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I have to solve the following exercise:

Use the Wronskian to check whether the given set of functions is linearly dependent:

$$f_1(t) = 2t -3, \,f_2(t) = 2t^2 +1, \,f_3(t) = 3t^2 + t$$

How should I solve this exercise? I only know how to use the Wronskian for a system that is at least $2\times 2$!

Edit: I understand that if $f_1$ and $f_2$ are of dimension $n$, $n>1$, that in order to check whether they're linearly dependent we can look at the determinant of the resulting system. What I don't understand is that if $f_1$ and $f_2$ are one-dimensional we can just look at the determinants of $f_1$ and $f_2$. I only need to show linear (in)dependence between $f_1$ and $f_2$, not between $f = [f_1, f'_1]$ and $g = [f_2,f'_2]$, right?

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Form the matrix of derivatives,

$$\begin{bmatrix}f_1(t) & f_2(t) & f3(t)\\ f'_1(t) & f'_2(t) & f'_3(t)\\ f''_1(t) & f''_2(t) & f''_3(t)\end{bmatrix}$$

and then take the determinant. If the determinant of this matrix, which is what the wronskian actually is, is not identically zero on an interval then the functions are linearly independent on this interval.

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  • $\begingroup$ Thanks for your reply! Why does this work though? What do the derivatives of $f_1, f_2,$ and $f_3$ have to do with the linear dependence of $f_1,f_2$ and $f_3$? $\endgroup$ – Mr. President Mar 6 '18 at 14:44
  • $\begingroup$ If $f_1, f_2, f_3$ are linearly dependent on an interval then so are their derivatives and thus the columns of the matrix above will be linearly dependent. Then, the determinant will be zero on this interval. Do you undersatnd why the determinant tells us about linear independence for ordinary column vectors? $\endgroup$ – TSF Mar 6 '18 at 14:48
  • $\begingroup$ Yes I do. If these columns are linearly dependent then we have that there is no non-trivial solution and the determinant of the system will be zero. What I don't understand is why linear (in)dependence between $f_1$ and $f_2$ is necessarily accompanied by linear (in)dependence between $f'_1$ and $f'_2$ $\endgroup$ – Mr. President Mar 6 '18 at 15:03
  • $\begingroup$ Assume that $f_1$ and $f_2$ are linearly dependent (and to avoid trivial cases, neither function is identically 0). Then, there are nonzero constants $\alpha, \beta$ such that $\alpha f_1(t) + \beta f_2(t) =0$ identically. This means that $f_1 = -\frac{\beta}{\alpha}f_2(t)$. Taking the derivative gives that $f'_1 = -\frac{\beta}{\alpha}f'_2(t)\implies \alpha f'_1(t) + \beta f'_2(t) =0$ identically as well. This ultimately stems from the fact that the derivative is a linear operator. $\endgroup$ – TSF Mar 6 '18 at 15:13
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It's just the same except it has three lines and three columns $$W=\left |\pmatrix{f_1 &f_2 & f_3 \\f_1' &f_2' & f_3' \\f_1'' &f_2'' & f_3'' }\right |=\left |\pmatrix{2t-3 &2t^2+1 & 3t^2+t\\2 &.. & ... \\0 & 4 & 6 }\right |$$

Note that:

$$f_3(t) = \frac 3 2 f_2+\frac 12f_1$$

So you already now the answer

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