MHB How can the Wronskian be used to determine linear independence?

[Guest2]

I'm asked to check whether $\left\{1, e^{ax}, e^{bx}\right\}$ is linearly independent over $\mathbb{R}$ if $a \ne b$, and compute the dimension of the subspace spanned by it. Google said the easiest way to do this is something called the Wronskian. Is this how you do it? The matrix is:

$ \begin{aligned} \begin{bmatrix}1 & e^{ax} & e^{bx} \\ 0 & a e^{ax} & be^{bx} \\ 0 & a^2 e^{ax} & b^2e^{bx}\end{bmatrix} =\begin{bmatrix}1 & e^{ax} & e^{bx} \\ 0 & a e^{ax} & be^{bx} \\ 0 & 0 & b^2e^{bx}-abe^{bx}\end{bmatrix}\end{aligned}$

Which is in the upper triangular form, therefore $\mathcal{W}(1, e^{ax}, e^{bx}) =ae^{ax}(b^2e^{bx}-ab e^{bx})$ and

$ae^{ax}(b^2e^{bx}-ab e^{bx}) =0 \implies a=0, ~b=0, ~ a = b$ Thus $\left\{1, e^{ax}, e^{bx}\right\}$ is linearly independent if $a \ne b$.

The subspace spanned by $\left\{1, e^{ax}, e^{bx}\right\}$ is $l(x) = \left\{\lambda_1 +\lambda_2 e^{ax}+\lambda_3 e^{bx}: \lambda_1, \lambda_2, \lambda_3 \in \mathbb{R}\right\}$

It's a basis for this subspace since it's linearly independent and it spans it. So $\text{dim}(l(x)) = 3$.

 

[Euge]

Hi Guest,

There should be another condition, namely, $a$ and $b$ are nonzero. Otherwise, the set $\{1,e^{ax},e^{bx}\}$ will contain two of the same elements (the element $1$), which means that the set would be linearly dependent.

The matrix is:

$ \begin{aligned} \begin{bmatrix}1 & e^{ax} & e^{bx} \\ 0 & a e^{ax} & be^{bx} \\ 0 & a^2 e^{ax} & b^2e^{bx}\end{bmatrix} =\begin{bmatrix}1 & e^{ax} & e^{bx} \\ 0 & a e^{ax} & be^{bx} \\ 0 & 0 & b^2e^{bx}-abe^{bx}\end{bmatrix}\end{aligned}$

Did you mean determinant? Even so, why are the two equal?

To prove that $\{1,e^{ax}, e^{bx}\}$ is a linearly independent set of functions, I'll suppose there is a linear dependence relation

$$c_1 + c_2 e^{ax} + c_3 e^{bx} = 0\tag{*}$$

where $c_1,c_2,c_3\in \Bbb R$, and show that $c_1 = c_2 = c_3 = 0$. Evaluating at $x = 0$ gives

$$c_1 + c_2 + c_3 = 0.$$

Taking the derivative of $(*)$ with respect to $x$ and evaluating at $x = 0$, we get

$$ac_2 + bc_3 = 0,$$

or $ac_2 = -bc_3$. Finally, taking the second derivative of $(*)$ with respect to $x$ and evaluating at $x = 0$, we obtain

$$a^2 c_2 + b^2 c_3 = 0,$$

that is, $a^2 c_2 = -b^2 c_3$. Therefore

$$a^2 c_2 = -b^2 c_3 = b(-bc_3) = b(ac_2) = abc_2.$$

So $a(a - b)c_2 = (a^2 - ab)c_2 = 0$. Since $a\neq 0$ and $a\neq b$, we must have $c_2 = 0$. Now $bc_3 = -ac_2 = 0$, so as $b\neq 0$ we have $c_3 = 0$. Finally, $0 = c_1 + c_2 + c_3 = c_1 + 0 + 0 = c_1$. We have now shown that $c_1 = c_2 = c_3 = 0$.

It's a basis for this subspace since it's linearly independent and it spans it. So $\text{dim}(l(x)) = 3$.

That's right.

 

[Guest2]

Hi, Euge. Thanks for such a nice way of doing this.

Did you mean determinant? Even so, why are the two equal?

I meant the matrix. I should have used $\to$, not $=$. I row reduced the matrix (basically subtracted $a$ times row $2$ from row $3$) to echelon form so that I could extract the determinant as product of entries of the diagonal: $1 \cdot ae^{ax} \cdot (b^2e^{bx}-ab e^{bx})$ and this is only zero when $a=0, b = 0$ (which I missed earlier), and when $a=b$. My book failed to mention the condition that $a,b$ are nonzero.