MATH 323 Lecture 11

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Linear Independence

1. If $v_{1},\ldots ,v_{n}$ span $V$ and one of these vectors can be written as a linear combination of the $n-1$ others, then those $n-1$ vectors span $V$.
2. Given $n$ vectors $v_{1},\ldots ,v_{n}\in V$, it is possible to write one of the vectors as a linear combination of the other $n-1$ vectors iff there exist scalars $c_{1},\ldots ,c_{n}$ (not all zero!) such that $c_{1}\,v_{1}+\dots +c_{n}\,v_{n}=0$.

Recall that $c_{1}\,v_{1}+\dots +c_{n}\,v_{n}$ is a linear combination of $v_{1},\ldots ,v_{n}$.

The vectors $v_{1},\ldots ,v_{n}\in V$ are said to be linearly independent if $c_{1}\,v_{1}+\dots +c_{n}\,v_{n}=0$ implies that all the scalars $c_{1},\dots ,c_{n}$ must all be zero.

Example

For two vectors $v_{1},v_{2}\in V$ such that $c_{1}\,v_{1}+c_{2}\,v_{2}=0$

1. If $c_{1}\neq 0$, then $v_{1}=-{\frac {c_{2}}{c_{1}}}\,v_{2}$
2. If $c_{2}\neq 0$, then $v_{2}=-{\frac {c_{1}}{c_{2}}}\,v_{1}$

Thus the two vectors are linearly dependent

Example

Which of the following collectios are linearly independent?

1. ${\begin{pmatrix}1\\1\\1\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\0\end{pmatrix}}$ Yes.
2. ${\begin{pmatrix}1\\0\\1\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}$ Yes.
3. ${\begin{pmatrix}1\\2\\4\end{pmatrix}},{\begin{pmatrix}2\\1\\3\end{pmatrix}},{\begin{pmatrix}4\\-1\\1\end{pmatrix}}$ No. The matrix of system $c_{1}\,v_{1}+c_{2}\,v_{2}+c_{3}\,v_{3}=0$ is singular, therefore there are nontrivial solutions $\left\langle c_{1},c_{2},c_{3}\right\rangle$ that will satisfy the equation. (See #Theorem)

Theorem

Let $x_{1},\ldots ,x_{n}$ be $n$ vectors in $\mathbb {R} ^{n}$ and let $X=(x_{1},\ldots ,x_{n})$ be the $n\times n$ matrix formed by using the $x$ vectors as columns.

The vectors $x_{1},\ldots ,x_{n}$ will be linearly dependent iff $X$ is singular and linearly independent iff $X$ is nonsingular

Proof

Let ${\vec {c}}=\left\langle c_{1},\ldots ,c_{n}\right\rangle$, then $X\,{\vec {c}}={\vec {0}}$ has a nontrivial solution iff $X$ is singular.

For Non-Square Matrix

For $x_{1},\ldots ,x_{k}\in \mathbb {R}$, the system $c_{1}\,x_{1}\ldots c_{k}\,x_{k}=0$ can be written as $X\,{\vec {c}}={\vec {0}}$, where $X=(x_{1},\ldots ,x_{k})$ is an $n\times k$ matrix. Therefore if $n\neq k$, the determinant of $X$ is not defined and #Theorem does not apply.

However, the system has nontrivial solutions (i.e. $x_{1},\ldots ,x_{n}$ is linearly dependent) iff the row echelon form of $X$ has at least one free variable.

Example

$x_{1}=\left\langle 1,-1,2,3\right\rangle$, $x_{2}=\left\langle -2,3,1,-2\right\rangle$, and $x_{3}=\left\langle 1,0,7,7\right\rangle$

${\begin{bmatrix}1&-2&1&0\\-1&3&0&0\\2&1&7&0\\3&-2&7&0\end{bmatrix}}\longrightarrow {\begin{bmatrix}1&-2&1&0\\0&1&1&0\\0&0&0&0\\0&0&0&0\end{bmatrix}}$

Therefore, since $c_{3}$ is a free variable, $\{x_{1},x_{2},x_{3}\}$ is linearly dependent.

Theorem

Let $v_{1},\ldots ,v_{n}\in V$. A vector $w\in \mathrm {Span} (v_{1},\ldots ,v_{n})$ can be written uniquely as a linear combination of $v_{1},\ldots ,v_{n}$ iff $v_{1},\ldots ,v_{n}$are linearly independent.

Proof

$w\in \mathrm {Span} (v_{1},\ldots ,v_{n})\iff v=c_{1}\,v_{1}+\dots c_{n}\,v_{n}$

Assume the solution is not unique, that is, $v=\alpha _{1}\,v_{1}+\dots +\alpha _{n}\,v_{n}=\beta _{1}\,v_{1}+\dots +\beta _{n}\,v_{n}$ where $\alpha _{i}\neq \beta _{i}$ for some $i$. This would mean that $0=(\alpha _{1}-\beta _{1})v_{1}+\dots +(\alpha _{n}-\beta _{n})v_{n}$, where $\alpha _{i}-\beta _{i}\neq 0$, would be linearly dependent.

Therefore, if $v_{1},\ldots ,v_{n}$ are linearly dependent, then there exist $c_{1},\ldots ,c_{n}$ (not all zero) such that

${\begin{aligned}0&=c_{1}\,v_{1}+\dots +c_{n}\,v_{n}\\+w&=\alpha _{1}\,v_{1}+\dots \alpha _{n}\,v_{n}\\=w&=(\alpha _{1}+c_{1})v_{1}+\dots +(\alpha _{n}+c_{n})v_{n}\end{aligned}}$

The second and third lines are two different representations for $w$

Example

Let $p_{1}(x)=x^{2}-2x+3$, $p_{2}(x)=2x^{2}+x+8$, and $p_{3}(x)=x^{2}+8x+7$ be in $P_{3}$.

${\begin{aligned}c_{1}\,p_{1}(x)+c_{2}\,p_{2}(x)+c_{3}\,p_{3}(x)&=0\\c_{1}(x^{2}-2x+3)+c_{2}(2x^{2}+x+8)+c_{3}(x^{2}+8x+7)&=0\\(c_{1}+2c_{2}+c_{3})x^{2}+(-2c_{1}+c_{2}+8c_{3})x+(3c_{1}+8c_{2}+7c_{3})&=0x^{2}+0x+0\end{aligned}}$

Therefore, since coefficients of terms must be equal,

${\begin{aligned}c_{1}+2c_{2}+c_{3}&=0\\-2c_{1}+c_{2}+8c_{3}&=0\\3c_{1}+8c_{2}+7c_{3}&=0\end{aligned}}$

${\begin{vmatrix}1&2&1\\-2&1&8\\3&8&7\end{vmatrix}}=0$, so the matrix is singular, and therefore the polynomials $p_{1}(x),p_{2}(x),p_{3}(x)$ are linearly dependent.

Wronskian Theorem

The following determinant of a matrix of functions and derivatives

$W[f_{1},\ldots ,f_{n}](x)={\begin{vmatrix}f_{1}(x)&f_{2}(x)&\dots &f_{n}(x)\\f_{1}'(x)&f_{2}'(x)&\dots &f_{n}'(x)\\\vdots &\vdots &\ddots &\vdots \\f_{1}^{(n-1)}(x)&f_{2}^{(n-1)}(x)&\dots &f_{n}^{(n-1)}(x)\end{vmatrix}}$

Is called the Wronskian

Let $f_{1},\ldots ,f_{n}$ be $n-1$-th differentiable functions along the interval $[a,b]$. If there exists a point $x_{0}$ in $[a,b]$ such that $W[f_{1},\ldots ,f_{n}](x)\neq 0$, then $f_{1},\ldots ,f_{n}$ are linearly independent.