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The Fundamental Theorem of Linear Algebra

The following theorem, which I present without proof, is one of the most important results from linear algebra.

Theorem 2.1 (The projection theorem)   Suppose V is any inner product space (that is, vector space with an inner product) and W is a finite-dimensional subspace of V. Given any $v\in V$, there exists a unique vector $w\in W$ closest to v. In other words, there is a unique solution to

\begin{displaymath}\min_{z\in W}\Vert v-z\Vert.
\end{displaymath}

Moreover, this closest vector w is characterized by the following orthogonality condition:

 \begin{displaymath}
(v-w,z)=0\ \mbox{for all}\z\in W.
\end{displaymath} (5)

In the above theorem, (x,y) denotes the inner product of two vectors $x,y\in V$. The vector w is called the orthogonal projection of v onto W, or the best approximation to v from W. It is sometimes denoted by $w=\mbox{proj}_Wv$.

Given any matrix $A\in{\bf {\rm R}}^{m\times n}$, the range (or column space) of A is defined by

\begin{displaymath}\r(A)=\left\{Ax\,:\,x\in{\bf {\rm R}}^n\right\}
\end{displaymath}

and the null space (or kernel) of A is

\begin{displaymath}{\cal N}(A)=\left\{x\in{\bf {\rm R}}^n\,:\,Ax=0\right\}.
\end{displaymath}

I wish to discuss the relationships that exist between the ranges and null spaces of A and AT. The following concept is crucial.

Definition 2.2   Suppose S is a nonempty subset of ${\bf {\rm R}}^n$. The orthogonal complement of S is the set

\begin{displaymath}S^{\bot}=\left\{y\in{\bf {\rm R}}^n\,:\,x\cdot y=0\ \forall\ x\in S\right\}.
\end{displaymath}

The basic properties of orthogonal complements are summarized in the following theorem.

Theorem 2.3   Suppose S is a nonempty subset of ${\bf {\rm R}}^n$. Then
1.
$S^{\bot}$ is a subspace of ${\bf {\rm R}}^n$.
2.
$\left(S^{\bot}\right)^{\bot}=\mbox{span}(S)$, that is, $\left(S^{\bot}\right)^{\bot}$ is the smallest subspace containing S. In particular, if S is a subspace of ${\bf {\rm R}}^n$, then $\left(S^{\bot}\right)^{\bot}=S$.
3.
If S is a subspace of ${\bf {\rm R}}^n$, then ${\bf {\rm R}}^n=S\oplus S^{\bot}$, that is, every $x\in{\bf {\rm R}}^n$ can be written uniquely as x=y+z, where $y\in S$ and $z\in S^{\bot}$.



 
next up previous
Next: Proof: Up: First-order necessary conditions for Previous: Optimality conditions
Mark S. Gockenbach
2003-03-07