# null space and rank

• Jan 9th 2009, 07:49 PM
Kat-M
null space and rank
let A be mxn matrix and B be the transpose of A. let P=BA. prove that A and P have the same null space and the same rank.
• Jan 10th 2009, 11:30 AM
NonCommAlg
Quote:

Originally Posted by Kat-M

let A be mxn matrix and B be the transpose of A. let P=BA. prove that A and P have the same null space and the same rank.

we only need to prove that the null space of A and P are equal, because then by the rank-nullity theorem they must also have the same rank.

if $Ax=0,$ then $Px=0$ and hence $\text{null}(A) \subseteq \text{null}(P).$ conversely, if $x \in \text{null}(P),$ then $Px=A^TAx=0,$ i.e. $Ax \in \text{null}(A^T)=(\text{col}(A))^{\perp}.$ hence $Ax \in \text{col(A)} \cap (\text{col}(A))^{\perp}=(0). \ \Box$
• Jan 10th 2009, 01:18 PM
Opalg
Quote:

Originally Posted by NonCommAlg
if $x \in \text{null}(P),$ then $Px=A^TAx=0,$ i.e. $Ax \in \text{null}(A^T)=(\text{col}(A))^{\perp}.$ hence $Ax \in \text{col(A)} \cap (\text{col}(A))^{\perp}=(0). \ \Box$

Alternatively, if $Px=A^{\textsc t}Ax=0$ then $(Ax)^{\textsc t}(Ax)= x^{\textsc t}A^{\textsc t}Ax = 0$ and so $Ax = 0$.
• Jan 10th 2009, 01:33 PM
NonCommAlg
Quote:

Originally Posted by Opalg

Alternatively, if $Px=A^{\textsc t}Ax=0$ then $(Ax)^{\textsc t}(Ax)= x^{\textsc t}A^{\textsc t}Ax = 0$ and so $Ax = 0$.

that's certainly an easier way!
• Jan 10th 2009, 01:43 PM
Kat-M
Quote:

Originally Posted by NonCommAlg
we only need to prove that the null space of A and P are equal, because then by the rank-nullity theorem they must also have the same rank.

if $Ax=0,$ then $Px=0$ and hence $\text{null}(A) \subseteq \text{null}(P).$ conversely, if $x \in \text{null}(P),$ then $Px=A^TAx=0,$ i.e. $Ax \in \text{null}(A^T)=(\text{col}(A))^{\perp}.$ hence $Ax \in \text{col(A)} \cap (\text{col}(A))^{\perp}=(0). \ \Box$

i am trying to see why $\text{null}(A^T)=(\text{col}(A))^{\perp}.$. could you also help me with this?
• Jan 10th 2009, 03:05 PM
HallsofIvy
Then look at Opalg's response.