Question

Hello. Can someone explain the MT' matrix in the recitation video on Linear Transformations ? Shouldn't that matrix be (w1;w2;w3;-w4) or (1,0,0,0; 0,0,0,1; 0,1,1,0; 0-1,1,0) ? Thanks for your time. http://www.youtube.com/watch?feature=player_embedded&v=-X04WJoTDBc

Answer 1

Let's see:

\[ \left[\begin{matrix}w1&w2&w3&w4\end{matrix}\right] \left[\begin{matrix} 1&0&0&0\\ 0&0&1&-1\\0&0&1&1\\0&1&0&0 \end{matrix}\right] = \left[\begin{matrix}w1\\w3-w4\\w3+w4\\w2\end{matrix}\right] \] So that does not jive with the \[w1^T = w1\\w2^T = w2\\ w3^T=w3\\w4^T=-w4\]

I'm not sure if your MT' had column or row entries, but if I switched them, it would the multiplication would yield\[\left[\begin{matrix}w1\\w4\\w2+w3\\w3-w2\end{matrix}\right] \] So correct answer is the one given in the recitation video.

\[ \left[\begin{matrix}w1&w2&w3&w4\end{matrix}\right] \left[\begin{matrix} 1&&&\\ &1&&\\&&1&\\&&&-1 \end{matrix}\right] = \left[\begin{matrix}w1\\w2\\w3\\-w4\end{matrix}\right] \]

Answer 2

Thanks for your reply datanewb. I watched prof. Strang's lecture on Linear Transformations and their Matrices and got a little confused when applying it to solve the problems on the cited video.
Anyway, I think I got it now. It's pretty obvious that the transformation matrix expressed in basis W as to be equal to:
\begin{matrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1 \\ \end{matrix}

because it's [ w1 w2 w3 -w4 ] (column vectors) written in W vectors.


What's bugging me is that I managed to get the some result by multiplying my original wrong matrix with a matrix whose columns are the standard vectors expressed in basis W:


\[\left[\begin{matrix} 1&0&0&0 \\ 0&0&0&1 \\ 0&.5&.5&0 \\ 0&.5&-.5&0 \\ \end{matrix}\right] \left[\begin{matrix} 1&0&0&0 \\ 0&0&1&-1 \\ 0&0&1&1 \\ 0&1&0&0 \\ \end{matrix}\right] = \left[\begin{matrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1 \\ \end{matrix}\right]\]
Let the product above be A B = MT'.
If A is a change of basis matrix from V to W (not sure on this one) then what is B?
Or maybe it's just a coincidence and in that case B wouldn't represent anything in this context (unlikely, I guess).

Answer 3

Ahh, I think I see now. I had yet to study the section on linear transformations. I'm just about to take quiz 2, so I'm a bit behind you, but I watched lecture 30 this morning.

So your answer: \[\left[\begin{matrix} 1&0&0&0 \\ 0&0&1&-1 \\ 0&0&1&1 \\ 0&1&0&0 \\ \end{matrix}\right]\] is in the standard basis, or basis V, whereas the recitation video answer is in the basis W. So either answer would be correct depending on which basis you were talking about?

And you found the matrix: \[\left[\begin{matrix} 1&0&0&0 \\ 0&0&0&1 \\ 0&.5&.5&0 \\ 0&.5&-.5&0 \\ \end{matrix}\right]\]
that converts from basis V (standard basis) to W (basis defined with vectors w1,w2,w3,w4).

I agree with all of that.

Answer 4

You've summarized it well datanewb. It seems that the initial matrix that I found ( which I called B in my 2nd post) was the transformation matrix MT' but with the vector columns w1 to w4 written with respect to V.
It may seem confusing but maybe it becomes simpler if we try to understand what that matrix does to vectors (after a phew hours of thought I think I'm now able to answer my own question). The matrix B takes vectors from W, transposes them, and presents the output vector with respect to V. Essentially B performs two operations:
1) Transposes in W;
2) Changes basis from W->V.

So, to finalize, B can be written as (using my 2nd post notation):
\[B = A^{-1} M _{T}' , \]
where A inverse is the change of basis matrix W->V and MT' is the matrix of the transformation in W basis (it makes sense because operations are performed starting from the right).

Just a final note. Linear transformations ( in this case T(A) = transpose of A) are independent from the basis in which they're performed.