Proving change of basis matrix from $\mathcal B$ to $\mathcal B'$ is $\mathbf P$=$(\mathbf Y^T \mathbf Y)^{-1}\mathbf Y^T\mathbf X$

Question

If $\mathcal B=${${x_1,x_2,...,x_n}$} and $\mathcal B'=${${y_1,y_2,...,y_n}$} are bases for an $n$-dimensional subspace of $\mathcal V \subset R^m$.

Say I let $X_{mxn}$ and $Y_{mxn}$ be the matrices whose columns are the vectors $\mathcal B$ and $\mathcal B'$ respectively.

How would I go about proving that the change of basis matrix from $\mathcal B$ to $\mathcal B'$ is $\mathbf P$=$(\mathbf Y^T \mathbf Y)^{-1}\mathbf Y^T\mathbf X$?

If I'm not mistaken, I would first have to show that $\mathbf Y^T \mathbf Y$ is indeed invertible? How do I go about doing this?

score 1 · Accepted Answer · answered Apr 08 '18 at 09:20

1

Note that $A^TA$ is invertible if A has independent columns, then for $v\in\mathcal V$ we have $$v=Yy=Xx\implies Y^TYy=Y^TXx\implies y=(Y^TY)^{-1}Y^TXx$$

answered Apr 08 '18 at 09:20

user

154,566

@MarissaB Yes of course! – user Apr 08 '18 at 11:08
And in this case the columns have to be independent because $A$'s columns are bases? But, why is $Yy=Xx$? – MarissaB Apr 08 '18 at 11:08
Just another thought, but would there be another particular outcome of $P$ when $m=n$? – MarissaB Apr 09 '18 at 02:11

Proving change of basis matrix from $\mathcal B$ to $\mathcal B'$ is $\mathbf P$=$(\mathbf Y^T \mathbf Y)^{-1}\mathbf Y^T\mathbf X$

1 Answers1