# how I do I find the matrices of these linear transformattions

I asked this questions two days ago on Stackoverflow, and was advised to migrate it here. The problem is the way the bases are given is not in the usual way. It is rather confusing. Help will be appreciated.

Going through the text on Linear Algebra by A. O. Morris (2nd edition) I am trying to understand something on linear transformations. There is a problem where the R-bases of $U$ and $V$ are given as

$\{u_1, u_2\}$ and $\{v_1,v_2,v_3\}$ respectively and the linear transformation from U to V is given by $$T(u_1)=v_1+2v_2-v_3$$ $$T(u_2)=v_1-v_2$$ The problem is to: a) find the matrix of $T$ relative to these bases,

b) the matrix relative to the R-bases $\{-u_1+u_2,2u_1-u_2\}$ and $\{v_1,v_1+v_2,v_1+v_2+v_3\}$

and

c) the relationship between the two matrices.

From the very good treatment of the subject here Determine the matrix relative to a given basis I figured out that the first matrix $T$ has columns

\begin{bmatrix} 1 & 1\\ \ 2 & -1\\ -1 & 0 \end{bmatrix}

then for part b) I figure out Matrix $A$ which I take to be the transform of ordered basis of $U$ to standard basis of $U$ as \begin{bmatrix} -1 & 2\\ 1 & -1 \end{bmatrix}

Matrix $B$ which I take to be the transform of ordered basis of $V$ to standard basis of $V$ as

\begin{bmatrix} 1 &1 &1\\ 0 & 1& 1\\ 0 & 0 & 1 \end{bmatrix}

I then find inverse of $B$ \begin{bmatrix} 0 & -1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{bmatrix}

and form the product

$B^{-1}CA$ as the answer to part c).

\begin{bmatrix} 4 &-7\\ -2 & 3\\ 1 & -2 \end{bmatrix}

Somehow I do not seem to get it. I am completely at sea. I would appreciate help to understand this. My sincere apologies for not being able to use Latex.

• write the question in standard format. It's problematic to understand your problem.@Zilore Mumba – adember Mar 15 '15 at 9:19
• Problem is I don't know how to write a matrix on a forum like this – Zilore Mumba Mar 15 '15 at 9:23
• Hint: your inverse of $B$ seems wrong. – Emilio Novati Mar 15 '15 at 9:40
• A sanity check for inverses - any matrices inverse can not have a zero row or column - since that would mean it has not full rank. And it must have full rank to be invertible. – mathreadler Mar 15 '15 at 10:11

Your work seems correct, but you have done a mistake in the inverse of $B$.
$$B^{-1}= \begin{bmatrix} 1 & -1 &0\\ 0 & 1 &-1\\ 0 & 0 & 1 \end{bmatrix}$$
here is one way to do the second part. observe that $$T(-u_1 + u_2) = Tu_1 +Tu_2 = v_1 + 2v_2 - v_3 +v_1 - v_2= 2v_1+v_2 -v_3\\=1 (v_1) + 2(v_1+v_2) -1(v_1+v_2 + v_3)\\ T(2u_1 - u_2 )= 2Tu_1 - Tu_2 = 2(v_1 + 2v_2 -v_3)-(v_1-v_2)=v_1+5v_2-2v_3\\=-4v_1+7(v_1+v_2)-2(v_1+v_2+v_3)$$ the matrix representing $T$ in the bases $\{-u_1+u_2, 2u_1 -u_2\}$ and $\{v_1, v_1+v_2, v_1+v_2 + v_3 \}$ is $$\pmatrix{1&-4\\2&7\\-1&-2}$$
• @ZiloreMumba, just like you did the first part where you represented $T$ by the matrix with base $\{u_1, u_2\}$ and $\{v_1,v_2,v_3\}.$ the first column of the matrix is $Tu_1$ expressed as a linear combination of the $v's$. i did the same with the new sets of bases. the bases were simple enough that you could do hand. when i computed the coefficients of $T = av_1+bv_2 + c_v3$ i started from the back, i.e. $av_1 + bv_2 + cv_3 = (a-b+c-c)v_1 + (b-c)(v_1+v_2)+c(v_1+v_2+v_3)$ – abel Mar 15 '15 at 11:09