So, I asked a similar question earlier today but still quite don't get it. I don't know why but I just can't see the intuition. Anyway, heres the problem:

Decide the standard transformation matrix for $f: R^2\to R^2 $ represented by, with respect to the basis B {$ \pmatrix{1\\1}\pmatrix{0\\2}$}, $T=\pmatrix{-1&11\\2&-8}$.

So, if I want to represent this in standard basis I assumed that I could just multiply it with the change of coordinate matrix from the basis B to the standard basis S which is (right?): $\pmatrix{1 & 0\\1 & 2}$ which transforms T into: $\pmatrix{-1 & 11\\3 & -5}$.

I think I'm right so far but don't know where to go from here. To be honest, I'm not completely sure on what I've done so far. Any tips/hints would be greatly appreciated. Really trying to learn and see the concept behind this.

  • $\begingroup$ What do you mean by Decide the standard matrix? $\endgroup$ – P. Siehr Apr 4 '18 at 15:35
  • $\begingroup$ @P.Siehr If I haven't misunderstood the question I would assume that it's the transformation matrix T with respect to basis {$\pmatrix{1 \\ 0}$ $\pmatrix{0 \\ 1}$} $\endgroup$ – gbgult Apr 4 '18 at 15:37
  • $\begingroup$ @gbgult You should try to understand deeply the concept behind the method and how and why it works. If in the previuos OP you still had some doubts you should ask for that. The key point is the role of Matrix $M$ you need to convince yourself that this matrix is crucial to determine the matrix T in any basis. $\endgroup$ – gimusi Apr 4 '18 at 15:49

Let condider $M=[b_1\, b_2]$ then matrix $M$ transforms a vector from basis $B\,$ $v_B$ to the standard basis that is

$$v_S=M\cdot v_B\implies v_B=M^{-1}\cdot v_S$$

then for

$$w_B=T\cdot v_B\implies M^{-1}\cdot w_S=T\cdot M^{-1}\cdot v_S\implies w_S=M\cdot TM^{-1}\cdot v_S$$

thus the matrix transformation with respect to the standard basis is $$M\cdot T\cdot M^{-1}$$

  • $\begingroup$ Okay, so I get what M is. What do you mean by $Tv_B$? Just the transformation matrix with respect to the matrix B? $\endgroup$ – gbgult Apr 4 '18 at 15:56
  • $\begingroup$ Since T is given in basis B with $w_B=T\cdot v_B$ I'm indicating the tranformed of vector $v$ by $T$ in $w$ that is $w=T(w)$. I use $v_B$ and $w_B$ to emphatize that (the coordinates of) those vectors are expressed in the basis $B$. $\endgroup$ – gimusi Apr 4 '18 at 15:59
  • $\begingroup$ Alright, so just to clarify (obviously I need it without any possible doubts) - $w_B=Tv_B $ means the transformation of some arbitrary vector with respect to basis B and $w_S$ is the transformation of some arbitrary vector with respect to the standard basis? $\endgroup$ – gbgult Apr 4 '18 at 16:02
  • $\begingroup$ @gbgult Yes exactly and $v_S\, w_S$ are the same vectors expressed in the standard basis. $\endgroup$ – gimusi Apr 4 '18 at 16:04
  • $\begingroup$ Ah, I didn't really think about that you need a vector as input (That's $v_B$, right?) to get the output (a.k.a. transformed vector $W_b$). Thank's alot! I think I'm starting to get the concept now. I'll re-read the previous answer too and see. $\endgroup$ – gbgult Apr 4 '18 at 16:09

Let me try to explain the intuition, and then you can go back and read the answer to your prior question again. The given matrix of the transformation T shows how to transform a vector, expressed in terms of the basis B, into another another vector, also expressed in terms of B. We want a matrix that accomplishes the same transformation T, but expresses the vectors in terms of S. So, if we're given a vector in terms of S, we first express it in terms of B, and then we can use the given matrix to apply the transformation.

But that expresses the result in terms of B, and we want it in terms of S, so we have to apply the inverse of the change-of-basis transformation.

This kind of transform, solve, transform approach is ubiquitous in math. It's like changing variables in an indefinite integral. You change from a function of $x$ to a function of $u$ to get something easy to integrate, and then you change back from $u$ to $x$ to get the answer in terms of the original question. Or like analytic geometry -- you translate to the origin and rotate the axes to make the problem easy to solve, but then you have to undo the transformation to state the answer in terms of the original problem.

  • $\begingroup$ Thanks! You explained it very well, I think I kind of looked over the fact that the transformation matrix requires an input vector to transform and that it might be in a different basis. Gonna do some re-reading now and see. $\endgroup$ – gbgult Apr 4 '18 at 16:10
  • $\begingroup$ I might’ve added that the input and output bases don’t have to be the same, but otherwise +1. $\endgroup$ – amd Apr 4 '18 at 20:11

If i have understood your problem correctly. Then what you mean to say is that the matrix $T$ of the transformation $f:\mathbf{R}^2\to\mathbf{R}^2$ with respect to the basis $B = \{(1,1),(0,2)\}$ is as follows $$T=\pmatrix{-1&11\\2&-8}$$

and that you wish to determine the matrix $T'$ for $f$ corresponding to the standard basis that is $B' = \{(1,0),(0,1)\}$. If so then please consider the following.

First you need to express $(1,0)$ and $(0,1)$ as linear combinations of $(1,1)$ and $(0,2)$ like so $$(1,0) = 1\cdot(1,1)-0.5\cdot(0,2)$$ $$(0,1) = 0\cdot(1,1)+0.5\cdot(0,2)$$

Now to determine $T'$ you need to determine $f(1,0)$ and $f(0,1)$ like this $$f(1,0) = \pmatrix{-1&11\\2&-8}\pmatrix{1\\-0.5} = \pmatrix{-6.5\\6}$$ $$f(1,0) = \pmatrix{-1&11\\2&-8}\pmatrix{0\\0.5} = \pmatrix{5.5\\-4}$$ then your desired matrix that is $T'$ as follows $$T' = \pmatrix{-6.5&5.5\\6&-4}$$

Just to check your understanding try to guess what you would get by computing $f(1,1)$ and $f(0,2)$ with $T'$.


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