# How would one prove that the row space and null space are orthogonal compliments of each other?

It makes intuitive sense to me that the very definition of the null space - all x's that produce the zero vector when multiplied by the rows of a matrix A - would coincide with the conditions for orthogonality (dot product of two vectors = 0). But how would one show that this applies for all combinations of A? I guess i'm trying to wrap my head around a more rigorous way of understanding the notion that the row space is the orthogonal compliment to the null space of a matrix... I hope this makes sense and I am looking forward to hearing some insights!

Note that matrix multiplication can be defined via dot products. In particular, suppose that $A$ has rows $a_1$, $a_2, \dots, a_n$, then for any vector $x = (x_1,\dots,x_n)^T$, we have: $$Ax = (a_1 \cdot x, a_2 \cdot x, \dots, a_n \cdot x)$$ Now, if $x$ is in the null-space, then $Ax = \vec 0$. So, if $x$ is in the null-space of $A$, then $x$ must be orthogonal to every row of $A$, no matter what "combination of $A$" you've chosen.
• @Hunle consider any vector $x$ which is perpendicular to the column space. Note that $Ax$, which contains the dot products $a_i \cdot x$, will be $0$. Thus, $x$ is in the null space of $A$. – Ben Grossmann Apr 4 '18 at 21:43
Proof: Suppose $$\newcommand{\v}{\vec v} \v\in\N(A)$$ and $$\v\in\R(A)$$. Recall that, as Omnomnom has shown, if $$\n\in\N(A)$$ then $$\n\cdot\r=0$$ where $$\r\in\R(A)$$. Now $$\v\cdot\v=0=\left\|\v\right\|^2$$ Therefore $$\left\|\v\right\|=0$$, so any vector $$\v$$ in both $$\R(A)$$ and $$\N(A)$$ must equal $$\vec0$$. Therefore $$\R(A)\cap\N(A)=\left\{\vec0\right\}$$, and so by definition $$\R(A)$$ and $$\N(A)$$ are complementary subspaces as well as orthogonal. $$\blacksquare$$