Question

Question 23: Let $A$ be an $m\times n$. Prove that every vector x in ${\mathbb{R}}^n$ can be written in the form $\mathrm{x}=\mathrm{p}+\mathrm{u}$, where $\mathrm{p}$ is in Row A and u is in $\mathrm{Nul}A$. Also, show that if the equation $A\mathrm{x}=\mathrm{b}$ is consistent, then there is a unique p in Row A such that $A\mathrm{p}=\mathrm{b}$.

Short answer

It is proved that there is a unique $\mathbf{p}$ in $\mathrm{Row}A$ such that $A\mathrm{p}=\mathrm{b}$.

Step-by-step solution

The Orthogonal Decomposition Theorem

Suppose that $W$ is a subspace of ${\mathbb{R}}^n$. Then, each $\mathbf{y}$ in ${\mathbb{R}}^n$ can be expressed uniquely in the form:

$\mathbf{y}=\hat{\mathbf{y}}+\mathbf{z}$ … (1)

With $\hat{\mathbf{y}}$ is in $W$ and $\mathbf{z}$ is in $W^{\mathrm{\perp }}$. In particular, when $\left\{{\mathrm{u}}_1,\dots ,{\mathrm{u}}_p\right\}$ is anorthogonal basis of $W$, then;

$\hat{\mathbf{y}}=\frac{\mathrm{y}\cdot {\mathrm{u}}_1}{{\mathrm{u}}_1\cdot {\mathrm{u}}_1}{\mathrm{u}}_1+\dots +\frac{\mathrm{y}\cdot {\mathrm{u}}_p}{{\mathrm{u}}_p\cdot {\mathrm{u}}_p}{\mathrm{u}}_p$ … (2)

And, $\mathbf{z}=\mathbf{y}-\hat{\mathbf{y}}$.

Show that if the equation $A\mathrm{x}=\mathrm{b}$ is consistent, then there is a unique p in Row A

Theorem 3states that consider A as a $m\times n$ matrix, and thenull spaceofA is theorthogonal complement of the row space ofA.The orthogonal complement of the column space of A is known as thenull spaceof $A^T$.

${\left(\mathrm{Row}A\right)}^{\mathrm{\perp }}=\mathrm{Nul}A$ and ${\left(\mathrm{Col}A\right)}^{\mathrm{\perp }}=\mathrm{Nul}{A}^T$

According to the Orthogonal Decomposition Theorem, every $\mathbf{x}$ in ${\mathbb{R}}^n$ can be expressed uniquely as $\mathbf{x}=\mathbf{p}+\mathbf{u}$, where $\mathbf{p}$ in $\mathrm{Row}A$, and $\mathbf{u}$ is in ${\left(\mathrm{Row}A\right)}^{\mathrm{\perp }}$. According to theorem 3, ${\left(\mathrm{Row}A\right)}^{\mathrm{\perp }}=\mathrm{Nul}A$, therefore, $\mathbf{u}$ is in $\mathrm{Nul}A$.

Assume that $A\mathrm{x}=b$ is consistent. Consider $\mathbf{x}$ as a solution and write $\mathbf{x}=\mathbf{p}+\mathbf{u}$ as shown before. Then;

$\begin{array}{c}A\mathrm{p}=A\left(\mathrm{x}-\mathrm{u}\right)\\ =A\mathrm{x}-A\mathrm{u}\\ =\mathrm{b}-0\\ =\mathrm{b}\end{array}$

Therefore, the equation $A\mathrm{x}=\mathrm{b}$ contains at least one solution p in $\mathrm{Row}A$.

Furthermore, assume that $\mathbf{p}$ and ${\mathbf{p}}_1$ are both in $\mathrm{Row}A$ and both satisfied $A\mathrm{x}=\mathrm{b}$. Therefore, $\mathbf{p}-{\mathbf{p}}_1$ is in $\mathrm{Nul}A={\left(\mathrm{Row}A\right)}^{\mathrm{\perp }}$, because $A\left(\mathbf{p}-{\mathbf{p}}_1\right)=A\mathbf{p}-A{\mathbf{p}}_1=\mathrm{b}-\mathrm{b}=0$.

The equation $\mathbf{p}={\mathbf{p}}_1+\left(\mathbf{p}-{\mathbf{p}}_1\right)$ and $\mathbf{p}=\mathbf{p}+0$ are decomposed into a sum of a vector in $\mathrm{Row}A$ and a vector in ${\left(\mathrm{Row}A\right)}^{\mathrm{\perp }}$. According to the uniqueness of the orthogonal decomposition, $\mathbf{p}={\mathbf{p}}_1$ and $\mathbf{p}$ is unique.

Thus, it is proved that there is a unique $\mathbf{p}$ in $\mathrm{Row}A$ such that $A\mathrm{p}=\mathrm{b}$.