1. ## invertible matrix

Let A be an nxn matrix (not assumed to be invertible). Show that there exists an invertible nxn matrix P such that
APA = A

if such matrix existed then wouldn't it mean PA = I ( nxn identity matrix) and
AP = I hence A is invertible.
but anyways here are my ideas so far on this.
I am guessing P would be an elementary matrix (i.e. the ones concerned with row / column operations) if you left multiple by these elementary matrices you are doing row operations and if you right multiply you are doing column operations.
any sort of helpful hints would be much appreciated.

2. Originally Posted by bubble86
Let A be an nxn matrix (not assumed to be invertible). Show that there exists an invertible nxn matrix P such that
APA = A

if such matrix existed then wouldn't it mean PA = I ( nxn identity matrix) and
AP = I hence A is invertible.
but anyways here are my ideas so far on this.
I am guessing P would be an elementary matrix (i.e. the ones concerned with row / column operations) if you left multiple by these elementary matrices you are doing row operations and if you right multiply you are doing column operations.
any sort of helpful hints would be much appreciated.
This problem is much easier if you think in terms of linear transformations rather than matrices.

Let $\{f_1, f_2,\ldots,f_r\}$ be a basis for the range of A, where $f_i = Ae_i\;\,(1\leqslant i\leqslant r)$. Then we can extend $\{e_1,\ldots,e_r\}$ to a basis of $\mathbb{R}^n$ by adding vectors $e_{r+1},\ldots,e_n$ say, and we can extend $\{f_1,\ldots,f_r\}$ to a basis of $\mathbb{R}^n$ by adding vectors $f_{r+1},\ldots,f_n$. Now define P to be (the matrix of) the linear transformation defined by $Pf_i = e_i\;\,(1\leqslant i\leqslant n)$. Then P is invertible, and APA = A.