, 0 0 1).
But, when I'm about to solve these exercises, I simply cannot solve them. I don't know why, because I know how to do them theoretically.
Can you help me to solve this exercise?
Best Answer
$\operatorname{ker}A$ is the set of vectors $\vec x$ that satisfy $A\vec x=0.$ And you are given that $x-2y+z = 0.$ So, if we let $\alpha=z,\ \beta =y$ then $x=2\beta-\alpha$ and so $\vec x=(2\beta-\alpha,\beta,\alpha)^T.$ Collecting this information in a slightly different way, $\begin{pmatrix} x\\y \\ z \end{pmatrix}=\begin{pmatrix} -1\\0 \\ 1 \end{pmatrix}\alpha+\begin{pmatrix} 2\\1 \\ 0 \end{pmatrix}\beta.$ Set $\vec v_1 = (-1,0,1)^T$ and $\vec v_2=(2,1,0)^T$ and extend to a basis for $\mathbb R^3:$ we may take $\vec v_3=(0,0,1)^T.$
So, in the basis $\{\vec v_1,\ \vec v_2,\ \vec v_3\}$, we have $A\vec v_1=A\vec v_2=0$ and $A\vec v_3=(a,b,c,d)^T$ for some numbers $a,b,c,d.$ The matrix of $A$ in this basis relative to the standard basis of $\mathbb R^4$ is therefore
$A=\begin{pmatrix} 0 &0 &a\\ 0 &0 &b \\ 0&0 &c \\ 0 &0 &d \end{pmatrix}.$
Now we compute the change of basis matrix $[\vec v_1,\vec v_2,\vec v_3]\to [\vec i,\vec j,\vec k]:$
$B=\begin{pmatrix} -1 & 2 &0 \\ 0& 1 &0 \\ 1 & 0 &1 \end{pmatrix}$
The matrix of $A$ relative to the standard bases for $\mathbb R^3$ and $\mathbb R^4$ is therefore $AB$.