Linear Algebra with functions

Basically my question is - How to check for linear independence between functions ?!

Let the group $\mathcal{F}(\mathbb{R},\mathbb{R})$ Be a group of real valued fnctions.

i.e $\mathcal{F}(\mathbb{R},\mathbb{R})=\left\{ f:\mathbb{R}\rightarrow\mathbb{R}\right\}$

Let 3 functions $f_{1},f_{2},f_{3}$ be given such that

$\forall x\in\mathbb{R}\,\,\,f_{1}=e^{x},\,\,f_{2}=e^{2x},\,\,f_{3}=e^{3x}$

$W=sp(f_{1},f_{2},f_{3})$ what is $dim(W)$ ?

How to approach this question ? (from a linear algebra perspective)

I know that $\forall x\in\mathbb{R}\,\,\,W=\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}$

And to get the dimension I need to find the base of $W$

so I need to check whether the following holds true :

$\forall x\in\mathbb{R}\,\,\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\,\Leftrightarrow\,\alpha,\beta,\gamma=0$

However when $x=0$ I get $\alpha+\beta+\gamma=0$ which leads to infinite amount of solutions.

How to approach this question ?

• Hint: You want it to be zero for EVERY $x$, not only for $x = 0$. – amcalde Apr 27 '16 at 19:27
• I know that I want it, I dont know whether I can reach it, or how to verify that I can never reach it ? – Pavel Penshin Apr 27 '16 at 19:28
• I like to use this method: math.stackexchange.com/a/269694/8157 but the linked Q&A contains many others. – Giuseppe Negro Apr 27 '16 at 20:22

You need to check if the functions are independent, as you said.

A way to go about this, which that ties it in with things you likely know is to evaluate it at several points, as you did for $x=0$.

You get one condition for $x=0$. You get another condition for $x=1$ and still another one for $x=2$.

Each will allow more than one solution, but they'll only have one common solution, which is what you are after.

• $$\begin{array}{cc} x=0 & \alpha+\beta+\gamma=0\\ x=1 & \alpha e+\beta e^{2}+\gamma e^{3}=0\\ x=2 & \alpha e^{2}+\beta e^{4}+\gamma e^{6}=0 \end{array}$$ $\rightarrow\begin{bmatrix}1 & 1 & 1\\ e & e^{2} & e^{3}\\ e^{2} & e^{4} & e^{6} \end{bmatrix}\rightarrow$ I found this matrix, the determinant is not zero thus there is only 1 solution which means that $\alpha,\beta,\gamma=0$ for $x=1,2,3$ how does that helps ? – Pavel Penshin Apr 27 '16 at 19:54
• Note that you need $\alpha, \beta, \gamma$ that work for all $x$ at the same time (they must not depend on $x$). You just showed that for $\alpha, \beta, \gamma$ to work for $x=0,1,2$ you already only have the unique choice all $0$. So you are done. – quid Apr 27 '16 at 20:23

Write $$\alpha e^x + \beta e^{2x} + \gamma e^{3x} = 0$$ You can go ahead and cancel out a positive number like $e^x$ so: $$\alpha + \beta e^{x} + \gamma e^{2x} = 0$$ Suppose you have some solution for this with $\alpha$, $\beta$, $\gamma$ not all zero. Then, as you say $$\alpha + \beta + \gamma = 0\qquad \qquad (1)$$ Because this must be true at $x = 0$ but it must also be true at $x = \ln n$ which gives: $$\alpha + \beta n + \gamma n^2= 0\qquad \qquad (2)$$ for every $n > 1$. It should be clear that this is unsolvable except when they are all zero. But to press the point I'll continue. Substituting in $(1)$ gives $\alpha = -\beta - \gamma$, which we can plug into $(2)$ to get $$\beta (n-1) + \gamma (n^2 - 1)= 0$$ which must be true for all $n > 1$. Now put, say, $n = 2$ and $n = 3$ to get the pair of equations: $$\beta + 3 \gamma = 0 \qquad 2\beta + 8\gamma = 0$$ This solves for $\beta = \gamma = 0$.

So your functions are proved to be linearly independent.

Hint:

let $e^x=y$, $e^{2x}=y^2$, $e^{3x}=y^3$ you have:

$\alpha y +\beta y^2+ \gamma y^3=0$

where the $0$ at RHS is the zero polynomial.

Now: when a polynomial is the zero polynomial?

In general:

The $0$ at RHS is the neutral element for the sum of functions in the vector space, not simply the number $0$ and this means that it is the function $f(x)=0\quad \forall x \in \mathbb{R}$.

• $$\begin{array}{cc} x=0 & \alpha+\beta+\gamma=0\\ x=1 & \alpha e+\beta e^{2}+\gamma e^{3}=0\\ x=2 & \alpha e^{2}+\beta e^{4}+\gamma e^{6}=0 \end{array}$$ $\rightarrow\begin{bmatrix}1 & 1 & 1\\ e & e^{2} & e^{3}\\ e^{2} & e^{4} & e^{6} \end{bmatrix}\rightarrow$ I found this matrix, the determinant is not zero thus there is only 1 solution which means that $\alpha,\beta,\gamma=0$ for $x=1,2,3$ how does that helps ? what is RHS ? googling didnt help :( – Pavel Penshin Apr 27 '16 at 19:57
• The key fact is that in $\alpha f_1+\beta f_2+\gamma f_3=0$ The $0$ is the zero function i.e. a function that is null for all values of $x$ in the domain. Your linear system shows that you can find values for $\alpha, \beta, \gamma$ such that $\alpha f_1+\beta f_2+\gamma f_3=0$ is true for some value of $x$ but not for all the possible values. – Emilio Novati Apr 27 '16 at 20:06

Hint: Use Wronskian and show that the Wronskian-Determinant does not vansish.

You have to prove $$\forall x\in\mathbb{R}:\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\Leftrightarrow\alpha,\beta,\gamma=0,$$

but I think the quantifier applies only to the part on the left side of the $\Leftrightarrow$, like this: $$\left(\forall x\in\mathbb{R}:\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\right) \Leftrightarrow\,\alpha,\beta,\gamma=0.$$

So for example $\alpha = -1, \beta = 1, \gamma = 0$ satisfies $\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0$ when $x=0$, but it doesn't satisfy the equation for all values of $x$.

If you had to prove $$\forall x\in\mathbb{R}:\left(\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0 \Leftrightarrow\,\alpha,\beta,\gamma=0\right)$$ then you would be in trouble, because that statement is not true; but that's not how we prove independence of the functions, so you don't need to worry about that.

You need to show the three vectors are linearly independent. In this case I would use this trick; so that you don't need to worry about them being functions and the equality to hold for every value of $x$.

If you consider $D: \mathcal{F} \rightarrow \mathcal{F}$, the derivative operator, is an endomorphism in $\mathcal F$ (i.e. a linear map from $\mathcal{F}$ to itself). The derivatives of the three functions are $$Df_1=De^x=e^x=f_1$$ $$Df_2=De^{2x}=2e^{2x}=2f_2$$ $$Df_3=De^{3x}=3e^{3x}=3f_3$$ So $f_1,f_2,f_3$ are eigenvectors of $D$, with eigenvalues $\lambda_1=1, \lambda_2=2, \lambda_3=3$, respectively. Since $f_1, f_2, f_3$ are eigenvectors with distinct eigenvalues of the same endomorphism $D$, they are linearly independent so they form a base for $W$ and $\text{dim}W=3$.

If you have $$\alpha e^{x} + \beta e^{2x} + \gamma e^{3x} \equiv 0,$$ Then you can apply the derivative operator $D$ to obtain \begin{align} 0 & \equiv (D-2)(D-3)\{\alpha e^{x} + \beta e^{2x} + \gamma e^{3x}\} \\ & = (1-2)(1-3)\alpha e^{x}. \end{align} Therefore $\alpha=0$. Then you can apply $(D-1)(D-3)$ in order to conclude that $\beta=0$. Similarly $\gamma =0$. So $\{ e^x,e^{2x},e^{3x} \}$ is a linearly independent set of functions, which means that the dimension of $W$ is $3$.

• Thanks for your answer I did not quite understand your notation for the derivative operator. also, can this answer be obtained in another way ? (without Wronskian as well ) – Pavel Penshin Apr 27 '16 at 21:16
• @user313448 Have you studied differential equations where they use the annihilator method? That's what I'm using. If not, you can do this with limits. Multiply by $e^{-3x}$ and let $x\rightarrow\infty$ in order to obtain $\gamma =0$. Then you can isolate $\beta=0$ and, finally, you isolate $\alpha=0$ with no limits. – DisintegratingByParts Apr 27 '16 at 21:34