# Thread: Help! Proof of Rank A transpose = Nul A perp.

1. ## Help! Proof of Rank A transpose = Nul A perp.

Prove.

$Rank(A^T ) = (Null(A))^\perp$

Where, if V is a subspace,

$V^\perp = \{\vec{w} : \vec{w}$ is perpendicular to $\\\vec{v} \ \ \forall \ \ \vec{v} \ \ \varepsilon \ \ V \}$

Now, I know the rank is number of linearly independent columns in a matrix. So the rank of $A^T$ would be the number of linearly independent rows of A.

I know that the nullspace is the solutions to $A\vec{x}=\vec{0}$
However, the "perpendicular compliment" is a term I am not familiar with. And the simple definition does not really help me. Can someone help me with this proof?

2. Originally Posted by Lovemachinzero
If A is a subspace of a vectorspace, prove that

$Rank(A^T ) = Null(A)^\perp$

Where, if V is a subspace,

$V^\perp = \{\vec{w} : \vec{w}$ is perpendicular to $\\\vec{v} \ \ \forall \ \ \vec{v} \ \ \varepsilon \ \ V \}$

Now, I know the rank is number of linearly independent columns in a matrix. So the rank of $A^T$ would be the number of linearly independent rows of A.
But you started by saying that A is a subspace of some vector space, not a matrix! And if A is a matrix, I don't know what you mean by " $Null(A)^\perp$".

I know that the nullspace is the solutions to $A\vec{x}=\vec{0}$
However, the "perpendicular compliment" is a term I am not familiar with. And the simple definition does not really help me. Can someone help me with this proof?

3. I apologize. A is a matrix, and the perp should be inclusive of the subspace Null(A). I edited the first post!

Now, I think that since Null(A) is the set of vectors $\vec{x}$ such that $A\vec{x}=\vec{0}$ The perpendicular compliment of Nul(A) would mean that there is a set of vectors $\vec{v}$ such that $<\vec{v},\vec{x}> = 0$

Is the $Range(A^T) = Col(A^T) = Row(A)$? Am I even on the right track?! =]