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2-norm Pseudoinverse Upper Bound

  1. Jan 8, 2012 #1

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

    [itex]\left\|(AB)^+\right\|_2\leq\left\|A^+\right\|_2 \left\|B^+\right\|_2[/itex]

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

    Any hint how I can show it?

    Thanks in advance!

    Last edited: Jan 8, 2012
  2. jcsd
  3. Jan 10, 2012 #2
    Are you sure that [itex]m\ge p[/itex] and not [itex]p\ge m[/itex]?
    Also I assume that by pseudoinverse you mean the Moore-Penrose pseudoinverse.

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

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

    The minimal left inverse [itex] A^{min}_L [/itex] has the property that for any other left inverse [itex]A_L [/itex] of [itex]A [/itex] we have [itex] A^{min}_L = P_{Ran (A)} A_L[/itex], where [itex] P_{Ran (A)} [/itex] is the orthogonal projection onto the range (column space) of [itex]A[/itex]. 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 [itex] A [/itex] and [itex]B [/itex], are left invertible, so [itex] A^+ [/itex] and [itex] B^+ [/itex] are the minimal left inverses of [itex] A [/itex] and [itex] B [/itex] respectively. Therefore, [itex] B^+A^+ [/itex] is a left inverse of [itex] AB [/itex]; generally it is not the minimal left inverse, but the minimality property for the norm of minimal left inverse implies

    \|(AB)^+\| \le \| B^+A^+\| \le\|B^+\|\cdot\|A^+\|
  4. Jan 12, 2012 #3

    Thanks a lot for the details of your illustration.

    Yes, [itex]m\geq p[/itex] was actually correct.

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

    Thanks again!

  5. Jan 12, 2012 #4
    Yes, it should be [itex]A_L^{min} = A_L^{min} P_{Ran(A)}[/itex].
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