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In $P_2$, find the change-of-coordinates matrix from the basis $B=\{1-2t+t^2, 3-5t+4t^2, 2t+3t^2\}$ to the standard basis $C=\{1, t, t^2\}$. Then find the B-coordinate vector for $-1+2t$ I know how to do the first part. $P$ from $B$ to $C: \begin{bmatrix}1& 3& 0\\ -2 &-5& 2\\ 1& 4& 3\end{bmatrix}$. I do not know what the process is for finding the B coordinate vector though. Can someone give me a place to start for doing that?

John Hughes
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mmm
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  • The change of basis matrix is invertible and gives to the change of coordinates from $B$ to $C$. You know the coordinates for the vector $-1+2t$ with respect to the basis $C$, so multiply that vector by the inverse of the change of basis matrix you found. – Alex Wertheim Mar 12 '14 at 00:09
  • So I found the inverse of the matrix to be [-23 -9 6, 8 3 -2, -3 -1 1]. When multiplied that matrix by [-1 2 0] I got [26 18 0] as my column vector. But the answer in my book says the answer is [5 -2 1]. Can anyone verify that that I have correctly done the inverse? – mmm Mar 12 '14 at 01:03
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    Megan: You have done the inverse correctly, but your multiplication is incorrect. See http://www.bluebit.gr/matrix-calculator/multiply.aspx – Alex Wertheim Mar 12 '14 at 01:08
  • @AlexWertheim Please consider converting your comments into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Aug 22 '15 at 08:45
  • @mmm Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Feb 01 '18 at 22:59

2 Answers2

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So, you have $P_{C \leftarrow B} = \begin{bmatrix}1&3&0\\-2&-5&2\\1&4&3\\ \end{bmatrix}$. So to get the second part, you need to row reduce the augmented matrix $ \begin{bmatrix} P_{C \leftarrow B} & [x]_{C} \\ \end{bmatrix}$.

So, what is $[x]_{C}$? Take $-1+2t$ and plug it in: $$[x]_{C} = \begin{bmatrix} ?\\ ?\\ ?\\ \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 0\\ \end{bmatrix}$$

If you don't see where the $0$ came from, think about the coefficient for $t^{2}$.

So now, we have

$$\begin{bmatrix} P_{C \leftarrow B}& [x]_{C}\\ \end{bmatrix} = \begin{bmatrix} 1&3&0&-1\\-2&-5&2&2\\1&4&3&0\\ \end{bmatrix}$$

Now row reduce, and you'll have your answer.

Michael
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The basis vectors of B expressed with respect to the standard basis are:

$$v_1=(1,-2,1), v_2=(3,-5,4), v_3=(0,2,3)$$

Thus, the following matrix M, which colums are the component of the basis vectors B with respect to the standard basis:

$$M=\begin{bmatrix}1& 3& 0\\ -2 &-5& 2\\ 1& 4& 3\end{bmatrix}$$

represent the change of coordinates from the basis B to the standard basis C, that is:

$$v_C=Mv_B\implies v_B=M^{-1}v_C$$

where

$$M^{-1}=\begin{bmatrix}-23& -9& 6\\ 8 &3& -2\\ -3& -1& 1\end{bmatrix}$$

represent the change of coordinates from the standard basis C to the basis B.

Thus for the vector $-1+2t=v_C=(-1,2,0)$ in the standard basis C, we obtain:

$$v_B=v_CM^{-1}=\begin{bmatrix}-23& -9& 6\\ 8 &3& -2\\ -3& -1& 1\end{bmatrix}\begin{bmatrix}-1\\ 2 \\ 0\end{bmatrix}=\begin{bmatrix}5\\ -2 \\ 1\end{bmatrix}$$

You can check indeed that:

$$v_C=5v_1-2v_2+v_3$$

user
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