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Original matrix is:

$$A=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$

original basis: $$\langle \mathbf{e_1},\mathbf{e_2},\mathbf{e_3} \rangle$$

basis where I have to find matrix: $$\langle f_1,f_2,f_3 \rangle$$

Where:

$$f_1=2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3} \\ f_2 = 3\mathbf{e_1}+4\mathbf{e_2}+\mathbf{e_3} \\f_3=\mathbf{e_1}+2\mathbf{e_2}+2\mathbf{e_3}$$

Therefore:

$$x_1\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3}=y_1f_1+y_2f_2+y_3f_3$$

Expanding that we get:

$$x_1\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3}=y_1(2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3})+y_2(3\mathbf{e_1}+4\mathbf{e_2}+\mathbf{e_3} )+y_3(\mathbf{e_1}+2\mathbf{e_2}+2\mathbf{e_3})$$

How should I group things at right side around unit vectors? for example for first coordinate $2\mathbf{e_1} \, !=\, 3\mathbf{e_1}$, how could I group coordinates?


${\Large \textbf {Solution:}}$

Opening brackets:

$$x\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3} = 2y_1\mathbf{e_1}+3y_1\mathbf{e_2}+y_1\mathbf{e_3}+3y_2\mathbf{e_1}+4y_2\mathbf{e_2}+y_2\mathbf{e_3}+y_3\mathbf{e_1}+2y_3\mathbf{e_2}+2y_3\mathbf{e_3}$$

Sorting by unit vectors:

$$x_1\mathbf{e_2}+x_2\mathbf{e_3}+x_3\mathbf{e_3} = (2y_1+3y_2+y_3)\mathbf{e_1}+(3y_1+4y_2+2y_3)\mathbf{e_2}+(y_1+y_2+2y_3)\mathbf{e_3} \tag{1}$$

Based on fotmula:

$$A'=T^{-1}AT$$

Matrix $T$ is based on right side of equation $(1)$, and we have:

$$T=\begin{pmatrix} 2&3&1\\3&4&2\\1&1&2 \end{pmatrix}$$

The rest is more less technical.

$$T^{-1}=\begin{pmatrix} -6&5&-2\\4&-3&1\\1&-1&1 \end{pmatrix}$$

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Call the first and second basis $E$ and $F$ respectively. You are given $$A_E=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$ and the change of coordinates matrix that changes $F$ coordinates to $E$ coordinates is $$Q=\begin{pmatrix}2&3&1\\3&4&2\\1&1&2 \end{pmatrix}$$ Then $$A_F=Q^{-1}A_EQ.$$

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    $\begingroup$ Thank you for your time, You confirmed that I am right, I've just added my expanded thoughts $\endgroup$ – M.Mass Jun 29 '17 at 12:41
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To compute the matrix in the other basis We need to compute $T(f_1), T(f_2), T(f_2)$, where $T$ is the linear map defined by $A$ in the basis $e$.

By definition of matrix of a linear map, the coordinates of $T(f_i)$ in the basis $e$ will be obtained if we multiply $A$ by the coordinates of $f_i$ in the basis $e$.

For example, the coordinates of $f_1$ in the basis $e$ is $(2,3,1)^{T}$ since $f_1=2e_1+3e_2+e_3$. Multiplying $A(2,3,1)^{T}=(2,3,1)^{T}$.

Therefore $T(f_1)=2e_1+3e_2+e_1=f_1=1f_1+0f_2+0f_3$. Therefore, the first column of the matrix of $T$ in the basis $f$ is $(1,0,0)^{T}$.

Doing the same for $f_2$ and $f_3$ gives you the other two columns.

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