Matteo's question at Yahoo Answers (change of basis)

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    Basis Change of basis
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The discussion centers on the calculation of the change-of-basis matrix for a linear map L: ℝ³ → ℝ² represented by the matrix A = [[7, 7, 7], [7, 7, 7]]. The bases B for ℝ³ and B' for ℝ² are provided, with B = [[0, 1, 3], [1, 0, 2], [0, 0, 1]] and B' = [[1, 1], [2, 0]]. The change-of-basis matrices P and Q are computed as P = [[0, 1, 0], [1, 0, 0], [3, 2, 1]] and Q = [[1, 2], [1, 0]]. The final change-of-basis matrix M_{B',B}(L) is derived using the formula M_{B',B}(L) = Q^{-1}AP.

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Here is the question:

Let L: IR^3 -> IR^2 the linear map represented by the matrix A:

7 7 7
7 7 7

with respect to the canonical bases in arriving and departing.

Are B:

0 1 3
1 0 2
0 0 1

and B':

1 1
2 0

respectively bases of R^3 and R^2.
Find the matrix of change of basis MB' ,B (L)

Here is a link to the question:

Matrix of change of basis? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Matteo,

The change-of-basis matrix in $\mathbb{R}^3$ from $B_c=\{(1,0,0),(0,1,0),(0,01)\}$ to $B=\{(0, 1, 3), (1, 0, 2), (0, 0, 1)\} $ is
$$P=\begin{bmatrix}{0}&{1}&{0}\\{1}&{0}&{0}\\{3}&{2}&{1}\end{bmatrix}$$
The change-of-basis matrix in $\mathbb{R}^2$ from $B^*_c=\{(1,0),(0,1)\}$ to $B'=\{(1, 1), (2, 0)\} $ is
$$Q=\begin{bmatrix}{1}&{2}\\{1}&{0}\end{bmatrix}$$
The matrix of $L$ with respect to $B_c$ and $B_c^*$ is
$$A=\begin{bmatrix}{7}&{7}&{7}\\{7}&{7}&{7} \end{bmatrix}$$

Accorging to a well-konwn theorem, the matrix of $L$ with respect to the new basis $B$ and $B'$ is $Q^{-1}AP$, so
$$M_{B',\;B}(L)=\begin{bmatrix}{1}&{2}\\{1}&{0} \end{bmatrix}^{-1}
\begin{bmatrix}{7}&{7}&{7}\\{7}&{7}&{7} \end{bmatrix}\begin{bmatrix}{0}&{1}&{0}\\{1}&{0}&{0}\\{3}&{2}&{1} \end{bmatrix}=\ldots$$
 

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