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\begin{align} f(t)= \begin{cases} t^2 & t\leq 0\\ 0 & t\geq 0 \end{cases} \end{align} \begin{align} g(t)= \begin{cases} 0 & t\leq 0\\ t^2 & t\geq 0 \end{cases} \end{align}

Consider the expression \begin{align} \lambda_{1} f+\lambda_{2} g=0 \end{align}

For $t<0$, $g(t)=0$ , so the expression has a non-trivial solution.

For $t\geq0$, $f(t) = 0$ , so the expression also has a non-trivial solution.

How do i show that $f$ and $g$ are indeed linearly independent?

What went wrong in my reasoning above?

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    If $f$ and $g$ are linearly dependent, then they must be scalar multiples of each other. – 2017-02-12
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    i know of this, but im asking what went wrong with my reasoning above – 2017-02-12
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    For $t<0$ we have $$0=\lambda_1f(t)+\lambda_2g(t)=\lambda_1t^2+\lambda_2\cdot 0=\lambda_1t^2$$ which means $\lambda_1=0$. Similarly $\lambda_2=0$. Thus $\lambda_1f+\lambda_2g=0$ if and only if $\lambda_1=\lambda_2=0$, i.e. $f,g$ are linearly independent. – 2017-02-12
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    Why couldn't $\lambda_{2} \neq 0$ ? Any scalar multiplication on the $0$ constant function has no effect. – 2017-02-12
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    Because for $t>0$ we have $$0=\lambda_1f(t)+\lambda_2g(t)=\lambda_1\cdot 0+\lambda_2t^2=\lambda_2t^2.$$ This forces $\lambda_2=0$. – 2017-02-12
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    I see, thanks :) – 2017-02-12
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    You're welcome:) – 2017-02-12
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    $f$ and $g$ are linearly dependent iff there are constants $A, B$ , not both $0,$ such that $Af(t)+Bg(t)=0$ for ALL $t.$ – 2017-02-12

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