# Change of basis

## Homework Statement

Recall that the matrix for $$T: R^{2} \rightarrow R^{2}$$ defined by rotation through an angle $$\theta$$ with respect to the standard basis for $$R^{2}$$ is

$$$A =\begin{array}{cc}cos \theta & -sin \theta \\sin \theta & cos\theta \\\end{array}$\right]$$

a) What is the matrix of T with respect to the basis (2,1),(1,-2)?

## The Attempt at a Solution

So, here's the deal, I missed this particular lecture last week. I'm trying to teach myself this concept of changing basis. Here's one of my homework problems. Can you please check it?

alpha = standard basis, beta = new basis

I(1,0) = a(2,1) + b(1,-2) --> a = .4, b = .2
I(0,1) = a(2,1) + b(1,-2) --> a = .2, b = -.4

$$\[[I^{\beta}_{\alpha}] = \begin{array}{cc} .4 & .2 \\ .2 & -.4 \\ \end{array}$$

$$\[[I^{\beta}_{\alpha}]^{-1} = (\frac{1}{.4(-.4)-.2(.2)})\begin{array}{cc} -.4 & -.2 \\ -.2 & .4 \\ \end{array} = -5(\begin{array}{cc} -.4 & -.2 \\ -.2 & .4 \\ \end{array}) = \begin{array}{cc} 2 & 1 \\ 1 & -2 \\ \end{array}$$

The new transformation matrix should be $$[I^{\beta}_{\alpha}]^{-1}A[I^{\beta}_{\alpha}] = (\begin{array}{cc} 2 & 1 \\ 1 & -2 \\ \end{array})(\begin{array}{cc}cos \theta & -sin \theta \\sin \theta & cos\theta \\\end{array})(\begin{array}{cc} .4 & .2 \\ .2 & -.4 \\ \end{array}) = \begin{array}{cc}cos \theta & sin \theta \\ -sin \theta & cos \theta$$

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