Shrinking and Linear Independence

This class will finish our discussion on Shrinking from last class, and then will begin the discussion on Linear Independence.

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Shrinking Continued

Let’s begin with a list of vectors:

$$\left(\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right)\left(\begin{array}{c}2\\ 0\\ -1\\ -1\end{array}\right)\left(\begin{array}{c}-3\\ -1\\ 2\\ 2\end{array}\right)\left(\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right)\left(\begin{array}{c}2\\ 0\\ -2\\ 0\end{array}\right)$$

The shrinking process involves taking the matrix of this list of vectors:

And then we find the RREF, nothing the values of $1$. (we will also note the values of -1 in the 3rd column)

$$\left(\begin{array}{ccccc}1& 2& -3& 1& 2\\ 1& 0& -1& -1& 0\\ -1& -1& 2& 1& -2\\ -1& -1& 2& -2& 1\end{array}\right){\to }^{RREF}\left(\begin{array}{ccccc}\ast 1\ast & 0& -1& 0& -1\\ 0& \ast 1\ast & -1& 0& 2\\ 0& 0& 0& \ast 1\ast & -1\\ 0& 0& 0& 0& 0\end{array}\right)$$

We are dumping this matrix into MATLAB with zero intent of actually solving the linear system.

Taking the 3rd vector out of the matrix:

$$\left(\begin{array}{c}-3\\ -1\\ 2\\ 2\end{array}\right)=-\left(\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right)-\left(\begin{array}{c}2\\ 0\\ -1\\ -1\end{array}\right)$$

Since the 5th vector also has negatives, we take that one out as well:

$$\left(\begin{array}{c}2\\ 0\\ -2\\ 0\end{array}\right)=-\left(\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right)+2\left(\begin{array}{c}2\\ 0\\ -1\\ -1\end{array}\right)-\left(\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right)$$

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We are left with the surviving 3 vectors:

$$\left(\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right)\left(\begin{array}{c}2\\ 0\\ -1\\ -1\end{array}\right)\left(\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right)$$

Finding the RREF of the matrix of these 3 vectors:

$$\left(\begin{array}{ccc}1& 2& 1\\ 1& 0& -1\\ -1& -1& 1\\ -1& -1& -1\end{array}\right){\to }^{RREF}\left(\begin{array}{ccccc}1& 0& 0& \mid & 0\\ 0& 1& 0& \mid & 0\\ 0& 0& 1& \mid & 0\\ 0& 0& 0& \mid & 0\end{array}\right)$$

This List is now UNSHRINKABLE

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Theorem

Given a list of vectors ${\vec{v}}_1,{\vec{v}}_2,\dots ,{\vec{v}}_n$ in ${\mathbb{R}}^n$:

1. ${\vec{v}}_1,{\vec{v}}_2,\dots ,{\vec{v}}_n$ is unshrinkable: no vector in a linear combination of previous vectors in the list
2. ${\vec{v}}_1,{\vec{v}}_2,\dots ,{\vec{v}}_n$ is irredundant: no vector in the list in a linear combination of the others
3. The homogeneous system $x_1{\vec{v}}_1+x_2{\vec{v}}_2+\cdots +x_n{\vec{v}}_n=\vec{0}$ only has the trivial solution $x_1=x_2=\cdots =x_n=0$,
   • linear independence

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Definition

A basis for a subsapce $W$ of ${\mathbb{R}}^n$ in a linear independent list of vectors that spans $W$

For example, the list of vectors:

$$\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$$

is a basis for ${\mathbb{R}}^3$

This rule is true for any list of vectors that spans ${\mathbb{R}}^n$.

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Example

Consider $W$ the subsapce of solutions to the homogeneous system:

$$x-y+2z=0$$

Let’s start by defining the variables

$$\begin{array}{rl}z& =a\\ y& =b\\ x& =b-2a\end{array}$$

Basic Solutions:

$$\begin{array}{c}a=1\\ \left(\begin{array}{c}-2\\ 0\\ 1\end{array}\right)\end{array},\begin{array}{c}b=1\\ \left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)\end{array}spanW$$

Check for independence:

$$a\left(\begin{array}{c}-2\\ 0\\ 1\end{array}\right)+B\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$$

Solving for these vectors:

$$\begin{array}{rl}-2a+b& =0\\ b& =0\end{array}$$