# linear transformation and standard basis

We have a linear transformation $$R^3 \rightarrow R^3$$ where in a standard basis there is a matrix:

$$A =\begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 2 & 0 & 1 \end{bmatrix}$$

1.)What matrix does belong to this linear transformation in a basis:

$$B = \left\{\quad\begin{bmatrix} 0 \\ 1\\ 2 \end{bmatrix} ,\ \begin{bmatrix} 1 \\ 0\\ -1 \end{bmatrix} ,\ \begin{bmatrix} 1 \\ 1\\ 0 \end{bmatrix}\quad \right\}$$

My understanding of this:

1.The standard basis consists of vectors that are independent of each other. How can then a matrix be in standard basis. Does the matrix A consist of these vectors?

2.What does the $$R^3\rightarrow R^3$$ have to do with this problem , what if the question would be $$R^3\rightarrow R^2$$ or $$R^2\rightarrow R^3$$.

1. So a linear transformation is a linear mapping that maps the zero vector to a zero vector. I read all the theory behind it, but still cant figure out the initial problem
• Am I missing something or you are trying to prove that a linear transformation is indeed a linear transformation? – José Carlos Santos Mar 6 '20 at 17:39
• @JoséCarlosSantos Sorry that was a misunderstanding, I added that myself and now see that its pointless. I will remove this subquestion. But the initial question persists – Bili Debili Mar 6 '20 at 17:40
• Based on these questions I think the best thing that you could do is review the course material leading up to this problem. You’re asking about the very basics of representing linear transformations by matrices. – amd Mar 6 '20 at 18:04

To find the matrix representing a linear transformation in a given basis, apply the linear transformation to each basis vector in turn and write the result as a linear combination of the basis vectors. The coefficients in that linear combination form a column of the matr8ix.

Here, the first basis vector is $$\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}$$. A applied to that is $$\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}= \begin{bmatrix}1 \\ 3 \\ 2 \end{bmatrix}= 0\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}- 2\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}+ 3\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}$$.

So the first column of the new matrix is $$\begin{bmatrix} 0 \\ -2 \\ 3\end{bmatrix}$$.