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Let $f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}, f(x) = Ax$ where matrix $$A= \begin{pmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{pmatrix}$$

Calculate the dimensions and basis of the kernel $\text{Ker}(f)$ and of the image $\text{Im}(f)$.

I tried this:

$$\text{Ker(f)}= \begin{pmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{pmatrix} \cdot x=\begin{pmatrix} 0\\ 0 \end{pmatrix} \Leftrightarrow$$

$$ \Leftrightarrow\begin{pmatrix} 0\\ 0 \end{pmatrix}= \begin{pmatrix} 1\\ 4 \end{pmatrix}x + \begin{pmatrix} 2\\ 5 \end{pmatrix}+ \begin{pmatrix} 3\\ 6 \end{pmatrix}$$

So kernel has 3 basis and 2 dimensions?

I don't know how to do it with the image and I'm not sure if this is correct at all. Please can you tell me if this is right and if not how to do it correctly? It's no homework, I took this task to understand it for more complicated tasks in exam.

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1 Answer 1

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Let $x=(x_1,x_2,x_3)^\top\in\ker(f)$ so $Ax=0$ i.e.

$$x_1+2x_2+3x_3=0\;(1)$$ and $$4x_1+5x_2+6x_3=0\;(2)$$ Now $4(1)-(2)$ gives $3x_2+6x_3=0\iff x_2=-2x_3$ and from $(1)$ we get $x_1=x_3$ hence $$x=(x_3,-2x_3,x_3)^\top=x_3(1,-2,1)^\top=x_3v\in \operatorname{span}(v)=\ker(f)$$ Finally by the rank-nullity theorem we get $$\dim(\operatorname{Im}(f))=\dim\Bbb R^3-\dim\ker f=3-1=2=\dim\Bbb R^2$$ which means that $\operatorname{Im}(f)=\Bbb R^2$.

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