I am trying to prove that $A{A}^{+}{\overrightarrow{a}}_{i}^{\phantom{\rule{thinmathspace}{0ex}}}={\overrightarrow{a}}_{i}^{\phantom{\rule{thinmathspace}{0ex}}}$ , $\mathrm{\forall}i=\mathrm{1..}n$ is equal to the more popular version $A{A}^{+}A=A$. I started by setting up a system of equations as follows:

$\{\begin{array}{l}A{A}^{+}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}={\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ A{A}^{+}{\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}={\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}$

Then I turned it into a matrix form:

$\left[\begin{array}{c}A{A}^{+}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ A{A}^{+}{\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]=\left[\begin{array}{c}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ {\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]$

$A{A}^{+}\left[\begin{array}{c}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ {\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]=\left[\begin{array}{c}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ {\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]$

Since ${\overrightarrow{a}}_{i}^{\phantom{\rule{thinmathspace}{0ex}}}$ is a column vector it has shape (m x 1) which seems to produce weird vector in the last equation. Is this part of a "proof" any good or is it a bad idea from the begining?

$\{\begin{array}{l}A{A}^{+}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}={\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ A{A}^{+}{\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}={\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}$

Then I turned it into a matrix form:

$\left[\begin{array}{c}A{A}^{+}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ A{A}^{+}{\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]=\left[\begin{array}{c}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ {\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]$

$A{A}^{+}\left[\begin{array}{c}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ {\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]=\left[\begin{array}{c}{\overrightarrow{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ \vdots \\ {\overrightarrow{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]$

Since ${\overrightarrow{a}}_{i}^{\phantom{\rule{thinmathspace}{0ex}}}$ is a column vector it has shape (m x 1) which seems to produce weird vector in the last equation. Is this part of a "proof" any good or is it a bad idea from the begining?