Unit 3 Continued

Basis of a Subspace

A Basis of a subspace W is a set ${\vec{v}}_1,...,{\vec{v}}_k$ of vectors such that

1. Span(${\vec{v}}_1,...,{\vec{v}}_k$ ) = W
2. ${\vec{v}}_1,...,{\vec{v}}_k$ are linearly independent

B is a basis of W if

1. Span(B) = W
2. B is $l.i.$

Theorem

Every vector z in W can be written in exactly one way as a linear combination of the vectors in a basis.

Proof

Suppose $B={\vec{v}}_1,...,{\vec{v}}_k$ is a basis of W.

Suppose $\vec{z}=a_1{\vec{v}}_1+...+a_k\overrightarrow{v_k}$

Suppose $\vec{z}=b_1{\vec{v}}_1+...+b_k\overrightarrow{v_k}$

Subtract:

$0=(a_a-b_1){\vec{v}}_1+...+(a_k-b_k)\overrightarrow{v_k}$

So:

$a_1-b_1=0\to a_1=b_1$

$⋮$

$a_k-b_k=0\to a_k=b_k$

The Reduction Algorithm

Suppose Span(${\vec{v}}_1,...,{\vec{v}}_k$) $=W$ if ${\vec{v}}_1,...,{\vec{v}}_k$ is not $l.i$. We can find one ${\vec{v}}_i$ that is a combination of the others. If we remove that vector from the set, the span does not change. Repeating this process until the set is $l.i$, we obtain a basis of W.

Extension Algorithm

Suppose ${\vec{v}}_1,...,{\vec{v}}_k$ are in W and they are linearly independent. Then the set can be extended to a basis of W.

Example

Find a basis for $W=$ span(${\vec{v}}_1,{\vec{v}}_2,{\vec{v}}_3,{\vec{v}}_4$) where

Build a matrix with variables as columns:

Put in REF:

${\vec{v}}_1$ and ${\vec{v}}_2$ are a basis for W because ${\vec{v}}_3$ and $\overrightarrow{v_4}$ are dependent.

The vectors corresponding to the columns of the REF containing leading entries form a basis for the span of the columns, which is called the column space: COL(A).

Theorem

Any two basis of a subspace W have the same number of elements.

Definition

The dimension of a subspace W is the number of elements in any base of W.

Let A be a matrix. Then DIM(COL(A)) = Rank(A)

Proof

If we reduce A to a REF, the columns of A corresponding to the columns of the REF with leading entries, form a basis.

To show Basis or not Basis

Put in REF, check rank.

The Null Space

The Null space of a matrix A is the set of solutions of the system $A{\vec{x}}_1=0$:

If $A\vec{v}=\vec{0}$

And $A\vec{w}=\vec{0}$

Then $A(\vec{v}+\vec{w})=A\vec{v}+A\vec{w}=\vec{0}+\vec{0}=0$

Suppose A is $m\times n$, then COL(A) is a subspace of of ${\mathbb{R}}^m$, and NULL(A) is a subspace of ${\mathbb{R}}^n$.

DIM(COL(A))+DIM(NULL(A))=n

Rank(A) = DIM(COL(A))

Nullity = DIM(NULL(A))