Let A be a n × m matrix with rank (A) = r < m. Then there exists a matrix B of order s × m such that rank (B) = m − r, and no overlap between their row spaces i.e. C(AT) ∩ C(BT)= {0} .

Show that:

(1) ATA + BTB is of full rank

(2) (ATA + BTB)−1 is a g-inverse of ATA,

i.e. ATA(ATA + BTB)−1ATA = ATA

(1) I understand that ATA will give me an m x m matrix and that BTB will give me an m x m matrix simply by looking at their dimensions when multiplying the transpose by the matrix, so if I add two matrices that are m x m together is that enough to "prove" (1)?

(2) Do I try to use single value decomposition to approach this one?


1 Answer 1


Regarding 1: no, the fact that $A^TA,B^TB$ are matrices with ranks whose sum is $n$ is not enough to deduce that $A^TA + B^TB$ has full rank. Here's a hint to one approach:

Let $M = A^TA + B^TB$. Because $M$ is square, $M$ has full rank iff $Mx = 0 \implies x=0$. In fact, because $Mx = 0 \implies x^TMx = 0$, it suffices to show that $x^TMx = 0 \implies x = 0$. Note that $x^TMx$ can be written in the form $v^Tv + w^Tw$ for some vectors $v,w$. argue that if $x \neq 0$, then we must have either $v \neq 0$ or $w \neq 0$. The conclusion follows.

Regarding 2, no: singular value decomposition is not useful here. I'm not sure how to prove this, but I suspect it is helpful to write $A^TA = [A^TA + B^TB] - B^TB$. Alternatively, it suffices to show that $A^TA(A^TA + B^TB)^{-1}x = x$ holds for all $x$ in the row-space of $A$.

  • $\begingroup$ Thanks, Ben! Your responses are always very helpful :) $\endgroup$
    – theGRUMBER
    Commented Sep 29, 2022 at 0:57

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