TITLE: Calculate kernel space of matrix map and its orthonormal space
QUESTION [5 upvotes]: The question asks:
a) What is kernel space of linear map defined by  $$
M =
\begin{bmatrix}
1 & 2 & 3 \\
2 & 4 & 6 \\
3 & 6 & 9 \\
\end{bmatrix}
$$
b) Give orthonormal basis for map's kernel from a)
My attempt of a)

I attempted to find the kernel space, but I am not sure if I am right, furthermore, I don't know how to approach b). Thank you for any help

REPLY [3 votes]: So since ${\rm ker}M=\left\{\begin{bmatrix}-2y-3z\\ y\\ z\end{bmatrix}\in \mathbb{R}^{3}: y,z\in \mathbb{R}\right\}$ hence a basis for ${\rm ker}M$ is given by $\beta=\left\{\begin{bmatrix}-2\\1\\0\end{bmatrix},\begin{bmatrix}-3\\0\\1\end{bmatrix}\right\}$. Define $v_{1}=\begin{bmatrix}-2\\1\\0\end{bmatrix}$ and $v_2=\begin{bmatrix}-3\\0\\1\end{bmatrix}$.
An orthonormal basis for ${\rm ker}M$ is given by $\beta^{*}=\{u_1,u_2\}$ where $u_1$ and $u_2$ is  obtained by Gram-Schmidt algorithm:

$\displaystyle u_1 =\frac{v_1}{||v_1||}$.

$\displaystyle u_2=\frac{v_{2}-\langle v_{2},u_{1}\rangle u_{1}}{||v_{2}-\langle v_{2},u_{1}\rangle u_{1}||}$.


Where $||\cdot||$ the norm induced by the inner product $\langle \cdot, \cdot \rangle$. If you haven't seen general inner products, then you may be working with the usual inner product of $\mathbb{R}^n$ which is just the dot product between vectors.
Using the dot product as the inner product we get $$\beta^{*}=\left\{\begin{bmatrix} -\frac{\sqrt{35}}{35}\\ \frac{\sqrt{35}}{7}\\-\frac{3\sqrt{35}}{35} \end{bmatrix}, \begin{bmatrix} -\frac{3\sqrt{10}}{10}\\0\\  \frac{\sqrt{10}}{10}\end{bmatrix} \right\}.$$
Recall the dot product in $\mathbb{R}^{3}$ is given by $$\left\langle \begin{bmatrix} a\\ b\\ c\end{bmatrix},\begin{bmatrix}x\\ y\\ z\end{bmatrix} \right\rangle=ax+by+cz$$ and the norm induced by the dot product is given by $$\left|\left|\begin{bmatrix} a\\b\\c\end{bmatrix}\right|\right|=\sqrt{a^{2}+b^2 + c^2 }.$$