Find matrix of T with respect to the standard basis of R^3

Collisionman
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Homework Statement



For this whole question let T be a linear transformation from R^3 to R^3 with
T(1,0,0) = (2,2,2),

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

T(0,0,1) = (0,0,1).

(a) Find the image of (1,1,2009)
(b) Find the matrix of T with respect to the standard basis in R^3

Homework Equations



Standard basis of R^3 = (e1,e2,e3) = (1,0,0) (0,1,0) (0,0,1)

The Attempt at a Solution



I actually don't know where to start with this question.
 
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Collisionman said:

Homework Statement



For this whole question let T be a linear transformation from R^3 to R^3 with
T(1,0,0) = (2,2,2),

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

T(0,0,1) = (0,0,1).

(a) Find the image of (1,1,2009)
(b) Find the matrix of T with respect to the standard basis in R^3

Homework Equations



Standard basis of R^3 = (e1,e2,e3) = (1,0,0) (0,1,0) (0,0,1)


The Attempt at a Solution



I actually don't know where to start with this question.
Two of the most basic properties of a linear transformation are that T(u + v) = T(u) + T(v) and T(ku) = kT(u). Apply these properties to find T(1, 1, 2009).

For the b part, your text must have some examples of finding the matrix of a linear transformation in terms of a particular basis.
 
Hint: the vectors you wrote for the standard basis are exactly the same as the ones T is acting on in the problem statement.
 
Notice that if
A= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}
then
Ae_1= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d \\ g\end{bmatrix}

Ae_2= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}b \\ e \\ h\end{bmatrix}

Ae_2= \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}= \begin{bmatrix}c \\ f \\ i\end{bmatrix}

That is, the results of applying the transformation to the standard basis vectors are the columns of the matrix.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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