# Matrix with respect to other basis

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

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?

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

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

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

• 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. – Yagger Aug 23 '19 at 7:44
• Yes that makes sense – khaled014z Aug 23 '19 at 7:47

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 $$B$$.