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I need help solving (b) in this math problem.

Determine the matrix of f in the standard basis (Linear Algebra)

What is the relationship between the matrix f and the transformation matrix in the standard basis?

Thanks for any suggestions!

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Your formulation of the question at the end appears to indicate some confusion about the objects involved. $f$ is a linear transformation, not a matrix. You can get a representation of $f$ by a matrix by choosing a basis, but none of the matrices you get in this way "is" $f$. Thus, the question at the end should read "What is the relationship between the matrix of $f$ in the basis $(\mathbf{g}_1,\mathbf{g}_2)$ and the matrix of $f$ in the standard basis?". Does that help? If not, please say some more about where you're stuck. –  joriki Mar 7 '11 at 16:00
    
Yeah I got that, only I am not very good at expressing myself in English (about math). I know I am supposed to find the matrix for the linear transformation in the standard basis, given another basis {g1 g2}. Though, I am not sure how to do this. How to switch basis, that is. I think I should be able to use the result from (a) to determine the matrix, but I'm not sure how so any pointers would be appreciated! –  Mickel Mar 7 '11 at 16:16

1 Answer 1

up vote 1 down vote accepted

Say $\gamma=[g_1,g_2]$, $\beta$ is the standard ordered basis, and $[T]_{\gamma}$ is the matrix from part $(a)$.

We can find the change-of-basis matrix from $\gamma$ to $\beta$ by writing $g_1$ and $g_2$ as linear combinations of ${1 \choose 0}, {0 \choose 1}$:

${5 \choose 4}=5{1 \choose 0}+ 4{0 \choose 1}$

${-4 \choose 5}=-4{1 \choose 0}+ 5{0 \choose 1}$

These coefficients form the columns of the change-of-basis-matrix from $\gamma$ to $\beta$.

$[Q]_{\gamma}^\beta= \begin{pmatrix} 5 & -4 \\ 4 & 5 \end{pmatrix} $

We can also find the change-of-basis-matrix from $\beta$ to $\gamma$ by reversing the process above (writing ${1 \choose 0}$ and ${0 \choose 1}$ as linear combinations of $g_1$ and $g_2$) or by taking the inverse of $[Q]_{\gamma}^\beta$.

In either case, $[Q]_{\beta}^\gamma=\begin{pmatrix} \frac{5}{41} & \frac{4}{41} \\ \frac{-4}{41} & \frac{5}{41} \end{pmatrix} $

Then we multiply the matrices in the proper order. (The matrix $[T]_{\beta}$ is the matrix representation of $f$ with respect to the standard ordered basis.)

$[T]_{\beta}=[Q]_{\gamma}^\beta[T]_{\gamma}[Q]_{\beta}^\gamma= \begin{pmatrix} 5 & -4 \\ 4 & 5 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \frac{5}{41} & \frac{4}{41} \\ \frac{-4}{41} & \frac{5}{41} \end{pmatrix}= \begin{pmatrix} \frac{9}{41} & \frac{40}{41} \\ \frac{40}{41} & \frac{-9}{41} \end{pmatrix}$

I hope that helps!

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Very helpful, thanks! –  Mickel Mar 7 '11 at 17:11

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