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I am studying linear transformations on my own and I have met the following theorem.

Suppose the vectors $v_1,\dots,v_n$ are the basis for the space $\mathbf{V}$, and vectors $w_1,\dots,w_m$ are basis for $\mathbf{W}$. Each linear transformation $T$ from $\mathbf{V}$ to $\mathbf{W}$ is represented by a matrix $\mathbf{A}$ whose $j$th column is found by applying $T$ to the $j$th basis vector $v_j$, and writing $T(v_j)$ as a combination of the $w$'s:

$$\text{Column}~~~\mathbf{j}~~~\text{of}~~~\mathbf{A}~~~T(v_j)=Av_j=a_{1j}w_1 + a_{2j}w_2 + \dots + a_{mj}w_m$$

Given this theorem we can compute the transformation matrix $\mathbf{A}$ of differentiation of polynomials. $\frac{d}{t}:\mathbf{P}_n \rightarrow \mathbf{P}_{n-1}$ where $\mathbf{P}_n$ are the polynomial of degree $n+1$ (consider constant term). Thus given the natural basis of third and second degree polynomials by $\mathbf{V} = \{1,t,t^2,t^3\}$ and $\mathbf{W} = \{1,t,t^2\}$ respectively, we may compute transformation matrix $\mathbf{A_{diff}}$. More specifically,

$$\frac{d}{dx}v_1 = 0~~~\text{or}~~~\mathbf{A_{diff}}v_1 =0w_1+0w_2+0w_3+0w_4$$, the coefficients $[0,0,0,0]^T$ form the first column of $\mathbf{A_{diff}}$.

$$\frac{d}{dx}v_2= 1~~~\text{or}~~~\mathbf{A_{diff}}v_2 =1w_1+0w_2+0w_3+0w_4$$, the coefficients $[1,0,0,0]^T$ form the second column of $\mathbf{A_{diff}}$. With similar way, we can compute the rest of the columns to obtain

$$\mathbf{A_{diff}}=\begin{bmatrix} 0&1&0&0\\ 0&0&2&0\\ 0&0&0&3\\ \end{bmatrix}$$

Similarly I have tried to compute the rotation transformation matrix $\mathbf{A_{rot}}$. Given the basis of the initial space $\mathbf{V}=\{[1, 0]^T, [0, 1]^T\}$, the basis of the rotated space is $\mathbf{W}=\{[cos\theta, sin\theta]^T, [-sin\theta, cos\theta]^T\}$

Could you please help to find the $\mathbf{A_{rot}}$ using the same steps as with $\mathbf{A_{diff}}$?

Thanks!

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  • $\begingroup$ If you're using that basis for $W$ then the matrix $A_{\text{rot}}$ will be the identity, since you're just mapping basis vectors of $V$ to basis vectors of $W$. $\endgroup$ Jan 23 '16 at 16:44
  • $\begingroup$ @A.P. Thanks for the comment, I can see that, but I have tried to follow the same logic as in differentiation. That is apply, the transformation in each basis of $\mathbf{V}$ to obtain the basis $\mathbf{W}$ and then try to express each basis of $\mathbf{V}$ with basis from $\mathbf{W}$. Is that logic not correct? $\endgroup$
    – Thoth
    Jan 23 '16 at 17:00
  • $\begingroup$ You are confused. The basis for $W$ is independent of the transformation $T$. You don't apply $T$ to basis vectors of $V$ to obtain basis vectors of $W$. (What if $T$ was the linear transformation that mapped everything to $0$? You wouldn't get any basis vectors.) What you do is fix bases for $V$ and $W$, and then compute the matrix of $T$ in these bases. This is actually what you do in the differentiation example: the basis for $W$ was fixed, you didn't use differentiation to obtain it. $\endgroup$ Jan 23 '16 at 17:05
  • $\begingroup$ I am trying to figure it out but still nothing. I have seen your answer. You are telling me that the basis of two spaces are considered fixed. Thus, because both transforms are on the $\mathbb{R}$ we may pick the same basis, $\{[1,0]^T,[0,1]^T\}$. So far clear. However, I am stucked with my last statements of my post which where made using the logic of differentiation. Could you please elaborate a little more, even in the differentiation example?I mean according to your comment, the basis of $\mathbf{P}_3$ are the columns, of $\mathbf{I}_{4\times 4}$. Base on the theorem what next? Thks! $\endgroup$
    – Thoth
    Jan 23 '16 at 17:33
  • $\begingroup$ The bases for your spaces of polynomials are whatever you choose them to be. The choice is up to you and has nothing to do with differentiation. In your example, you chose the standard bases $(1,t,t^2,t^3)$ and $(1,t,t^2)$. If you chose the basis $(1,2t,3t^2)$ for $W$ instead, your matrix would be $A_\text{diff} = \begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$. $\endgroup$ Jan 23 '16 at 17:49
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Let $T:V \to W$ be rotation by $\theta$ in $\mathbb{R}^2$. Pick the standard basis for $V = \mathbb{R}^2$. We have that $T([1,0]) = [\cos \theta,\sin \theta]$ and $T([0,1]) = [-\sin\theta, \cos\theta]$.

If you choose the standard basis $([1,0],[0,1])$ for $W = \mathbb{R}^2$ then the matrix that you get is $$A_\text{rot} = \begin{pmatrix} \cos \theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.$$

If on the other hand you choose the basis $([\cos \theta,\sin \theta],[-\sin\theta, \cos\theta])$ for $W = \mathbb{R}^2$ then the matrix that you get is the identity $$A_\text{rot} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$

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  • $\begingroup$ Something last and I think I almost got it, with some doubts (more practice needed). So, more analytically, according to our discussion above, if we choose $\{[1,0]^T,[0,1]^T\}$ as the basis for both $\mathbf{V}$ and $\mathbf{W}$, which is actually the standard basis of both, $T_{\theta}([1,0]^T)=cos\theta [1,0]^T + sin\theta [0,1]^T$. Thus, $[cos\theta, sin\theta]^T$ is the first column of $\mathbf{A}_{rot}$. Siimilarly, for $T_{\theta}([0,1]^T)$ we get $[-sin\theta, cos\theta]$ as the second column of $\mathbf{A}_{rot}$. Am I right? $\endgroup$
    – Thoth
    Jan 24 '16 at 9:54
  • $\begingroup$ @Thoth Yes, that is exactly the right way to look at it. The columns of the transformation matrix are the coordinate vectors of the transformed $V$-basis vectors $T(v_j)$ with respect to the $W$-basis. $\endgroup$ Jan 24 '16 at 16:44

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