2-norm Pseudoinverse Upper Bound

[baggiano]

Hello

I'm trying to show that the following upper bound on the matrix 2-norm is true:

\left\|(AB)^+\right\|_2\leq\left\|A^+\right\|_2 \left\|B^+\right\|_2

where + is the matrix pseudoinverse and A\in\Re^{n\times m} and B\in\Re^{m\times p} are full-rank matrices with n\geq m\geq p.

Any hint how I can show it?

Thanks in advance!

Bag

 

[Hawkeye18]

Are you sure that m\ge p and not p\ge m?
Also I assume that by pseudoinverse you mean the Moore-Penrose pseudoinverse.

If n\ge m and p\ge m, then (AB)^+ =B^+ A^+ holds, and this identity gives your estimate.

If n\ge m\ge p, then A, B and so AB are left invertible. For a left invertible matrix A, the Moore-Penrose pseudoinverse A^+ is the minimal left inverse, A^{min}_L = (A^*A)^{-1}A^*.

The minimal left inverse A^{min}_L has the property that for any other left inverse A_L of A we have A^{min}_L = P_{Ran (A)} A_L, where P_{Ran (A)} is the orthogonal projection onto the range (column space) of A. In particular, this implies that the norm of the minimal left inverse is the minimal possible norm of a left inverse.

Now let us gather all this information together: both A and B, are left invertible, so A^+ and B^+ are the minimal left inverses of A and B respectively. Therefore, B^+A^+ is a left inverse of AB; generally it is not the minimal left inverse, but the minimality property for the norm of minimal left inverse implies

<br /> \|(AB)^+\| \le \| B^+A^+\| \le\|B^+\|\cdot\|A^+\|<br />

 

[baggiano]

Hello

Thanks a lot for the details of your illustration.

Yes, m\geq p was actually correct.

In the fourth paragraph of your reply you say that A_L^{min}=P_{Ran(A)}A_L. Shouldn't it be A_L^{min}=A_LP_{Ran(A)} instead?

Thanks again!

Bag

 

[Hawkeye18]

Yes, it should be A_L^{min} = A_L^{min} P_{Ran(A)}.