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If $\phi$ belongs to $\text{Hom}\left(\mathbb{R}^{3}\right)$ and has in respect to $E$ (the standard basis of $\mathbb{R}^{3}$) has a matrix $$f= \begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 3 & 5 & 3 \\ \end{bmatrix},$$ Find the matrix $J$ of $\phi$ respecting the basis $A$, where $$A=\left\{a_1=(9,0,9),a_2=(0,3,5),a_3=(3,-3,-1)\right\}.$$

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  • $\begingroup$ I would like some instructions... $\endgroup$ – john Apr 15 '17 at 18:57
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Hint:

Consider the matrix that has as columns the vectors of the basis $A$ $$ A=\begin{bmatrix} 9&0&3\\ 0&3&-3\\ 9&5&-1 \end{bmatrix} $$ If $\vec x$ is a vector that has components in the standard basis $[x_i]_S^T$ than its components in the basis $A$ are $[x_i]_A^T=A^{-1}[x_i]_S^T$ so that we have $[x_i]_S^T=A[x_i]_A^T$.

In the standard basis the linear transformation $\phi$ is represented by the matrix $F$ ( your $f$) that acts as: $$ [\phi(\vec x)_i]_S^T=F[x_i]_S^T=FA[x_i]_A^T $$ and if you want $\phi(\vec x)$ in the base $A$ you have to transform as: $$ [\phi(\vec x)_i]_A^T=A^{-1}[\phi(\vec x)_i]_S^T $$

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  • $\begingroup$ i'm confused... can I see the calculations? $\endgroup$ – john Apr 15 '17 at 20:24
  • $\begingroup$ You have: $J=A^{-1}FA$ $\endgroup$ – Emilio Novati Apr 15 '17 at 20:35

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