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a)Define the matrix of linear transformation that maps any vector in $R^2$ to it's correspondent counter clockwise rotation of $\frac{\pi}{4}$ vector, and then find the matrix of that transformation in standard basis ($\gamma$) for $R^2$.

b)Find another matrix that represents the change of basis from standard basis to the basis: $\beta = \left\{ (1,2),(0,-2)\right\}$.

c)Find a the transformation $T_{\beta}^{\gamma}$, that is, the transformation $T(X)$ that starts in the basis $\gamma$ and ends in the basis $\beta$.

My answer:

a)$$ A = \left[\begin{matrix} cos(\frac{\pi}{4}) & -sin(\frac{\pi}{4}) \\ sin(\frac{\pi}{4}) & cos(\frac{\pi}{4}) \end{matrix}\right] $$ Now the matrix of that transformation in standard basis: $$ \gamma = \left\{(1,0),(0,1)\right\}\\ A\gamma_1 = A \left[\begin{matrix} 1 \\ 0 \end{matrix}\right] = \left[\begin{matrix} \frac{\sqrt{2}}{2} \\ 0 \end{matrix}\right] \\ A\gamma_2 = A \left[\begin{matrix} 1 \\ 0 \end{matrix}\right] = \left[\begin{matrix} 0\\ \frac{\sqrt{2}}{2} \end{matrix}\right] \\ T_{\gamma}^{\gamma} = \left[\begin{matrix} \frac{\sqrt{2}}{2} & 0\\ 0 & \frac{\sqrt{2}}{2}\\ \end{matrix}\right] \\ $$

b)Matrix of change of basis, from $\gamma$ to $\beta$: $$ (1,0) = a(1,2)+b(0,-2) \Rightarrow a = 1 , b = 1\\ (0,1) = a(1,2)+b(0,-2) \Rightarrow a = 0 , b = -\frac{1}{2}\\ I_{\beta}^{\gamma} = \left[\begin{matrix} 1 & 1\\ 1 & -\frac{1}{2} \end{matrix}\right] $$

c)$$ T_{\beta}^{\gamma} = I_{\beta}^{\gamma}\cdot T_{\gamma}^{\gamma}\\ \left[\begin{matrix} 1 & 1\\ 1 & -\frac{1}{2} \end{matrix}\right] \cdot \left[\begin{matrix} \frac{\sqrt{2}}{2} & 0\\ 0 & \frac{\sqrt{2}}{2}\\ \end{matrix}\right] = \left[\begin{matrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{4}\\ \end{matrix}\right] $$

I'm new to change of basis and linear transfromations... All I need in the moment is someone to check if my approach is good enough, and if everything I did is correct and makes sense... Thanks a lot!!

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  • $\begingroup$ A counter clockwise rotation matrix by $\theta$ is of the form $$\left(\begin{matrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{matrix}\right)$$ $\endgroup$
    – Ian Miller
    Oct 2 '16 at 0:04
  • $\begingroup$ But in the case for the standard basis, isn't correct what I've done? $\endgroup$
    – Bruno Reis
    Oct 2 '16 at 0:06
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(a) In the standard basis, you can simply evaluate the cosines and sines in your matrix $A$: $$T_\gamma^\gamma = \frac12\begin{bmatrix} \sqrt2 & -\sqrt2\\ \sqrt2 & \sqrt2\end{bmatrix}.$$ Your method will give the same result if you multiply $A\gamma_1$ and $A\gamma_2$ correctly.

(b) Your calculations here are correct, but you accidentally filled in the matrix incorrectly. The correct matrix is $$I_\beta^\gamma = \begin{bmatrix} 1 & 0\\ 1 & -\frac12\end{bmatrix}.$$

(c) $$\begin{align} T_\beta^\gamma & = I_\beta^\gamma \cdot T_\gamma^\gamma\\\\ & = \begin{bmatrix} 1 & 0\\ 1 & -\frac12\end{bmatrix} \cdot \frac12\begin{bmatrix} \sqrt2 & -\sqrt2\\ \sqrt2 & \sqrt2\end{bmatrix}\\\\ & = \frac14\begin{bmatrix} 2 \sqrt2 & -2 \sqrt2\\ \sqrt2 & -3 \sqrt2\end{bmatrix}.\end{align}$$

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