# Show full rank and g-inverse for matrices with no overlap between row spaces.

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?

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 $$A$$.