1. ## linear transformation

Let $L: R^{2}->R^{2}$be the linear transformation defined by $L(\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix})=\begin {pmatrix}1&2\\2&4\end{pmatrix}\begin{pmatrix}a_{1} \\a_{2}\end{pmatrix}$
a) Is $\begin{pmatrix}1\\2\end{pmatrix}$ in ker L?
b) Is $\begin{pmatrix}2\\-1\end{pmatrix}$ in ker L?
c) Is $\begin{pmatrix}3\\6\end{pmatrix}$ in range L?
c) Is $\begin{pmatrix}2\\3\end{pmatrix}$ in range L?

2. Originally Posted by antman
Let $L: R^{2}->R^{2}$be the linear transformation defined by $L(\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix})=\begin {pmatrix}1&2\\2&4\end{pmatrix}\begin{pmatrix}a_{1} \\a_{2}\end{pmatrix}$
a) Is $\begin{pmatrix}1\\2\end{pmatrix}$ in ker L?
b) Is $\begin{pmatrix}2\\-1\end{pmatrix}$ in ker L?
c) Is $\begin{pmatrix}3\\6\end{pmatrix}$ in range L?
c) Is $\begin{pmatrix}2\\3\end{pmatrix}$ in range L?

Just compute $L \left[ \begin{pmatrix}1\\2\end{pmatrix} \right]$ and see if it is $\begin{pmatrix}0\\0\end{pmatrix}$ for part a. Then do the same for the other ones.

3. For range, I would just have to find a vector v such that L(v)=w, so w is in the range of L. So a is not in kernel L but b is and c is in range L but d is not? For the range, I set up a 2x3 matrix with the vector in question as column 3 then found rref. Is this correct? d would not be in range because in the reduced matrix -1 does not equal 0.

4. Originally Posted by antman
For range, I would just have to find a vector v such that L(v)=w, so w is in the range of L. So a is not in kernel L but b is and c is in range L but d is not? For the range, I set up a 2x3 matrix with the vector in question as column 3 then found rref. Is this correct? d would not be in range because in the reduced matrix -1 does not equal 0.
Notice that $\begin{pmatrix}1&2\\2&4\end{pmatrix}\begin{pmatrix }a_{1}\\a_{2}\end{pmatrix} = a_1 \begin{pmatrix}1\\2\end{pmatrix} + a_2 \begin{pmatrix}2\\4\end{pmatrix}$.

Thus, to be in the range it needs to be a linear combination of $\left\{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}2\\4\end{pmatrix} \right\}$.

There are of course many other ways to solve this problem, I just happen to like the approach that uses linear combinations. I looks like that approach uses least amount of computation since these are just $2\times 2$ matrices.

5. I understand that. Thank you! The way I did it is still correct then, just more work?

6. Originally Posted by antman
The way I did it is still correct then, just more work?
Yes, setting up the matrix and reducing it by Gauss-Jordan elimination to see if it gets solved will tell you how to find a v so that L(v) = w. If you are more comfortable doing it that way then do it that way.