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
$$A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}$$

$$A^{T}A$$ 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 $$\lambda \in R^{k}$$, [tex]\alpha , r \in R^{n}[/tex].

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.