# Basis of Kernel of Transformation Matrix

Find the matrix of the given linear transformation T with respect to the given basis.

Determine whether T is an isomorphism. If T isn't an isomorphism find bases of the kernel and image of T, and thus determine the rank of T.

T(M) = M$\begin{bmatrix}1&2\\0&1\end{bmatrix}$ - $\begin{bmatrix}1&2\\0&1\end{bmatrix}$M from U^2x2 to U^2x2

with respect to the basis

$\mathfrak{B}$ = ($\begin{bmatrix}1&0\\0&1\end{bmatrix}$,$\begin{bmatrix}0&1\\0&0\end{bmatrix}$,$\begin{bmatrix}1&0\\0&-1\end{bmatrix}$)

I already found the matrix of the linear transformation

T(M) = $\begin{bmatrix}0&0&0\\0&0&4\\0&0&0\end{bmatrix}$

since the rref does not reduce to the identity matrix I know that it is not an isomorphism so I have to find the kernel, image and rank

I know how to do image and got im(T) = $\begin{bmatrix}0\\4\\0\end{bmatrix}$

I know the answer for the kernel is $\begin{bmatrix}1\\0\\0\end{bmatrix}$,$\begin{bmatrix}0\\1\\0\end{bmatrix}$ but I am unclear on how they arrived at this answer. I have looked up several sources to try and learn how to do the kernel but I am still not understanding the process since none of the examples have looked like mine. Can anyone explain how to go about finding the basis of the kernel for a problem that looks like this? Thank you

And the rank would be 1 because the rref has one non-zero row?

• What is $U^{2\times2}$? – Bernard Nov 20 '15 at 1:17
• The space of upper triangular 2x2 matrices – Lindsey G Nov 20 '15 at 1:22

## 1 Answer

To find the kernel, you just have to put the matrix in row echelon form, which is already the case, and solve. The solutions have to satisfy the only equation $z=0$, hence the solutions are isomorphic to $K^2$ (I denote $K$ your base field), by the isomorphism \begin{align*} K^2&\longrightarrow U^{2\times 2}\\ (x,y)&\longmapsto xI+yE_{12}=\begin{bmatrix}x&y\\0&x\end{bmatrix}. \end{align*}

• ok! I think what I was doing wrong was misunderstanding how to set up the equations from the rref (it's been awhile). Am I correct with the rank being 1? – Lindsey G Nov 20 '15 at 1:43
• Yes. But the image does not reduce to one vector, but it's the subspace generated by this vector, and eventually by the second vector of your basis (that I denote $E_{12}$). – Bernard Nov 20 '15 at 1:48
• I don't think I understand what you mean by the second equation you wrote – Lindsey G Nov 20 '15 at 2:11
• Not the second equation I wrote, the second vector you wrote – the matrix $\;\begin{bmatrix}0&1\\0&0\end{bmatrix}$. – Bernard Nov 20 '15 at 2:16