MATH 323 Lecture 11

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Linear Independence

1. If $v_1,\dots ,v_n$ 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 $v_1,\dots ,v_n\in V$, it is possible to write one of the vectors as a linear combination of the other $n-1$ vectors iff there exist scalars $c_1,\dots ,c_n$ (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 $v_1,\dots ,v_n$.

The vectors $v_1,\dots ,v_n\in V$ 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 $c_1\ne 0$, then $v_1=-\frac{c_2}{c_1}{v}_2$
2. If $c_2\ne 0$, 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. $\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$ Yes.
2. $\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$ Yes.
3. $\left(\begin{array}{c}1\\ 2\\ 4\end{array}\right),\left(\begin{array}{c}2\\ 1\\ 3\end{array}\right),\left(\begin{array}{c}4\\ -1\\ 1\end{array}\right)$ 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 $⟨c_1,c_2,c_3⟩$ that will satisfy the equation. (See #Theorem)

Theorem

Let $x_1,\dots ,x_n$ be $n$ vectors in ${\mathbb{ℝ}}^n$ and let $X=(x_1,\dots ,x_n)$ be the $n\times n$ matrix formed by using the $x$ vectors as columns.

The vectors $x_1,\dots ,x_n$ will be linearly dependent iff $X$ is singular and linearly independent iff $X$ is nonsingular

Proof

Let $\overrightarrow{c}=⟨c_1,\dots ,c_n⟩$, then $X\overrightarrow{c}=\overrightarrow{0}$ has a nontrivial solution iff $X$ is singular.

For Non-Square Matrix

For $x_1,\dots ,x_k\in \mathbb{ℝ}$, the system $c_1{x}_1\dots c_k{x}_k=0$ can be written as $X\overrightarrow{c}=\overrightarrow{0}$, where $X=(x_1,\dots ,x_k)$ is an $n\times k$ matrix. Therefore if $n\ne k$, the determinant of $X$ is not defined and #Theorem does not apply.

However, the system has nontrivial solutions (i.e. $x_1,\dots ,x_n$ is linearly dependent) iff the row echelon form of $X$ has at least one free variable.

Example

$x_1=⟨1,-1,2,3⟩$, $x_2=⟨-2,3,1,-2⟩$, and $x_3=⟨1,0,7,7⟩$

$\left[\begin{array}{cccc}1& -2& 1& 0\\ -1& 3& 0& 0\\ 2& 1& 7& 0\\ 3& -2& 7& 0\end{array}\right]⟶\left[\begin{array}{cccc}1& -2& 1& 0\\ 0& 1& 1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

Therefore, since $c_3$ is a free variable, $\{x_1,x_2,x_3\}$ is linearly dependent.

Theorem

Let $v_1,\dots ,v_n\in V$. A vector $w\in \mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(v_1,\dots ,v_n)$ can be written uniquely as a linear combination of $v_1,\dots ,v_n$ iff $v_1,\dots ,v_n$are linearly independent.

Proof

$w\in \mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(v_1,\dots ,v_n)⟺v=c_1{v}_1+\dots c_n{v}_n$

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 ${\alpha }_i\ne {\beta }_i$ for some $i$. This would mean that $0=({\alpha }_1-{\beta }_1)v_1+\dots +({\alpha }_n-{\beta }_n)v_n$, where ${\alpha }_i-{\beta }_i\ne 0$, would be linearly dependent.

Therefore, if $v_1,\dots ,v_n$ are linearly dependent, then there exist $c_1,\dots ,c_n$ (not all zero) such that

$\begin{array}{rl}0& =c_1{v}_1+\dots +c_n{v}_n\\ +w& ={\alpha }_1{v}_1+\dots {\alpha }_n{v}_n\\ =w& =({\alpha }_1+c_1)v_1+\dots +({\alpha }_n+c_n)v_n\end{array}$

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 $P_3$.

$\begin{array}{rl}{c}_1{p}_1(x)+c_2{p}_2(x)+c_3{p}_3(x)& =0\\ c_1(x^2-2x+3)+c_2(2x^2+x+8)+c_3(x^2+8x+7)& =0\\ (c_1+2c_2+c_3)x^2+(-2c_1+c_2+8c_3)x+(3c_1+8c_2+7c_3)& =0x^2+0x+0\end{array}$

Therefore, since coefficients of terms must be equal,

$\begin{array}{rl}{c}_1+2c_2+c_3& =0\\ -2c_1+c_2+8c_3& =0\\ 3c_1+8c_2+7c_3& =0\end{array}$

$\left|\begin{array}{ccc}1& 2& 1\\ -2& 1& 8\\ 3& 8& 7\end{array}\right|=0$, 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

$W[f_1,\dots ,f_n](x)=\left|\begin{array}{cccc}{f}_1(x)& f_2(x)& \dots & f_n(x)\\ {f_1}^{\prime }(x)& {f_2}^{\prime }(x)& \dots & {f_n}^{\prime }(x)\\ ⋮& ⋮& \ddots & ⋮\\ f_1^{(n-1)}(x)& f_2^{(n-1)}(x)& \dots & f_n^{(n-1)}(x)\end{array}\right|$

Is called the Wronskian

Let $f_1,\dots ,f_n$ be $n-1$-th differentiable functions along the interval $[a,b]$. If there exists a point $x_0$ in $[a,b]$ such that $W[f_1,\dots ,f_n](x)\ne 0$, then $f_1,\dots ,f_n$ are linearly independent.