# Is this a possible coordinate change matrix?

$$B_1=\{v_1,v_2,v_3\}$$ is a basis for $$\mathbb R^3$$.

$$B_2=\{u_1,u_2,u_3\}$$ is an orthonormal basis the we got after applying gram schmidt procces on $$B_1$$. is it possible that $$P= \begin{bmatrix} 1 & 0 & 1 \\ 2 & 3 & 1 \\ 2 & 0 & 0 \\ \end{bmatrix}$$ is the transition matrix fron $$B_1$$ to $$B_2$$ ? I don't know how to even approach the question. I tried to find $$P^{-1}= \begin{bmatrix} 0 & 0 & 0.5 \\ -1/3 & 1/3 & -1/6 \\ 1 & 0 & -0.5 \\ \end{bmatrix}$$ but also then I didnt get to any conclusion

If you do Gram-Schmidt on $$(v_1, v_2, v_3)$$ and get $$(u_1,u_2,u_3)$$, then $$u_1$$ will always end up as a multiple of $$v_1$$.

This means that $$P$$ must have the form $$\begin{bmatrix} * & * & * \\ 0 & * &* \\ 0 &*&* \end{bmatrix}$$ But it doesn't. So it cannot be the corresponding coordinate change.

An extension of this argument shows that the transition matrix will always be upper triangular.

These conditions may be relaxed if your description of the Gram-Schmidt process don't require processing the basis vectors in order, but merely says "pick one of them you haven't processed before". But even so, the first basis vector to be produced will always be parallel to one of the initial basis vectors, so your $$P$$ can still only be the transition matrix if you scrambled $$B_2$$ after producing it ...

• hi , so why in your answer its not upper triangular? Commented Jun 24, 2019 at 10:54
• @KIMKES1232 How do you know that it isn’t? Only the form of the first column has been specified, which is the only one that’s relevant for this particular problem.
– amd
Commented Jun 24, 2019 at 23:03