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I want to find the matrix representation of a rotation T in a plane with respect to a basis of $\bf{a_1} = \bf{e_1}$, $\bf{a_2} = \bf{e_1 + e_2}$. I found the following, but I'm having trouble understanding why it worked. Note: I'm avoiding the use of inverse matrices and trying to do it simply by definitions of linear transformations.

First, note that the rotation in the canonical basis is: $$R = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix} = \begin{bmatrix} c & -s \\ s & c\end{bmatrix}$$

Applying it to the new basis,

$$Ra_1 = \begin{bmatrix} c \\ s \end{bmatrix}$$ $$Ra_2 = \begin{bmatrix} c-s \\ c+s \end{bmatrix}$$

So apparently the new transformation matrix will also transform the new basis in the same manner?

$$Ta_1 = \begin{bmatrix} c \\ s \end{bmatrix} = \alpha \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \beta \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ $$Ta_2 = \begin{bmatrix} c-s \\ c+s \end{bmatrix} = \gamma \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \delta \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ We can solve to find the coefficients, and then they form the new matrix but by placing them in columns? $$\alpha = c-s,\space \beta = s,\space \gamma = -2s,\space \delta = c+s$$ $$T = \begin{bmatrix} c-s & -2s \\ s & c+s\end{bmatrix}$$ This seems to be the correct answer - if we take the angle to be $\pi/4$ and rotate $\bf{a_1}$ we get a vector in the $\bf{a_2}$ direction but magnitude $1/\sqrt{2}$ since it's originally of magnitude 1 and $\bf{a_2}$ is of magnitude $\sqrt{2}$. If we rotate $\bf{a_2}$ similarly we get $\begin{bmatrix} -2\sqrt{2} \\ 3\sqrt{2} \end{bmatrix}$ which also looks correct.

So could somebody explain what happened here? Why can we set $Ra_j = Ta_j = \sum_{i=1}^2 T_{ij}a_j$?

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  • $\begingroup$ I think you meant a rotation by $\pi/4$ rather than $\pi/2$; in either case when I construct and apply $T$ according to your formula I do not get the same answer as you do for rotating $\mathbf a_2,$ but I do get what appears to be a correct answer. $\endgroup$ – David K Oct 8 '17 at 19:19
  • $\begingroup$ You are correct, I meant $\pi /4$ $\endgroup$ – wobertson Oct 12 '17 at 20:04
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The key thing here is that the columns of the matrix of a transformation are the images of the basis vectors, expressed in the “output” basis. If we call the $(\mathbf a_1,\mathbf a_2)$ basis $\mathcal B$, this means that the columns of the matrix $[T]_{\mathcal B}^{\mathcal B}$ of the rotation are the $\mathcal B$-coordinates of $T\mathbf a_1$ and $T\mathbf a_2$. That is, $$[T]_{\mathcal B}^{\mathcal B}=\begin{bmatrix} [T\mathbf a_1]_{\mathcal B} & [T\mathbf a_2]_{\mathcal B} \end{bmatrix}.$$ By definition, these coordinates are the coefficients of the linear combinations of $\mathbf a_1$ and $\mathbf a_2$ that produce $T\mathbf a_1$ and $T\mathbf a_2$. Using your notation, then, $$[T]_{\mathcal B}^{\mathcal B}=\begin{bmatrix}\alpha&\gamma \\ \beta&\delta \end{bmatrix}$$ where the elements of the matrix are the solutions to the system of equations $$\begin{align} \alpha+\beta &= c \\ \beta &= s \\ \gamma+\delta &= c-s \\ \delta &= c+s \end{align}$$ that you set up.

You can verify that you got the correct answer by computing this matrix using the usual change-of-basis formula. Calling the standard basis $\mathcal E$, $$\begin{align} [T]_{\mathcal B}^{\mathcal B} &= [id]_{\mathcal B}^{\mathcal E} \, [T]_{\mathcal E}^{\mathcal E} \, [id]_{\mathcal E}^{\mathcal B} \\ &= ([id]_{\mathcal E}^{\mathcal B})^{-1} \, [T]_{\mathcal E}^{\mathcal E} \, [id]_{\mathcal E}^{\mathcal B} \\ &= \begin{bmatrix}[\mathbf a_1]_{\mathcal E}&[\mathbf a_2]_{\mathcal E} \end{bmatrix}^{-1}\,R\,\begin{bmatrix}[\mathbf a_1]_{\mathcal E}&[\mathbf a_2]_{\mathcal E} \end{bmatrix} \\ &= \begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}c&-s\\s&c\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix} \\ &= \begin{bmatrix}c-s&-2s\\s&c+s\end{bmatrix}. \end{align}$$ These matrix multiplications and inversions are just a different way of solving the system of linear equations for the coordinates of $T\mathbf a_1$ and $T\mathbf a_2$.

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Consider these two equations from the question: \begin{align} \begin{bmatrix} c \\ s \end{bmatrix} &= \alpha \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \beta \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \tag1\\ \begin{bmatrix} c-s \\ c+s \end{bmatrix} &= \gamma \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \delta \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \tag2 \end{align}

This is fundamentally a linear system of four equations with four unknowns. And you can certainly solve it in the way you did.

Another way to write Equations $(1)$ and $(2)$ together is $$\begin{bmatrix} c & c-s \\ s & c+s \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \alpha & \gamma \\ \beta & \delta \end{bmatrix}. $$ It follows that $$\begin{bmatrix} c & c-s \\ s & c+s \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \alpha & \gamma \\ \beta & \delta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \tag3 $$ for any real numbers $x,y.$

Observe that the matrix on the left side of the equation is the matrix that transforms the coordinates of a vector relative to the basis $\mathbf a_1, \mathbf a_2$ into the coordinates of the rotated vector relative to the basis $\mathbf e_1, \mathbf e_2$. That is, if we have a vector $\mathbf v = x \mathbf a_1 + y \mathbf a_2,$ and the rotation of this vector produces $\mathbf v',$ then the coordinates of $\mathbf v'$ in the standard basis are given by $$\begin{bmatrix} c & c-s \\ s & c+s \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. $$

The matrix $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix},$ on the other hand, transforms the coordinates of a vector relative to the basis $\mathbf a_1, \mathbf a_2$ into the coordinates of the same vector relative to the basis $\mathbf e_1, \mathbf e_2$, without rotating it. So if $\mathbf v' = x' \mathbf a_1 + y' \mathbf a_2,$ then $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix}. $$ gives the coordinates of $\mathbf v'$ in the canonical basis. That is, $$\begin{bmatrix} c & c-s \\ s & c+s \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix}. \tag4 $$

But the matrix $T$ for the rotation with respect to the basis $\mathbf a_1, \mathbf a_2$ is just the matrix by which we multiply any $\begin{bmatrix} x \\ y \end{bmatrix}$ in order to get the corresponding $\begin{bmatrix} x' \\ y' \end{bmatrix}.$ That is, $$ \begin{bmatrix} x' \\ y' \end{bmatrix} = T \begin{bmatrix} x \\ y \end{bmatrix}. $$ Making this substitution in Equation $(4),$ we get $$\begin{bmatrix} c & c-s \\ s & c+s \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} T \begin{bmatrix} x \\ y \end{bmatrix}. \tag5 $$

Now compare Equation $(5)$ with Equation $(3).$ One equation has the matrix $\begin{bmatrix} \alpha & \gamma \\ \beta & \delta \end{bmatrix}$ where the other has $T$; otherwise the equations are identical. And Equation $(3)$ does exactly the job that we want Equation $(5)$ to do. So it seems plausible that the matrix $\begin{bmatrix} \alpha & \gamma \\ \beta & \delta \end{bmatrix}$ is a suitable representation of $T.$

To actually prove that $T = \begin{bmatrix} \alpha & \gamma \\ \beta & \delta \end{bmatrix},$ you might multiply both sides of Equation $(3)$ and $(5)$ on the left by the matrix $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}.$ This would transform the coordinates on the left side of either equation back into coordinates with respect to the basis $\mathbf a_1, \mathbf a_2,$ while transforming the right side of $(3)$ to $\begin{bmatrix} \alpha & \gamma \\ \beta & \delta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}.$ You said you did not want to use inverse matrices; did that apply to the method of finding the rotation, or to the method and the proof of the method? There might be a way to do the proof without invoking the existence of the inverse of $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix},$ but I have not tried it.

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You have to express $Ta_1$ under the form $\alpha a_1+\beta a_2$ and to express $Ta_2$ as $\gamma a_1+\delta a_2$. Then, the matrix of $T$ with respect to $\{a_1,a_2\}$ is $\left(\begin{smallmatrix}\alpha&\beta\\\gamma&\delta\end{smallmatrix}\right)$.

There is another way of doing this. Let$$B=\begin{pmatrix}1&1\\0&1\end{pmatrix},$$ which is the change of basis matrix from $\{a_1,a_2\}$ to $\{e_1,e_2\}$. Then the matrix which you're after is$$B^{-1}.\begin{pmatrix}c&-s\\s&c\end{pmatrix}.B=\begin{pmatrix}c-s&-2s\\s&c+s\end{pmatrix},$$which is (not surprisingly) the same matrix that you got.

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  • $\begingroup$ You first asked why I introduced R to find T, but without R, how do I find T then? I tried to use geometry but it got complicated because of the different vector lengths. And you also used R in applying $B^{-1}RB$ - is this like taking a vector in basis $\{a_1,a_2\}$, transforming it into $\{e_1,e_2\}$, rotating it, and then transforming it back into $\{a_1,a_2\}$? $\endgroup$ – wobertson Oct 8 '17 at 15:10
  • $\begingroup$ @wobertson I see. You called $R$ to the matrix. I thought that it was a linear transformation. I will edit my answer. $\endgroup$ – José Carlos Santos Oct 8 '17 at 15:15
  • $\begingroup$ I just edited my question to note that I'm trying to accomplish this using the definitions of linear transformations and thus without explicit inverses - which the original work I did seems to accomplish, although I'm trying to figure out why. It's good to have your answer to verify the solution though, thank you. $\endgroup$ – wobertson Oct 8 '17 at 15:25

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