Q:

Change of basis matrix from a basis to a orthonormal equivalent basis

Let $B_1 = \{v_1, v_2, v_3\} $ be a basis for $\mathbb R^3$
Let $B_2 = \{o_1, o_2, o_3\}$ be an orthonormal basis after executing the Gram-Schmidt algorithm on $B_1$
Let $$ P =
    \begin{pmatrix}
    1 & 0 & 1 \\
    2 & 3 & 1 \\
    2 & 0 & 0 \\
    \end{pmatrix}
$$
Can $P$ be the change of basis matrix from $B_1$ to $B_2$?
It feels a bit tricky and i was not able to figure out how to approach this question... any help much appreciated!

A:

Yes, this is possible. Firstly notice that $P$ is invertible, so the columns are linearly independent. This is necessary. Let us try to construct an example of where this is the case. Using the standard `change of base diagram' with the identity transformation we can obtain the following formula, for all vectors $x \in \mathbb{R}^3$:
$$
Id_{B_1}^{B_2} \cdot co_{B_1}(x) = co_{B_2}(x). 
$$
If we now use $x = v_1, x=v_2, x=v_3$ respectively, and using the matrix $P = Id_{B_1}^{B_2}$ that you provided, we obtain:
\begin{align*}
Id_{B_1}^{B_2} \cdot e_1 = co_{B_2}(v_1) = [1,2,2]^\top \\
Id_{B_1}^{B_2} \cdot e_2 = co_{B_2}(v_2) = [0,3,0]^\top \\
Id_{B_1}^{B_2} \cdot e_3 = co_{B_2}(v_3) = [1,1,0]^\top \\
\end{align*}
Thus, this gives 
$$
\begin{cases}
v_1 = o_1 + 2o_2 + 2o_3 \\
v_2 = 3o_2 \\
v_3 = o_1+o_2. 
\end{cases}
$$
Now pick any orthonormal basis, for example pick $\{o_1,o_2,o_3\} = \{e_1,e_2,e_3\}$. Then according to the formulas we obtain 
$$
v_1 = [1,2,2]^\top, v_2 = [0,3,0]^\top, v_3 = [1,1,0]^\top. 
$$
You can check that this is a basis of $\mathbb{R}^3$ and that the matrix for change of base from $\{v_1,v_2,v_3\}$ to $\{o_1,o_2,o_3\}$ is given by $P$. As an example consider the vector 
$$
x = [3,14,4]^\top
$$
This has $co_{B_1}(x) = [2,3,1]^\top$. Now we multiply with $P$, this indeed gives 
$$
P \cdot co_{B_1}(x) = [3,14,4]^\top = co_{B_2}(x)
$$

