Solving for Surjective Matrix: A Possible Typo in the Theorem Statement?

[lauratyso11n]

I saw this in a book as a proposition but I think it's an error:

Assume that the (n-by-k) matrix, A, is surjective as a mapping,

A:R^{k}\rightarrow R^{n}.

For any y \in R^{n}, consider the optimization problem

min_{x \in R^{k}}\left{||x||^2\right}

such that Ax = y.

Then, the following hold:
(i) The transpose of A, call it A^{T} is injective.
(ii) The matrix A^{T}A is invertible.
(iii) etc etc etc...

I have a problem with point (ii), take as an example the (2-by-3) surjective matrix
A = \begin{pmatrix}<br /> 1 &amp; 0 &amp; 0\\<br /> 0 &amp; 1 &amp; 0<br /> \end{pmatrix}

A^{T}A in this case is not invertible.

What am I doing wrong ?

 

[Dick]

I don't think you are doing anything wrong. But I think there must be a typo in the theorem statement and they meant to write AA^(T) instead of A^(T)A.

 

[lauratyso11n]

I don't think you are doing anything wrong. But I think there must be a typo in the theorem statement and they meant to write AA^(T) instead of A^(T)A.

That would be fine, but the result is used to carry some analysis. The crux of it is:

\sigma\lambda = \alpha - \bf{r}

where \sigma \in R^{n-by-k} is surjective, and \lambda \in R^{k}, \alpha , <b>r</b> \in R^{n}.

How would you solve for \lambda ? Isn't is critical that the 'typo' has to be correct to be able to solve for this ?

The author's solution as you might expect is \lambda = \left(\sigma^{T}\sigma\right)^{-1}\sigma^{T}[\alpha - \bf{r}].

BTW, thanks for making the effort to look at the problem. Much appreciated.