Kernel and Image of 3 space

Question

nd the dimensions of $\operatorname{Im}(A)$ and $\ker(A)$, and find their bases for the linear transformation $A$ on $\Bbb R^{3}$ defined by $A(x, y, z) = (x − 2z, y + z, 0)$.

Answer 1

The kernel of $A$ is the subspace in $\mathbb{R}^3$ that gets mapped to the zero vector. So if $(x,y,z)\in\mathrm{ker}(A)$, it satisfies

$$\left\{\begin{array}{l} x-2z=0,\\ y+z=0. \end{array}\right.$$

These two equations define a one-dimensional subspace of $\mathbb{R}^3$ along the direction $(x,y,z)\propto$ $(2,-1,1)$. So we have

$$\mathrm{ker}(A)=\{(2t,-t,t)_{\,}|_{\,}t\in\mathbb{R}\}.$$

The image of $A$ is the range of values that $(x-2z,y+z,0)$ can take, with arbitrary $(x,y,z)\in\mathbb{R}^3$. So $\,x-2z\,$ and $\,y+z\,$ are independent and can take all numbers in $\mathbb{R}$. Therefore

$$\mathrm{Im}(A)=\{(u,v,0)_{\,}|_{\,}u,v\in\mathbb{R}\}.$$

This problem shows that the kernel and image need not be orthogonal complements in general. All we know is $\,\dim\,\mathrm{ker}(A)+\dim\,\mathrm{Im}(A)=3\,$ for a general linear transformation $A\,$ over $\,\mathbb{R}^3$, which is because $\,\mathbb{R}^3=\mathrm{ker}(A)\oplus\mathrm{Im}(A)\,$ can be written as their direct sum.