# Show that Gram-Schmidt$(Y)$ = $B$Gram-Schmidt$(X)$ such that $B$ is an orthogonal matrix

Let \begin{align} V_{k,n} &= \Bigl \{ \begin{pmatrix} v_{1} \\ v_{2} \\ \vdots \\ v_{k} \end{pmatrix} : v_i\in \mathbb{R}^n\text{ and } \{v_1,\cdots,v_k\}\text{ is linearly independent}\Bigr \} \end{align}. We say $X\sim Y$ if and only if $\exists A\in GL_k(\mathbb{R}^n)$ such that $Y = AX$ for $X,Y\in V_{k,n}$. Let us denote Gram-Schmidt orthonormalization process as gs.

Now, I need to show that gs$(Y) = B$gs$(X)$, such that $B$ is an orthogonal matrix. I understand that $B$ is an orthogonal matrix as gs$(Y)$ and gs$(X)$ are orthogonal matrices. It is that I cannot determine the matrix $B$ explicitly.

I would like somebody to help me computing the matrix $B$ explicitly. Thanks in advance.

• If $U$ and $V$ are orthogonal matrices and $U = BV$ then $B = UV^{-1} = UV^T$. – Arin Chaudhuri Apr 25 '17 at 2:09