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.

## Best Answer

The following might be useless, but it may give you some conceptual clarification:

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.