# Let $B_1$ be some basis of $\mathbb{R^3}$ and $B_2$ be the ONB produced by $B_1$ after the Gram-schmidt process

Let $B_1=\{v_1,v_2,v_3\}$ be some basis of $\mathbb{R^3}$ and $B_2=\{o_1,o_2,o_3\}$ be the orthonormal basis produced by $B_1$ after the Gram-schmidt process. Is it possible that the change of basis matrix from $B_1$ to $B_2$ satisfies $$P = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 3 & 1 \\ 2 & 0 & 0 \\ \end{pmatrix}$$

Hey everyone. First, i am aware of this post, but I’m not sure the answer given there is correct, since the user does not acknowledge the fact that $B_2$ is a specific basis, obtained by excecuting the Gram Schmidt process on $B_1$ (which in this case is not orthonormal, else $P$ must equal the identity matrix).

What I’ve tried:

$P=[Id]^{B_1}_{B_2}= \begin{pmatrix} | & | & | \\ [v_1]_{B_2} & [v_2]_{B_2} & [v_3]_{B_3} \\ | & | & | \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 3 & 1 \\ 2 & 0 & 0 \\ \end{pmatrix}$

and since $B_2$ is an orthonormal basis, we have $P_{ij}=\langle v_j,o_i \rangle$ for every $1\le i,j \le 3$. Therefore $\langle v_2, o_1\rangle =0, \langle v_2,o_3 \rangle =0$ and $\langle v_3,o_3\rangle =0$

Since $B_2$ is produced by $B_1$ we get; Without loss of generality we can assume $o_1=v_1$ (And then we can normalize each vector in the new basis after we finish constructing the orthogonal basis). Hence, $o_2=v_2-\frac{\langle v_2,o_1\rangle o_1}{\|o_1\|^2}=v_2-0=v_2 \implies \langle v_1,v_2\rangle =0.$ But $\langle v_1,v_2\rangle =0 \iff \langle [v_1]_{B_2},[v_2]_{B_2}\rangle =0$ since the coordinate map is an isomorphism of an inner product. But we can see $\langle [v_i]_{B_2},[v_j]_{B_2}\rangle \neq 0$ for all $1\le i,j\le 3$ since the columns of $P$ are not pairwise orthogonal. Thus, $P$ cannot be the given matrix.

Having said that, I’m still not sure this is the right direction.

I would be happy to hear your thoughts, thanks :)