# Solving for a Surjective Matrix

1. Sep 10, 2010

### 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,

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

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

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

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.

Can anyone confirm that part (ii) of this Proposition is indeed incorrect ?

Last edited: Sep 10, 2010
2. Sep 10, 2010

### ThirstyDog

I would agree with you that part (ii) of the proposition is incorrect (unless matrices are not acting on vectors from the left).

If you look at bijection part
http://en.wikipedia.org/wiki/Bijection,_injection_and_surjection
it reads: "If g o f is a bijection, then it can only be concluded that f is injective and g is surjective."

Working right to left with matrices and composition of functions says if A^{T}A was invertible (i.e. a bijection) then A would be injective and A^{T} would be surjective. Thus something is wrong!

P.S. I didn't see the bit where it clearly said the matrices were acting from the left so I would say that it is definitely wrong.

3. Sep 10, 2010

### Landau

Yes, you are right. (i) is true but (ii) is false. But (ii) is true if A^TA is replaced by AA^T, so maybe it's a typo? I don't understand what "the optimization problem" has to do with this, or are the other parts of the proposition about this?

4. Sep 10, 2010

### lauratyso11n

The full Proposition is as follows:

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

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

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

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

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) The unique optimal solution of the minimum norm problem is given by
$$(A^TA)^{-1}A^Ty$$

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.

Last edited: Sep 10, 2010
5. Sep 10, 2010

### lauratyso11n

I don't think it's a typo as he uses this result to carry some other analysis further. The crux of it is:

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

where $$\sigma \in \mathbb{R}^{n-by-k}$$ is surjective, and $$\lambda \in \mathbb{R}^{k}$$, $$\alpha , \bf{r} \in \mathbb{R}^{n}$$.

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 $$\lambda = \left(\sigma^{T}\sigma\right)^{-1}\sigma^{T}\left[\alpha - \bf{r}\right]$$.

So it's either a HUGE mistake on his part or I'm missing something. The author is actually quite insightful, and this error would be quite out of character for him.

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

Last edited: Sep 10, 2010
6. Sep 10, 2010

### Office_Shredder

Staff Emeritus
Looking at dimension counting with your example, AtA is a 3x3 matrix, which means (AtA)-1 A wouldn't make sense even if the inverse was defined because the sizes of the matrices don't match up. On the other hand A At and A are compatible matrices which suggests that he just put the transpose on the wrong one and carried the error through.

7. Sep 10, 2010

### Landau

This is correct provided that $\sigma^{T}\sigma$ is invertible, i.e. provided that $\left(\sigma^{T}\sigma\right)^{-1}$ makes sense.

The least square solution of Ax=y satisfies the normal equation:
$$A^TAx=A^Ty.$$
This solution is unique if and only if $A^TA$ is invertible. In this case, it is given by
$$x=(A^TA)^{-1}A^Tb$$, just like the author asserts.

But $A^TA$ is not necessarily invertible, contrary to the proposition. Note that $A^TA:\mathbb{R}^k\to\mathbb{R}^k$ is bijective if and only if its rank equals k. But since $A^TA$ and A always have equal rank, this happens if and only if A has rank k. Since $A:\mathbb{R}^k\to\mathbb{R}^n$ is assumed to be surjective, it has rank n and we must have $k\geq n$. So in fact, $A^TA$ is invertible if and only if k=n! Indeed, in your counter-example k and n are not equal.
This is not the expression the author uses; it is (AtA)-1 At.

Last edited: Sep 10, 2010
8. Sep 10, 2010

### Landau

Perhaps you could post a link to the book?

9. Sep 10, 2010

### Office_Shredder

Staff Emeritus
I assumed the * was just referring to matrix multiplication

10. Sep 10, 2010

### Landau

I assumed * means adjoint, so (in this real case) the transpose. This is also what lauratyso11n writes in post #5 (where A is called sigma).

11. Sep 10, 2010

### lauratyso11n

SORRYYYYYY, the $$A^*$$ is actually an $$A^T$$. I've corrected it.

12. Sep 12, 2010

### lauratyso11n

The correct statement is that if $$A$$ is surjective then $$A^T$$ is injective and $$AA^T$$ is invertible.

The formula for the optimal $$x$$ is

$$\hat{x}=A^T(AA^T)^{-1}y$$

13. Sep 12, 2010

### Landau

So it was a typo :)