reducing invertible matrix in an equation of matrix

Assume that $A^\mathrm{T}A$ is invertible and that $A$ is $3\times 3$, $B$ is $3\times 1$, $\alpha$ is $3\times 1$. How does the expression above reduce?

just to $A^\mathrm{T}B$ ?

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The information you've given is contradictory: $A$ cannot be both invertible and non-square. – Hurkyl Oct 12 '12 at 22:41
A $\,3\times 1\,$ matrix (or vector, as you wish to look at it) – DonAntonio Oct 12 '12 at 22:42
@Hurkyl We can, however, assume that the $3\times 3$ matrix $A^TA$ is invertible (which is what we're inverting anyways). Unless I've forgotten some important theorem. – Arthur Oct 12 '12 at 22:47
If $A$ is rectangular, it doesn't reduce further. – Ganesh Oct 12 '12 at 22:50
hmm you guys are right, that makes no sense. If we assume that ATA is invertible, then what does this reduce to? – mugetsu Oct 12 '12 at 23:13

If $A$ is square and $A^tA$ is invertible then of course $A$ is invertible, and your expression simplifies to $A^{-1}B$.

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could you explain the steps for me? thanks – mugetsu Oct 14 '12 at 3:25
You know that $(XY)^{-1}=Y^{-1}X^{-1}$? Apply that to $(A^tA)^{-1}$. – Gerry Myerson Oct 14 '12 at 4:18

The expression itself does not reduce (though there are more complicated methods that are efficient at solving it). I believe your equation may be based on the desire to solve $$A\vec{\alpha}=B$$ given $A$ and $B$, with $A$ underdetermined, meaning it has fewer columns than rows, which is consistent with your problem statement of $5$ rows and $3$ columns for $A$. This equation has no exact solution, thus the next best thing is sought.

As there is no exact solution, consider that a solution that is the "closest" of all possible. With the notion of "close" as the Euclidean distance, it becomes what is called a Least Squares equation. To give you the very abbreviated version of Least Squares, left multiply both sides of the equation to get $$A^TA\vec{\alpha}=A^TB$$ If you are not yet familiar with Least Squares, for now know that the reason for this is to have something invertible to work with ($A^TA$ will have an inverse) and to keep things in terms of the range of $A$. Thus, to solve it, again left multiply both sides of the equation to get $$(A^TA)^{-1}A^TA\vec{\alpha}=(A^TA)^{-1}A^TB$$ which reduces to $$\vec{\alpha}=(A^TA)^{-1}A^TB$$

This equation then has a solution, and it can be shown to be the closest solution of $\vec{\alpha}$ to the original equation $A\vec{\alpha}=B$

Back again to your original question if it is reduceable; the answer is no-- though there are ways of going about solving this without calculating every middle step as I have shown here, and that would take quite a bit more explaining than I have done thus far.

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thanks for the help. I think i misinterpreted the problem and now I'm going with A as a 3x3 matrix – mugetsu Oct 13 '12 at 0:08
The only way for the equation to reduce is if $A$ actually is invertible, then it would reduce to $\alpha=A^{-1}B$ – adam W Oct 13 '12 at 0:29