Linear Dependence of Three Functions.

Three functions y₁(t), y₂(t)and y₃(t)are said to be linearly dependent on an intervalif, on l, at least one of these functions is a linear combination of the remaining two e.g., if y₁(t) = c₁y₂(t) + c₂y₃(t). Equivalently (compare Problem), y₁,y₂and y₃are linearly dependent on lif there exist constants C₁,C₂and C₃, not all zero, such that C₁y₁(t) + C₂y₂(t) + C₃y₃(t) = 0for all tin l. Otherwise, we say that these functions are linearly independent on. For each of the following, determine whether the given three functions are linearly dependent or linearly independent on $\left(-\infty ,\infty \right)$:

role="math" localid="1655805193940" $\begin{array}{rcl}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{=}& \mathbf{1}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{t}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{y}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{.}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{=}& \mathbf{-}\mathbf{3}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{5}{\mathbf{sin}}^{\mathbf{2}}\mathbf{t}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{y}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{cos}}^{\mathbf{2}}\mathbf{t}\mathbf{.}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{=}& {\mathbf{e}}^{\mathbf{t}}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{te}}^{\mathbf{t}}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{y}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{.}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{}{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{=}& {\mathbf{e}}^{\mathbf{t}}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{y}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{cosht}\mathbf{.}\end{array}$

Three functions y₁,y₂ and y₃ are linearly dependent on an interval I if there are some constants c₁,c₂ such that y_i(t) = c₁y_j(t) +c₂y_k(t) for $\mathrm{t}\in \mathrm{l}$, where $i,j,k\in \{1,2,3\}$and $\mathrm{i}\ne \mathrm{j},\mathrm{j}\ne \mathrm{k}$and $\mathrm{k}\ne \mathrm{i}$. Three functions are linearly independent if they are not linearly dependent.

Assume that those three functions are linearly dependent, i.e., there are some constants c₁,c₂ such that ${\mathit{y}}_{\mathbf{3}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}}\mathbf{\iff }{\mathit{t}}^{\mathbf{2}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{t}$

But in the last equation, we have a polynomial of degree 2 on the left side and a polynomial of degree 1 on the right side, so there are no constants c₁,c₂ such that this equality is true for all $\in (-\infty ,\infty )$. We come to the same conclusion in cases t = c₁ + c₂t² and 1 = c₁t + c₂t². So, those functions are linearly independent.

From the trigonometry identity sin²t + cos²t = 1, we have that the function y(t) can be transformed to y₃(t) = 1-sin2t for all $\in (-\infty ,\infty )$.

Now,

$\begin{array}{rcl}{y}_3\left(t\right)& =& 1-{\sin }^{2}t\\ & =& \frac{-3}{-3}-\frac{5}{5}{\sin }^{2}t\\ & =& \frac{1}{-3}\times (-3)-\frac{1}{5}\times \left(5{\sin }^{2}t\right)\\ & =& -\frac{1}{3}{y}_1\left(t\right)-\frac{1}{5}{y}_2\left(t\right)\end{array}$

Therefore, if we take $c_1=-\frac{1}{3}$and$c_2=-\frac{1}{5}$ one has that${\mathit{y}}_{\mathbf{3}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$ for $\in (-\infty ,\infty )$, so the given functions are linearly dependent.

Assuming that those functions are linearly dependent, without loss of generality, one has that there are some constants c₁,c₂ such that ${\mathit{y}}_{\mathbf{3}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{\iff }{\mathit{t}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{t}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{t}{\mathit{e}}^{\mathbf{t}}$.

Dividing the previous equation by $e^t\left(e^t\ne 0\right)$we will get t² = c₁ + c₂t, and as in part (a), one has that there are no constants c₁,c₂ such that the previous equality is true for all $\in (-\infty ,\infty )$, so our assumption that those functions are linearly dependent is wrong.

Those functions are linearly independent.

$\cos ht=\frac{e^t+e^{-t}}{2}$(Definition of cosht )

If one takes constant$c_1=c_2=\frac{1}{2}$ one will have that y₃(t) = c₁y₁(t) + c₂y₂(t),$\mathrm{t}\in (-\infty ,\infty )$so those three functions are linearly dependent.