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Solving for a Surjective Matrix 
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#1
Sep1010, 08:56 AM

P: 8

I saw this in a book as a Proposition but I think it's an error:
Assume that the (nbyk) matrix, [tex]A[/tex], is surjective as a mapping, [tex]A:\mathbb{R}^{k}\rightarrow \mathbb{R}^{n}[/tex]. For any [tex]y \in \mathbb{R}^{n} [/tex], consider the optimization problem [tex]min_{x \in \mathbb{R}^{k}}\left{x^2\right}[/tex] such that [tex] Ax = y[/tex]. Then, the following hold: (i) The transpose of [tex]A[/tex], call it [tex]A^{T}[/tex] is injective. (ii) The matrix [tex]A^{T}A[/tex] is invertible. (iii) etc etc etc.... I have a problem with point (ii), take as an example the (2by3) surjective matrix [tex]A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}[/tex] [tex]A^{T}A[/tex] in this case is not invertible. Can anyone confirm that part (ii) of this Proposition is indeed incorrect ? 


#2
Sep1010, 11:27 AM

P: 34

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/Bijecti...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! [edit] 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. [edit] 


#3
Sep1010, 12:22 PM

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#4
Sep1010, 12:31 PM

P: 8

Solving for a Surjective Matrix
Assume that the (nbyk) matrix, [tex]A[/tex], is surjective as a mapping, [tex]A:\mathbb{R}^{k}\rightarrow \mathbb{R}^{n}[/tex]. For any [tex]y \in \mathbb{R}^{n} [/tex], consider the optimization problem [tex]min_{x \in \mathbb{R}^{k}}\left{\leftx\right^2\right}[/tex] such that [tex] Ax = y[/tex]. Then, the following hold: (i) The transpose of [tex]A[/tex], call it [tex]A^{T}[/tex] is injective. (ii) The matrix [tex]A^{T}A[/tex] is invertible. (iii) The unique optimal solution of the minimum norm problem is given by [tex](A^TA)^{1}A^Ty[/tex] I have a problem with point (ii), take as an example the (2by3) surjective matrix [tex]A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}[/tex] [tex]A^{T}A[/tex] in this case is not invertible. 


#5
Sep1010, 12:39 PM

P: 8

[tex]\sigma\lambda = \alpha  \bf{r}[/tex] where [tex]\sigma \in \mathbb{R}^{nbyk}[/tex] is surjective, and [tex]\lambda \in \mathbb{R}^{k}[/tex], [tex]\alpha , \bf{r} \in \mathbb{R}^{n}[/tex]. How would you solve for [tex]\lambda[/tex] ? 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 [tex]\lambda = \left(\sigma^{T}\sigma\right)^{1}\sigma^{T}\left[\alpha  \bf{r}\right][/tex]. 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. 


#6
Sep1010, 01:22 PM

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PF Gold
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Looking at dimension counting with your example, A^{t}A is a 3x3 matrix, which means (A^{t}A)^{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 A^{t} and A are compatible matrices which suggests that he just put the transpose on the wrong one and carried the error through.



#7
Sep1010, 02:08 PM

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P: 905

The least square solution of Ax=y satisfies the normal equation: [tex]A^TAx=A^Ty.[/tex] This solution is unique if and only if [itex]A^TA[/itex] is invertible. In this case, it is given by [tex]x=(A^TA)^{1}A^Tb[/tex], just like the author asserts. But [itex]A^TA[/itex] is not necessarily invertible, contrary to the proposition. Note that [itex]A^TA:\mathbb{R}^k\to\mathbb{R}^k[/itex] is bijective if and only if its rank equals k. But since [itex]A^TA[/itex] and A always have equal rank, this happens if and only if A has rank k. Since [itex]A:\mathbb{R}^k\to\mathbb{R}^n[/itex] is assumed to be surjective, it has rank n and we must have [itex]k\geq n[/itex]. So in fact, [itex]A^TA[/itex] is invertible if and only if k=n! Indeed, in your counterexample k and n are not equal. 


#8
Sep1010, 02:28 PM

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Perhaps you could post a link to the book?



#9
Sep1010, 06:25 PM

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#10
Sep1010, 07:13 PM

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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
Sep1010, 08:14 PM

P: 8




#12
Sep1210, 01:37 PM

P: 8

The correct statement is that if [tex]A[/tex] is surjective then [tex]A^T[/tex] is injective and [tex]AA^T[/tex] is invertible.
The formula for the optimal [tex]x[/tex] is [tex]\hat{x}=A^T(AA^T)^{1}y[/tex] 


#13
Sep1210, 01:43 PM

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