I practice some exercises in linear algebra and suddendly I have to compute a orthonormal basis for the subspace $\mathbb{M}_{2,2}$ of the following matrices given below. $$V_1=\begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix},\;\;\; V_2=\begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix} \;\;\;\text{ and }\;\;\; V_3=\begin{bmatrix} 0 & 1\\ 0 & 1 \end{bmatrix}.$$ Here is my approach. I want to compute $V_1'$, $V_2'$, $V_3'$ as I would do when using this method for vectors. $$V_1=V_1'=\begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix}.$$ Second, $$V_2'=V_2-\frac{\langle V_2,V_1' \rangle}{||V_1'||^2}V_1'=\begin{bmatrix} 1/2 & 1\\ -1/2 & 0 \end{bmatrix}.$$ Third is $$V_3'=V_3-\frac{\langle V_3,V_1' \rangle}{||V_1'||^2}V_1'-\frac{\langle V_3,V_2' \rangle}{||V_2'||^2}V_2'=\begin{bmatrix} -1/2 & 0\\ 1/2 & 1 \end{bmatrix}.$$ I will now normalize them $$\tilde{V_1}=\frac{V_1}{||V_1||}=\begin{bmatrix} 1 & 0\\ 1 & 0 \end{bmatrix},\;\;\; \tilde{V_2}=\frac{V_2}{||V_2||}=\begin{bmatrix} 1/3 & 2/3\\ -1/3 & 0 \end{bmatrix} \;\;\; \text{ and }\;\;\; \tilde{V_3}=\frac{V_3}{||V_3||}=\begin{bmatrix} -1/3 & 0\\ 1/3 & 2/3 \end{bmatrix}$$ Therefore my orthonormal basis spanned by $V_1,V_2,V_3$ is $span\{\tilde{V_1},\tilde{V_2},\tilde{V_3}\}$.
I am not sure if my method are correct because I have never tried to use this method on matrices but just on vectors. Anyway if it's wrong, please correct me. Thanks
By using the suggestion in comment, I end with the following, $$\tilde{V_1}=\frac{V_1'}{||V_1'||}=\begin{bmatrix} \frac{\sqrt{2}}{2} & 0\\ \frac{\sqrt{2}}{2} & 0 \end{bmatrix},\;\;\; \tilde{V_2}=\frac{V_2'}{||V_2'||}=\begin{bmatrix} \frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{3}\\ -\frac{\sqrt{6}}{6} & 0 \end{bmatrix} \;\;\; \text{ and }\;\;\; \tilde{V_3}=\frac{V_3'}{||V_3'||}=\begin{bmatrix} -\frac{\sqrt{6}}{6} & 0\\ \frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{3} \end{bmatrix}$$ however computing the norm of, let's say $\tilde{V_1}$ in Maple I get something that is not equal to 1. But when normalize things I should end with something that has a norm equal to 1??