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In this example from Axler's Linear Algebra done right,

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The map $T(x,y,z) = (2x+y,5y+3z,8z)$ has a matrix with respect to the standard basis given by $T(1,0,0) = (2,0,0)$ , $T(0,1,0)=(1,5,0)$ , $T(0,0,1)=(0,3,8)$

And we are trying to find a diagonal matrix of T, it is done by finding the eigenvectors of the eigenvalues $2,5,8$ , the matrix is found to be

$\begin{pmatrix} 2 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 8 \end{pmatrix}$

What I'm confused about is, why when we substitute this new eigenvector basis in T, we don't get the diagonal matrix?

$T(1,0,0) = (2,0,0)$

$T(1,3,0) = (5,15,0)$

$T(1,6,6)= (8,48,48)$

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    $\begingroup$ If you somehow got $T(1,3,0)=(0, a, 0)$ then $T(1,3,0) \neq \alpha (1,3,0)$ and $(1,3,0)$ wouldn't even be an eigenvector. $\endgroup$ – Yagger Aug 23 '19 at 7:44
  • $\begingroup$ Yes that makes sense $\endgroup$ – khaled014z Aug 23 '19 at 7:47
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You DO get diagonal matrix: $$ T(1, 0,0) = (2,0,0) = 2(1, 0,0)\\ T(1, 3, 0) = (5, 15, 0) = 5(1, 3, 0)\\ T(1, 6, 6) = (8, 48, 48) = 8(1, 6, 6) $$ In other words, if $a, b, c$ are the three vectors given, then $Ta = 2a, Tb = 5b$ and $Tc = 5c$. This is exactly the same as saying that in the basis $a, b, c$, the linear transformation $T$ is represented by the matrix $$ \begin{pmatrix} 2&0&0\\0&5&0\\0&0&8 \end{pmatrix} $$ It is a general fact that given a linear transformation $S$ and a basis $B = (\vec b_1, \vec b_2, \vec b_3)$, the columns in the matrix representation of $S$ in the basis $B$ are the vectors $S\vec b_1, S\vec b_2$ and $S\vec b_3$, expressed in $B$.

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  • $\begingroup$ Oh I get it now! Thanks $\endgroup$ – khaled014z Aug 23 '19 at 7:47

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