Let $A: \mathbb{R}^3 \rightarrow \mathbb{R}^3$. Given: $$ f_1= \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}, f_2= \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}, f_3= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \\ g_1= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, g_2=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}, g_3=\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} $$ $$A(f_1)=g_1, A(f_2)=g_2, A(f_3)=g_3$$ What is the matrix A of that linear operator in basis ${e_1,e_2,e_3}.$ I know that $i$ column of matrix $A$ is $A(e_i)$. How can I understand what's happening with basis vectors through given transformaions? I have no clue. Please help.
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- $\begingroup$ Try to write $e_1, e_2$ and $e_3$ in the basis $\{ f_1,f_2,f_3 \}$. After that, use the linearity of $A$ to find $A(e_i)$. $\endgroup$Danuso Rocha– Danuso Rocha2017-03-14 20:45:23 +00:00Commented Mar 14, 2017 at 20:45
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1 Answer
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$F = [f_1, f_2, f_3], G = [g_1,g_2,g_3]$
$AF = G\\ A = AFF^{-1} = GF^{-1} $
