# I am trying to prove that AA+vec(a)_i=vec(a)_i, AA i=1..n is equal to the more popular version AA^+A=A

I am trying to prove that $A{A}^{+}{\stackrel{\to }{a}}_{i}^{\phantom{\rule{thinmathspace}{0ex}}}={\stackrel{\to }{a}}_{i}^{\phantom{\rule{thinmathspace}{0ex}}}$ , $\mathrm{\forall }i=1..n$ is equal to the more popular version $A{A}^{+}A=A$. I started by setting up a system of equations as follows:
$\left\{\begin{array}{l}A{A}^{+}{\stackrel{\to }{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}={\stackrel{\to }{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ ⋮\\ A{A}^{+}{\stackrel{\to }{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}={\stackrel{\to }{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}$
Then I turned it into a matrix form:
$\left[\begin{array}{c}A{A}^{+}{\stackrel{\to }{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ ⋮\\ A{A}^{+}{\stackrel{\to }{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]=\left[\begin{array}{c}{\stackrel{\to }{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ ⋮\\ {\stackrel{\to }{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]$
$A{A}^{+}\left[\begin{array}{c}{\stackrel{\to }{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ ⋮\\ {\stackrel{\to }{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]=\left[\begin{array}{c}{\stackrel{\to }{a}}_{1}^{\phantom{\rule{thinmathspace}{0ex}}}\\ ⋮\\ {\stackrel{\to }{a}}_{n}^{\phantom{\rule{thinmathspace}{0ex}}}\end{array}\right]$
Since ${\stackrel{\to }{a}}_{i}^{\phantom{\rule{thinmathspace}{0ex}}}$ 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?
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Paige Paul
I suppose ${\stackrel{\to }{a}}_{j}$ means the j-th column of A. Then A is equal to $\left[\begin{array}{ccc}{\stackrel{\to }{a}}_{1}& \cdots & {\stackrel{\to }{a}}_{n}\end{array}\right]$. So, the correct identity should be
$\left[\begin{array}{ccc}A{A}^{+}{\stackrel{\to }{a}}_{1}& \cdots & A{A}^{+}{\stackrel{\to }{a}}_{n}\end{array}\right]=A{A}^{+}\left[\begin{array}{ccc}{\stackrel{\to }{a}}_{1}& \cdots & {\stackrel{\to }{a}}_{n}\end{array}\right]=A{A}^{+}A=A=\left[\begin{array}{ccc}{\stackrel{\to }{a}}_{1}& \cdots & {\stackrel{\to }{a}}_{n}\end{array}\right].$
The matrix-vector product $A{A}^{+}\left[\begin{array}{c}{\stackrel{\to }{a}}_{1}\\ ⋮\\ {\stackrel{\to }{a}}_{n}\end{array}\right]$ does not make sense because the size of $A{A}^{+}$ (which is, for example, $n×n$ when A is $n×n$) does not match the size of the the column vector $\left[\begin{array}{c}{\stackrel{\to }{a}}_{1}\\ ⋮\\ {\stackrel{\to }{a}}_{n}\end{array}\right]$ (which is ${n}^{2}×1$ in this example).