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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?

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    $\begingroup$ Small detail: In your last example, $M$ is not linear ($M(\lambda X) \neq \lambda M(X)$) $\endgroup$ – Clement C. Jun 13 '15 at 14:25
  • $\begingroup$ @ClementC. You're right. I will fix that. $\endgroup$ – qmd Jun 13 '15 at 14:27
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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 $\mathbf R^2$.

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  • $\begingroup$ So in this case I have the matrix $ \begin{pmatrix}1&0&1\\-1&1&0\end{pmatrix}$ with columns $\begin{pmatrix}1\\1\end{pmatrix}$,$\begin{pmatrix}0\\1\end{pmatrix}$, $\begin{pmatrix}1\\0\end{pmatrix}$. Are these column-vectors the image of $L$? Is it always the case that the column-vectors are the image? $\endgroup$ – qmd Jun 13 '15 at 14:51
  • $\begingroup$ The column-vectors are not the image of $L$: they span the image of $L$. But $\begin{pmatrix}1\\-1\end{pmatrix}$ andt $\begin{pmatrix}0\\1\end{pmatrix}$ also generate the image. $\endgroup$ – Bernard Jun 13 '15 at 14:55
  • $\begingroup$ I see. So I have to look at that column vectors and calculate the span. In this case $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ span $\Bbb R^2$ and therefore Im($L$)=$\Bbb R^2$? $\endgroup$ – qmd Jun 13 '15 at 14:58
  • $\begingroup$ Yes. You also know that here without any calculation thanks to the rank-nullity theorem. $\endgroup$ – Bernard Jun 13 '15 at 15:16
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An option to characterize all $\vec{y}\in\mathrm{im}(L)$ is to express that there must be a $\vec{x}$ such that $L( \vec{x}) = \vec{y}$. That is, write $$ \begin{pmatrix}1&0&1\\-1&1&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix}y_1\\y_2\end{pmatrix} $$ and solve for $\begin{pmatrix}y_1\\y_2\end{pmatrix}$ as a function of of $x_1,x_2,x_3$. This will give you the set (when $x_1,x_2,x_3$ vary in $\mathbb{R}$ of all $\vec{y}$'s in the image.

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  • $\begingroup$ Thanks. Can you give me an intuitive explanation of what the image is? Is it, as I asked in my question, "the subspace of the co-domain that the linear transformation actually maps to?" or am I confusing something here? Solving the linear system in your answer I get: $x_1+x_3=y_1$ and $x_2-x_1=y_2$. I don't understand how that helps me. $\endgroup$ – qmd Jun 13 '15 at 14:43
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    $\begingroup$ The image is the set of all poin ts that are "reached" by the linear transformation. In your case, it's basically the set of all points that can be written $(a+b,c-a)$ for some $a,b,c\in\mathbb{R}$. (Also, one can show that here, this is actually the whole set $\mathbb{R}^2$: for any choice of $y_1,y_2$, you can solve the system for $x_1,x_2,x_3$ and find solutions: that is, any point in $\mathbb{R}^2$ is reached by $L$. $\endgroup$ – Clement C. Jun 13 '15 at 14:45

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