# Orthogonality of Row A and Nul A and vectors in $\mathbb{R}^n$

linear algebraproof-writing

Let $$A$$ be a $$m \times n$$ matrix. Prove that every vector $$\vec{x}$$ in $$\mathbb{R}^n$$ can be written in the form $$\vec{x} = \vec{p} + \vec{u}$$ where $$\vec{p}$$ is in $$Row A$$ and $$\vec{u}$$ is in $$Nul A$$.

This follows from the theorems:

$$(Row A)^\perp = Nul A$$
$$rank A + dim Nul A = n$$

How do you write a complete proof?

Also, show that if the equation $$A\vec{x}=\vec{b}$$ is consistent, then there is an unique $$\vec{p}$$ in $$Row A$$ such that $$A\vec{p} = \vec{b}$$.

Solution:

Suppose $$A\vec{p} = \vec{b}$$ and $$A\vec{p}_1 = \vec{b}$$. Then $$A\vec{p} = A(\vec{p}_1+(\vec{p}-\vec{p}_1))$$, but $$\vec{p}-\vec{p}_1$$ must equal $$\vec{0}$$, since $$\vec{p}-\vec{p}_1$$ is in $$Row A$$ and $$rank A + dim Nul A = n$$, so $$\vec{p}=\vec{p}_1$$ and therefore $$\vec{p}$$ must be unique.

I use $$R$$ for range and $$N$$ for null space. As you have noted, $$R(A^T)^{\perp} = N(A)$$. Now note generally that if $$U$$ is a subspace of $$\mathbb{R}^n$$, then $$\mathbb{R}^n = U \oplus U^{\perp}$$, i.e. every vector $$v \in \mathbb{R}^n$$ can be written uniquely as $$v = u + w$$ with $$u \in U$$, $$w \in U^{\perp}$$. This can be proven by taking an orthonormal basis of $$U$$ and extending it to an orthonormal basis of $$\mathbb{R}^n$$. See the exercises on page 90 of https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/linalg.pdf for more details.
This means that $$\mathbb{R}^n = R(A^T) \oplus N(A)$$, which is what you want to prove.
As for the second problem, you can rephrase it as "$$A : R(A^T) \to R(A)$$ is an linear bijection". This is simple to show based on the decomposition $$\mathbb{R}^n = N(A) \oplus R(A^T)$$, which you already proved before.