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Problem: Let $\alpha$ be the standard basis of $\mathbb{R}^3$ and let $\beta = \left\{(1,0,0), (1,1,0), (1,1,1)\right\}$ be another basis. Consider the linear map $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by \begin{align*} T(x,y,z) = (x+2y+z, -y, x+4z). \end{align*} Find the change of basis matrix from $\alpha$ to $\beta$ and the change of basis matrix from $\beta$ to $\alpha$.

Attempt at solution: For the change of basis matrix from $\alpha$ to $\beta$ I did: $T(1,0,0) = (1,0,1), T(1,1,0) = (3,-1,1), T(1,1,1) = (4,-1,5)$. Then I wrote this in the columns of a matrix: \begin{align*} P = \begin{pmatrix} 1 & 3 & 4 \\ 0 & -1 & -1 \\ 1 & 1 & 5 \end{pmatrix}. \end{align*} So this would be the change of basis matrix from $\alpha$ to $\beta$.

Now, I wanted to check if it was correct by verifying a theorem I recall: "Let $P$ be the change of basis matrix from $\alpha$ to $\beta$. Then for each vector $v$ we have \begin{align*} &(i) P[v]_{\beta} = [v]_{\alpha} \\ & (ii) P^{-1} [v]_{\alpha} = [v]_{\beta},\end{align*} where $[v]_{\beta}$ represents the coordinates (written as a column vector) of an arbitrary vector with respect to the basis $\beta$.

So I picked an arbitry vector $v = (10,8,3) = 2(1,0,0) + 5(1,1,0) + 3(1,1,1)$. Hence its coordinates with respect to $\beta$ are \begin{align*} [v]_{\beta} = \begin{pmatrix} 2 \\ 5 \\ 3 \end{pmatrix}. \end{align*} So I did \begin{align*} \begin{pmatrix} 1 & 3 & 4 \\ 0 & -1 & -1 \\ 1 & 1 & 5 \end{pmatrix} \begin{pmatrix} 2 \\ 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 29 \\ -8 \\ 10 \end{pmatrix}. \end{align*} But $(v) = (10,8,3) \neq 29(1,0,0) -8 (0,1,0) + 10 (0,0,1)$ ?? So now I'm confused and I don't know what I did wrong. Does the theorem hold only under special conditions, or is my change of basis matrix wrong?

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I'm not sure where exactly $T$ plays a role in this problem. To find $P$, write the vectors in $\alpha$ in terms of $\beta$

$$ \alpha_1 = \begin{pmatrix}1\\0\\0\end{pmatrix}_\beta, \alpha_2 = \begin{pmatrix}-1\\1\\0\end{pmatrix}_\beta, \alpha_3 = \begin{pmatrix}0\\-1\\1\end{pmatrix}_\beta $$

and so

$$ P = \begin{pmatrix}1&-1&0\\0&1&-1\\0&0&1\end{pmatrix} $$

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  • $\begingroup$ Ok, thanks. I guess I confused it with the matrixrepresentation. By the way, your $\alpha_3$ is wrong, it should be $(0,-1,1)^T$. But I also think my theorem is wrong. It is $P [v]_{\alpha} = [v]_{\beta}$ and not $P[v]_{\beta} = [v]_{\alpha}$. Am I right? $\endgroup$ – Kamil Jul 6 '15 at 19:08
  • $\begingroup$ @Kamil Yes, what we have found was a change of basis matrix from $\alpha$ to $\beta$. $\endgroup$ – user137794 Jul 6 '15 at 23:31
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You have calculated the change of basis matrix, that matrix give you the coordenates to that linear transformation. If you consider $\vec{v}=(10,8,3)_{\alpha}=(2,5,3)_{\beta}$ then you have that $T(\vec{v})=(29,-8,10)$

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