# Linear transformations - bases of kernel and image

Let $V$ be the vector space of all real $2 \times2$ matrices. Let $T: V \to \mathbb R^2$ be the map defined by $T(\left(\begin{array}{crc} a_{11} & a_{12}\\ a_{21} & a_{22} \\ \end{array}\right))=\left(\begin{array}{crc} 2a_{11}-a_{21} & 2a_{12}-a_{22}\\ \end{array}\right)$

Find bases for the kernel and image of $T$.

If the linear transformation was something like $T(x,y)=(2x-y,x+z)$ I'd be able to find bases for the kernel and image, but the format of the question seems different to me and I'm feeling very puzzled. Would someone mind showing me how to do this problem?

Edit: Also how would you find the matrix of $T$ with respect to the standard basis of $V$ and the standard basis of $\mathbb R^2$?

• A matrix $A$ is in the kernel if $T(A)=0$ i.e. $2a_{11}-a_{21}=2a_{12}-a_{22}=0$. – paf Aug 13 '16 at 13:47
• try to work with $T(w,x,y,z)=(2w-x,2y-z)$ – user190080 Aug 13 '16 at 13:54

$T$ is surjective, so a basis for the image is the standard basis of $\mathbb{R}^2$. $V$ is four-dimensional, so by the dimension theorem we know the kernel of $T$ must be of dimension $2$. So we only need two linearly independent matrices in $V$ that are linearly independent and map to zero to find a basis. A good choice is $$\begin{bmatrix}1&1\\2&2\end{bmatrix}$$ and $$\begin{bmatrix}0&1\\0&2\end{bmatrix}.$$ It is easily seen these are linearly independent. To find the matrix representation of $T$, just express the images of the standard basis of $V$ in standard coordinates in $\mathbb{R}^2$. For example, the first column of this matrix would be $$T(\begin{bmatrix} 1&0\\0&0 \end{bmatrix})=\begin{bmatrix} 1\\0 \end{bmatrix}.$$