Show that the range of the 2 matrices are the same

[charlies1902]

Homework Statement

$P=A(A^*A)^{-1}A^*$
where A is a mxn real/complex matrix and $A^*A$ is invertible.
$A^*$ means the conjugate transpose of A.

Homework Equations

The Attempt at a Solution

Let y be in the range(A), such that
$y = Ax$ for some $x$.
We can see that $PA = A(A^*A)^{-1}A^*A = A$
Then
$y = PAx = P(Ax)$
Does this expression above alone show that y is also in the range(P)?

 

[andrewkirk]

Let y be in the range(A), such that
$y = Ax$ for some $x$.
We can see that $PA = A(A^*A)^{-1}A^*A = A$
Then
$y = PAx = P(Ax)$

That last expression is not correct.
It should be $Py = PAx = P(Ax)$.

 

[charlies1902]

That last expression is not correct.
It should be $Py = PAx = P(Ax)$.

But previous I had shown $PA=A$
so $y=Ax=PAx=P(Ax)$
Why is that not correct?