I have the following linear transformation
$$L:\Bbb R^3\rightarrow \Bbb R^2, (x_1,x_2,x_3)\mapsto(x_3+x_1,x_2-x_1)$$
And I want to determine the kernel and the image of $L$.
$\text{ker}(L):=v\in V \space | \space L(v)=0$
Is it accurate to say that I want to find the set of vectors that will map to the zero-vector when plugged into the linear transformation?
In matrix form:
$$\begin{pmatrix}1&0&1\\-1&1&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies x_1+x_3=0 \iff x_1=-x_3 \space \space \space \text{and} \space \space \space x_2-x_1=0 \iff x_2=x_1$$
Therefore:
$$ker(L)=\begin{pmatrix}t\\t\\-t\end{pmatrix}$$
Is this correct?
What is the image of a linear transformation? Is it the subspace of the co-domain that the linear transformation actually maps to? So for example, if I had the transformation:
$M:\Bbb R^2 \rightarrow \Bbb R, (x_1,x_2) \mapsto (2)$
The Image of $Q$ would be {$2$}? How do I determine the image for much more complicated linear transformations?
Best Answer
As soon as you have the matrix of a linear map $f$, you have a system of generators of the image of $f$.
Indeed the column-vectors of the matrix are the images of the vectors of the basis in the source-space, say $v_1,v_2,v_3$. Hence for any vector $v$ in $\mathbf R^3$, $v=\lambda_1 v_1+\lambda_2v_2+\lambda_3v_3$, we have: $$f(v)=\lambda_1 f(v_1)+\lambda_2f(v_2)+\lambda_3f(v_3).$$
Naturally, this system of generators is not minimal, i. e. it is not a basis of the image in general. But from this system, you can deduce a basis by column reduction.
It will not be necessary here, because the rank-nullity theorem ensures $\dim\operatorname{Im}f=2$ since you've found that $\dim\ker f=1$. Thus, $f$ is surjective, i. e. the image of $f$ is the whole of $\mathbf R^2$.