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!

Question

For   A  ∈            R              p      ×      n       and   B  ∈            R              m      ×      n      , show that       N    (  A  )  ⊆      N    (  B  ) iff   B      A    +    A  =  B. The &quot;+&quot; 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&#039;m not sure if I was able to do it or not. Tell me what I need to fix!

Haylee Krause · Accepted Answer

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  .

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.

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