- #1

- 70

- 0

## Homework Statement

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

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

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

## Homework Equations

## 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

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

[tex]\[[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}[/tex]

The new transformation matrix should be [tex][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[/tex]