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?

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

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*}

  • $\begingroup$ 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? $\endgroup$ – Lindsey G Nov 20 '15 at 1:43
  • $\begingroup$ 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 $E_{12}$). $\endgroup$ – Bernard Nov 20 '15 at 1:48
  • $\begingroup$ I don't think I understand what you mean by the second equation you wrote $\endgroup$ – Lindsey G Nov 20 '15 at 2:11
  • $\begingroup$ Not the second equation I wrote, the second vector you wrote – the matrix $\;\begin{bmatrix}0&1\\0&0\end{bmatrix}$. $\endgroup$ – Bernard Nov 20 '15 at 2:16

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