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Traveling without a paper visa, a series of foreigners crossed the US.. Ang cute ng ina mo torrent.Q:
How is $X^tX = A^tA$ equivalent to $A^tA = X^tX$?
How is it that $X^tX = A^tA$ is equivalent to $A^tA = X^tX$?
I have been trying to understand how this is true but I can't seem to wrap my head around the general concept.
I understand how to go from $X^tX = A^tA$ to $A^tA = X^tX$, so it's just a matter of computing an $X^t$ and then a $X$, but I can't seem to find an explanation of the converse.
A:
Consider the matrix $B = X^t$, or more generally $B = UXU^t$, where $U$ is an invertible matrix. Then $B^{ -1} A B = X$, by definition of the matrix inverse.
If $U$ is not invertible, then $B$ is not invertible either, but since $X=BU^{ -1}$, you can see why $B^{ -1} AB = X$ still holds.
The above is what you actually need to show. But let's give a second more tangible way, which is a bit more general.
Since
\begin{align*}
X^t X &= X^t U X U^t
\end{align*}
we have
\begin{align*}
(X^t X)^{ -1} &= X^{ -1} X^{ -t} = U^{ -t} U^{ -1}
\end{align*}
and hence
\begin{align*}
X^{ -1} (X^t X)^{ -1} &= U^{ -1}
\end{align*}
meaning that $X^{ -1}$ maps $X^t X$ to the identity. On the other hand, we have
\begin{align*}
(X^t X) X^{ -1} &= X X^{ -1 0b46394aab
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