# Thread: Finding a basis.

1. ## Finding a basis.

I have a question here that asks to find a basis, which states:

Find a basis for V, where V = {T belongs to L(R²,R³)| T(1,2) = (0,0)}

Could anyone offer any ideas on how to tackle this question? I appreciate any help. Thanks.

2. Originally Posted by GreenDay14
I have a question here that asks to find a basis, which states:

Find a basis for V, where V = {T belongs to L(R²,R³)| T(1,2) = (0,0)}

Could anyone offer any ideas on how to tackle this question? I appreciate any help. Thanks.
Are you sure you got the complete question?

3. Actually, the problem is not that there is not enough information but that what you give is contradictory. You say T is from $\displaystyle R^2$ to $\displaystyle R^3$ but then you say T((1, 2))= (0, 0) which is a transformation from $\displaystyle R^2$ to $\displaystyle R^3$.

Is the subspace the set of all linear transformation from $\displaystyle R^2$ to $\displaystyle R^3$ such that T((1,2))= (0,0,0)? Or did you mean that T is from $\displaystyle R^2$ to $\displaystyle R^2$?

Assuming you meant the former, any linear transformation from $\displaystyle R^2$ to $\displaystyle R^3$ can be written as a matrix:
$\displaystyle \begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}$
and your condition is
$\displaystyle \begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}\begin{bmatrix}1 \\ 2\end{bmatrix}= \begin{bmatrix}a+ 2b \\ c+ 2d \\ e+ 2f\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
so you must have a+ 2b= 0, c+ 2d= 0, and e+ 2f= 0. Those, of course, give a= -2b, c= -2d, and e= -2f. The matrix can be written
$\displaystyle \begin{bmatrix}-2b & b \\ -2d & d \\ -2f & f\end{bmatrix}= b\begin{bmatrix}-2 & 1 \\ 0 & 0 \\ 0 & 0\end{bmatrix}+ d\begin{bmatrix}0 & 0 \\ -2 & 1 \\ 0 & 0\end{bmatrix}+ f\begin{bmatrix}0 & 0 \\ 0 & 0 \\ -2 & 1\end{bmatrix}$.

If you mean $\displaystyle R^2$ to $\displaystyle R^2$ just follow the same ideas.

4. aghh, thanks both of your for the help and I aplogize. Yes I did make a mistake, i meant for it to be T is from R² to R². Sorry again.

5. Originally Posted by HallsofIvy
Actually, the problem is not that there is not enough information but that what you give is contradictory. You say T is from $\displaystyle R^2$ to $\displaystyle R^3$ but then you say T((1, 2))= (0, 0) which is a transformation from $\displaystyle R^2$ to $\displaystyle R^3$.

Is the subspace the set of all linear transformation from $\displaystyle R^2$ to $\displaystyle R^3$ such that T((1,2))= (0,0,0)? Or did you mean that T is from $\displaystyle R^2$ to $\displaystyle R^2$?

Assuming you meant the former, any linear transformation from $\displaystyle R^2$ to $\displaystyle R^3$ can be written as a matrix:
$\displaystyle \begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}$
and your condition is
$\displaystyle \begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}\begin{bmatrix}1 \\ 2\end{bmatrix}= \begin{bmatrix}a+ 2b \\ c+ 2d \\ e+ 2f\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
so you must have a+ 2b= 0, c+ 2d= 0, and e+ 2f= 0. Those, of course, give a= -2b, c= -2d, and e= -2f. The matrix can be written
$\displaystyle \begin{bmatrix}-2b & b \\ -2d & d \\ -2f & f\end{bmatrix}= b\begin{bmatrix}-2 & 1 \\ 0 & 0 \\ 0 & 0\end{bmatrix}+ d\begin{bmatrix}0 & 0 \\ -2 & 1 \\ 0 & 0\end{bmatrix}+ f\begin{bmatrix}0 & 0 \\ 0 & 0 \\ -2 & 1\end{bmatrix}$.

If you mean $\displaystyle R^2$ to $\displaystyle R^2$ just follow the same ideas.
@HallsofIvy - Thanks for the explanation. In the case you have illustrated, dimension of T such that T((1,2))= (0,0,0) is obviously 3. You in fact have shown us a basis. Is there an easy way to compute the dimension without actually solving equations?

6. I am very lost in this question lol.