Question

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Problem 2: In this problem, we explore another example of how the matrix associated with a inear transformation TAX)-Ax can be simplified when we change basis. Consider the linear equation given by: 17 61 in Tout We consider the change of variables given by: where the columns of u consistute our new basis. . Use this change of variable in the first equation to write it in terms of clout and Bout Bout Multiply both sides of this equation by the change of basis matrix u Explain how this shows that the linear transformation under this new basis becomes T제이 ) = B , where B = u-1 AU Compute BAll and show that it is diagonal

Answer 1

Given, and $\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}\begin{bmatrix} \alpha\\ \beta \end{bmatrix}$ .

a) Combinin we get,

$\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}\begin{bmatrix} \alpha_{out}\\ \beta_{out} \end{bmatrix}=\begin{bmatrix} 17 & 6\\ 6 & 8 \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}\begin{bmatrix} \alpha_{in}\\ \beta_{in} \end{bmatrix}$

Therefore, the required equation is : $\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}\begin{bmatrix} \alpha_{out}\\ \beta_{out} \end{bmatrix}=\begin{bmatrix} 17 & 6\\ 6 & 8 \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}\begin{bmatrix} \alpha_{in}\\ \beta_{in} \end{bmatrix}$

b) From the above equation we get,

$\begin{bmatrix} \alpha_{out}\\ \beta_{out} \end{bmatrix}=\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}^{-1}\begin{bmatrix} 17 & 6\\ 6 & 8 \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}\begin{bmatrix} \alpha_{in}\\ \beta_{in} \end{bmatrix}$

Then, B = $\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}^{-1}\begin{bmatrix} 17 & 6\\ 6 & 8 \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}$

c) Now, B = $\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}^{-1}\begin{bmatrix} 17 & 6\\ 6 & 8 \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}$

i.e., B = $\begin{bmatrix} 1/\sqrt{5} & 2/\sqrt{5}\\ -2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}\begin{bmatrix} 17 & 6\\ 6 & 8 \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}$

i.e., B = $\begin{bmatrix} 29/\sqrt{5} & 22/\sqrt{5}\\ -28/\sqrt{5} & -4/\sqrt{5} \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}$

i.e., B = $\begin{bmatrix} 73/5 & -36/5\\ -36/5 & 52/5 \end{bmatrix}$

Here, B is not a diagonal matrix.

B will be diagonal if U = $\begin{bmatrix} 1/\sqrt{5} & 2/\sqrt{5}\\ -2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}$ .