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.

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- Oct 21st 2009, 09:57 PMGreenDay14Finding 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. - Oct 21st 2009, 11:44 PMaman_cc
- Oct 22nd 2009, 04:52 AMHallsofIvy
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. - Oct 22nd 2009, 10:57 AMGreenDay14
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.

- Oct 22nd 2009, 11:07 AMaman_cc
- Oct 22nd 2009, 04:53 PMGreenDay14
I am very lost in this question lol.