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I have an assignment question that I am having some hard time working out and would appreciate any help understanding and solving this please. Here is the question

Let $L:\mathbb{R}^3\to\mathbb{R}^3$ be given by $$(x; y; z)^T \mapsto (x - y; y - z; z - x)^T$$ Find the matrix $A$ of $L$ with respect to the basis: $$(-1; 0; 1)^T ; (1; 1; 1)^T ; (0; 1; 0)^T$$ Then find the matrix $B$ of $L$ with respect to the basis: $$(0; 0; 1)^T , (0; 1; 1)^T , (1; 1; 1)^T$$ Finally, find a matrix $P$ so that $$B = P^{-1}AP$$

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Oops, sorry, I didn't see you edited it already. :) –  ae0709 Jul 15 '11 at 5:17
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To find the matrix $A$ of $L$ with respect to the basis $\beta=[(-1,0,1)^T, (1,1,1)^T, (0,1,0)^T]$, first write down the image of each basis vector under the linear transformation, and express it as a linear combination of the elements of $\beta$. The coefficients form the columns of $A$.

For example, evaluating $L$ at the second basis vector gives $$L\left(\begin{array}{c}1\\1\\1\end{array}\right) = \left(\begin{array}{c}1-1\\1-1\\1-1\end{array}\right) = \left(\begin{array}{c}0\\0\\0\end{array}\right) = 0\left(\begin{array}{r}-1\\0\\1\end{array}\right) + 0\left(\begin{array}{c}1\\1\\1\end{array}\right) + 0\left(\begin{array}{c}0\\1\\0\end{array}\right).$$ This tells you that the second column of $A$ is all $0$s.

Similarly, to find $B$, evaluate $L$ at each of the basis vectors of $\gamma=[(0,0,1)^T, (0,1,1)^T, (1,1,1)^T]$, and express the value in terms of the vectors of $\gamma$. This gives you the columns of $B$.

Finally, to find $P$, you need a matrix that "translates" from $\gamma$-coordinates to $\beta$-coordinates. So $P$ should be the matrix that tells you how to write each of the vectors in $\gamma$ in terms of the vectors in $\beta$. Express each vector in $\gamma$ as a linear combination of vectors in $\beta$, and use these expressions to give you the columns of $P$.

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