How can it be that those vectors form a basis?

Let $X$ and $Y$ be $m \times n$ matrices related by a linear transformation $P$ such that $Y=PX$. Is it true that rows of $P$ are a set of new basis vectors for expressing the columns of $X$? I find this statement quite weird, because obviously not always the rows of a $m \times n$ matrix form a basis. Source, page 3.

• What do you mean "related by a lin. trans."? – Timbuc Mar 22 '15 at 12:20
• It's worth noting that this comes from a paper called "A Tutorial on Principal Component Analysis" and that this particular step of the explanation has already been cast as a change of basis for an $m$-dimensional data set. This restricts the kinds of transformation that might be considered. – David K Mar 22 '15 at 13:03

$P$ is an $m\times m$ invertible matrix whose rows (columns) form a basis in $\mathbb{R}^m$ due to their linear independence.
• How do you know $P$ is invertible? It can just be any square matrix. Not all square matrices are invertible I guess. – user4205580 Mar 22 '15 at 12:31
• A linear transformation is always assumed invertible. If that is not the case then there is no one-to-one correspondence between $X$ and $Y$. – RTJ Mar 22 '15 at 12:33
• So ideally it should be in the text that $P$ is invertible. – user4205580 Mar 22 '15 at 12:43
• In section V rows of $P$ are made of eigenvectors of covariance matrix of $X$. Theorem 4 in appendix A proves these eigenvectors are orthogonal. If we have a matrix with rows made of orthogonal eigenvectors, is it necessarily an orthogonal matrix (definition says both rows and columns should be orthogonal). And therefore our matrix is orthogonal thus invertible. – user4205580 Mar 22 '15 at 13:12