For A in RR_(p xx n) and B in RR_(m xx n), show that ccN(A) sube ccN(b) iff BA^+A=B.

Teagan Huffman 2022-09-20 Answered
For A R p × n and B R m × n , show that N ( A ) N ( B ) iff B A + A = B. The "+" note indicates Moore-Penrose pseudo-inversion.
My attempt:
Suppose B A + A = B and x N ( A ). Then Ax=0. We have
B x = ( B A + A ) x = B A + ( A x ) = 0.
This shows that x N ( B )
Suppose N ( A ) N ( B ). Let x R n where x N ( A ) N ( B ). Thus we have Ax=0. But note that K A x = 0 for all K matrices with p columns. Thus assuming B A + has p columns we have
B x = A x = B A + ( A x ) = B A + A x .
In the second half, I'm not sure if I was able to do it or not. Tell me what I need to fix!
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Haylee Krause
Answered 2022-09-21 Author has 10 answers
If B A + A = B, then N ( A ) N ( B ) thus:
Suppose N ( A ) N ( B ). We will show that B A + A = B. Let x R n . Let us decompose x = x 1 + x 2 where x 1 N ( A ) and x 2 N ( A ). For x 2 we know that B A + A x 2 0 = B A + 0 = 0
Let us observe what B A + A does on x 1 . Observe that,
x 1 N ( A ) A x 1 R ( A ) A + A x 1 N ( A ) .
But since N ( A ) is isomorphic to R ( A ), this means that our x 1 that was since to R ( A ) and back to N ( A ) is still our x 1 we had starting out. Thus,
A + A x 1 = x 1 B A + A x 1 = B x 1 .
Lastley, let us observe what B does to x,
B x = B ( x 1 + x 2 ) = B x 1 + B x 2 0 = B x 1 .
We have,
B A + A x = B A + A x 1 = B x 1 B A + A = B .
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

New questions