# Changes of bases

I'm struggling with this linear algebra 4 part question. I believe I've managed to answer part (a) of the question by applying φ to the basis elements:

I worked out φ(1)=0, φ(x)=2+2x, φ($x^2$)=4x+6$x^2$ and φ($x^3$)=6$x^2$+12$x^3$ (hope that's correct) and by combining all the columns taken from the results of the transformations, I found that:

$[\phi]^{\beta}_{\beta} = \left[\begin{array}{rrrr} 0 & 2 & 0 & 0 \\ 0 & 2 & 4 & 0 \\ 0 & 0 & 6 & 6 \\ 0 & 0 & 0 & 12 \end{array}\right]$

Not sure this is correct but for (b) does $[id]^{\beta}_{\gamma}$ just take the columns of the basis vector of Y? So $[id]^{\beta}_{\gamma}=\left[\begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 6 \\ 0 & 0 & 3 & 15 \\ 0 & 0 & 0 & 15 \end{array}\right]$

but I doubt this is correct because the basis has addition signs. Any ideas on what I should do to calculate part (b)? (I think once $[id]^{\beta}_{\gamma}$ is calculated $[id]^{\gamma}_{\beta}$ is just the inverse)

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In general, when you compute the matrix representation you calculate what it does the basis and write the components of its images in the domain as columns in your matrix. So unless you made a mistake in computing what $\Phi$ does to the elements of $\beta$ that should be correct.

This same idea applies in (b), but in this case it simply amounts to expressing the basis $\gamma$ in terms of $\beta$ (for $[Id]^{\beta}_{\gamma}$), so this is correct. For the reverse direction you can either express $\beta$ in terms of $\gamma$ or take the inverse of $[Id]^{\beta}_{\gamma}$.

For (c): Notice that $[\Phi]^{\gamma}_{\gamma}=[Id]^{\gamma}_{\beta}[\Phi]^{\beta}_{\beta}[Id]^{\beta}_{\gamma}$. That is, given a vector written in Basis $\gamma$ you first transform it to basis $\beta$, then use the representation of $\Phi$ in that basis which you have from (a) and finally you transform back to basis $\gamma$.

For (d): Try to write it out in basis $\gamma$ and consider what happens when you multiply diagonal matrices.

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so are you saying my calculation for $[Id]^{\beta}_{\gamma}$ is correct? –  methuselah May 11 '12 at 22:19
Yes. You can also see it this way: The elements of $\gamma$ written in basis $\gamma$ have components $(1,0,0,0), (0,1,0,0),(0,0,1,0)$ and $(0,0,0,1)$. Multiplying your matrix by these will just select the corresponding column, giving you the components in terms of $\beta$ as they are given in the exercise. –  erlking May 11 '12 at 22:27
I got $[\phi]^{\gamma}_{\gamma} = \left[\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & 12 \end{array}\right]$ –  methuselah May 15 '12 at 17:53
so do I just raise the numbers in the diagonal to power 10 then multiply by q = \begin{array}{rrrr} 15 & 0 & 0 & 0 \\ 0 & 15 & 0 & 0 \\ 0 & 0 & 15 & 0 \\ 0 & 0 & 0 & 15 \end{array} –  methuselah May 15 '12 at 18:00