# Thread: Find a matrix transformation whose kernel is the given subspace

1. ## Find a matrix transformation whose kernel is the given subspace

The subspace of R^3 consisting of vectors whose second entry is zero.

So I made the kernel L={[a,0,b]^T }

I then let x = [1,0,1]^T be in L.

Now, I know I need to find a matrix A such that Ax=0. I'm not sure how to go about this...

I just wrote some stuff down:

0 1 0

0 1 0

0 1 0

Which does indeed make Ax=0 true, but it just doesn't seem right to me for reasons I can't explain..

What am I doing wrong?

2. ## Re: Find a matrix transformation whose kernel is the given subspace

It's not that you need to let be x = [1,0,1]^T but rather have x be of the form [a,0,b]^T

What you can do for slightly more rigor is let your unknown matrix transformation A be a 3x3 matrix with entries e_1 to e_9

ie. The matrix product Ax would be

[e_1 e_2 e_3] * [a 0 b]^T
[e_4 e_5 e_6]
[e_7 e_8 e_9]

= [e_1*a + e_3*b, e_4*a + e_6*b, e_7*a + e_9*b]^T

From here you want the result vector to have entries of all zeroes. Your matrix A with 1's in the second column satisfies this condition, with e_1, e_3, e_4, e_6, e_7, e_9 all equal to 0, so you are done.

3. ## Re: Find a matrix transformation whose kernel is the given subspace

As Macsters undead said, $\displaystyle \begin{bmatrix}a \\ 0 \\ b \end{bmatrix}$ does not reduce to $\displaystyle \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$ but is of the form $\displaystyle a\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}+ b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$ for any numbers a and b. What you need are matrices $\displaystyle \begin{bmatrix}e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33}\end{bmatrix}$ such that $\displaystyle \begin{bmatrix}e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33}\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix} e_{11} \\ e_{21} \\ e_{31}\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$ and $\displaystyle \begin{bmatrix}e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33}\end{bmatrix}\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}$$\displaystyle = \begin{bmatrix} e_{13} \\ e_{23} \\ e_{33}\end{bmatrix}$$\displaystyle = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$.

The set is a 3 dimensional subspace of the 9 dimensional space of all 3 by 3 matrices.