# Why a vector in both the null space and row space has to be $0$?

In this video https://www.youtube.com/watch?v=Go2aLo7ZOlU&index=34&list=PLE7DDD91010BC51F8, at about 26:00, the professor claims that a vector in both the null space and row space has to be $0$, and the following is his proof.

in which he says $x-y$ is in both the null space of $A$ and the row space of $A$, then $x-y=0$, but I don't know why this is true.

Thanks!

• What video?${}{}$ – martini Mar 1 '16 at 14:54

As dls points out, it must be that "if the transpose of a vector in the row space is in the right null space, it must be $0$".

Suppose that $x$ is in the row space of $A$: $x=uA$, and that $x^T$ is in the right null space of $A$: $Ax^T=0$, then $$AA^Tu^T=Ax^T=0\implies |uA|^2=uAA^Tu^T=0\implies x=uA=0$$

• The convention for the definition of nullspace is to multiply on the right. – dls Mar 1 '16 at 16:01
• @dls: That is not a universal convention; more of a convention within a given book, perhaps. I will amend my answer for "if the transpose of a vector in the row-space is in the right null-space, then that vector is $0$." – robjohn Mar 1 '16 at 16:07
• Agreed! And pedagogically I even like your answer because I find many instructors don't make it a point to specify their conventions precisely. But I would also point out that in the non-square case, if you mess up the conventions then the statement in the title doesn't make sense due to dimensionality. – dls Mar 1 '16 at 16:21

By the so-called fundamental theorem of linear algebra, the (right) nullspace is the orthogonal complement of the row space.

Let $V$ be a finite-dimensional vector space, $\:A\:$ a matrix (or linear transformation) from $V$ to $W$ (which sends a vector $v \in V$ to $Av \in W$) , $x,y \in Rowspace(A),\: x \neq y$. $$If Ax = Ay \:\text{then}\: Ax-Ay=A(x-y)=0, \text{then}\: x-y \in Nullspace(A)$$ Since $Rowspace(A)$ is a linear subspace of $V$, every linear combination of $x,y$ is in $Rowspace(A)$, in particular $x-y$, so $x-y \in Rowspace(A)$ and $x-y \in Rowspace(A) \cap Nullspace(A)$.

The last step is a consequence of the "Rank-Nullity theorem" (you can look for it in any linear Algebra book, for example, Friedberg's "Linear Algebra"). In terms of matrices it says that:

$$dimension(Rowspace(A)) \oplus dimension (Nullspace(A)) = dimension (V)$$

Now, $\oplus$ means direct sum of 2 linear subspaces of $V$, which means that $Rowspace(A) \cap Nullspace(A) = {0}$. By this last statement, $x-y=0$, so $x=y$, a contradiction of the hypothesis $x\neq y$.

If you want to really learn Linear Algebra I suggest you expand your learning experience from a book.