# 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

## 2 Answers

$P$ 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
• Yes if you look in the main paper this property is used e.g. in section V. – RTJ Mar 22 '15 at 12:55
• 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

yes,the rows of matrix are not always form the basis,but only when you consider the rows of matrix as basis,you can understand the multiplication of matrics better,namely,under this circumstance,we will ignore wheather those basis are linearly dependent or not