Linear Algebra: Linear Combinations, Span, Linear Independence

Linear Algebra: Linear Combination

Linear combinations of vectors are sums of a set of vectors multilpied by constants. For example, if we had $\overrightarrow{a}=\left(0,2\right)$ and $\overrightarrow{b}=\left(4,1\right)$, a linear combination of them could be $5\ast \overrightarrow{a}+-2\ast \overrightarrow{b}=\left(0,10\right)+\left(-8,-2\right)=\left(-8,8\right)$.

Linear Algebra: Linear Independence

We can imagine two vectors that point in the same direction, that their is a constant $c$ that when multiplied by one of the vectors gets us the other vector. $\overrightarrow{a}=\left(1,1\right),\overrightarrow{b}=\left(2,2\right),2\ast \overrightarrow{a}=\overrightarrow{b}$ is an example of linearly dependent vectors. They lie on the same line. Linearly independent vectors require more than just a scaling, and therefore define two lines. The test for linear independence/dependence is applicable in any number of dimensions: $\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots <!--FUNCTION\; APPLICATION-->,{\overrightarrow{v}}_n\right\}\iff c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+\cdots +c_n{\overrightarrow{v}}_n=\overrightarrow{0}$ where at least one $c$ is non-zero means the vectors are linearly dependant.

$$\overrightarrow{a}=\left(1,1\right),\overrightarrow{b}=\left(2,2\right)$$

$$c_1\overrightarrow{a}+c_2\overrightarrow{b}=\overrightarrow{0}=\left(0,0\right)$$

$$c_1+2c_2=0$$

$$c_1=-2c_2$$

$$c_1=5,\text{ arbitrary choice for }{c}_1$$

$$-2c_2=5$$

$$c_2=-\frac{5}{2}$$

$$c_1+2c_2=0$$

$$5+2\left(-\frac{5}{2}=0\right)$$

$$5+\left(-5\right)=0$$

Any non-zero constant can satisfy this equation $c_1+2c_2=0$, so we know $\overrightarrow{a},\overrightarrow{b}$ are linearly dependent.

$$\overrightarrow{c}=\left(2,1\right),\overrightarrow{d}=\left(3,2\right)$$

$$2c_1+3c_2=0$$

$$c_1+2c_2=0$$

$$2c_1=-3c_2$$

$$c_1=-\frac{3c_2}{2}$$

$$-\frac{3c_2}{2}+2c_2=0$$

$$\frac{c_2}{2}=0$$

$$c_2=0$$

$$c_1+2\left(0\right)=0$$

$$c_1+0=0$$

$$c_1=0$$

0’s are the only values that satisfy these equations, so we know $\overrightarrow{c}=\left(2,1\right),\overrightarrow{d}=\left(3,2\right)$ are linearly independent

Linear Algebra: Span and Basis

$C=\left\{\overrightarrow{a}\ast c1+\overrightarrow{b}\ast c2|c1,c2\in ℝ\right\}$ defines all the linear combinations of $\overrightarrow{a}$, $\overrightarrow{b}$ and is the ”span” of $\overrightarrow{a}$ and $\overrightarrow{b}$. A Span will either define a single line if the vectors are linearly dependent or a vector space ${ℝ}^n$ if all vectors $S=\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots <!--FUNCTION\; APPLICATION-->,{\overrightarrow{v}}_n\right\}$ are linearly independent. $Span\left(\left\{\overrightarrow{a},\overrightarrow{b}\right\}\right)={ℝ}^2\iff \overrightarrow{a},\overrightarrow{b}$ are linearly independent. If they are linearly independent, then the vectors $\overrightarrow{a},\overrightarrow{b}$ form a basis of ${ℝ}^2$, meaning the linear combinations of $\overrightarrow{a},\overrightarrow{b}$ describe EVERY vector in ${ℝ}^2$.