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Let $\mathcal{C}(\mathbb{R})$ be the $\mathbb{R}$-vector space of continuous functions from $\mathbb{R}$ to $\mathbb{R}$. Let $a_1$, $a_2$, $a_3$ be distinct real numbers. For $i = 1, 2, 3$, let $f_i \in \mathcal{C}(\mathbb{R})$ be the function $f_i(t)=e^{a_it}$ . Which of the following statement(s) is/are true?

(A) $f_1$, $f_2$ and $f_3$ are linearly independent

(B) $f_1$, $f_2$ and $f_3$ are linearly dependent.

(C) $f_1$, $f_2$ and $f_3$ form a basis of $\mathcal{C}(\mathbb{R})$.

My attempt: my answer is option A and C, because if we check the linearly independent that is,,,,, $c_1f_1 + c_2f_2 + c_3f_3 =0$ , $f_i(t)=e^{a_it} \neq 0$ where $c_1= c_2 = c_3= 0$,,,so $f_1$, $f_2$ and $f_3$ will form a basis of $\mathcal{C}(\mathbb{R})$ that mean $f_1$, $f_2$ and $f_3$ are linearly independent..so correct option is opton A and option C,,,

Is my answer is correct or incorrect. pliz verified my answer and tell me the solution,

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  • $\begingroup$ Consult this link. $\endgroup$
    – Mikasa
    Oct 28, 2017 at 9:06

1 Answer 1

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No, the correct answer is A. They don't for a basis of $\mathcal{C}(\mathbb{R})$. For instance, you cannot express the identity as a linear combination of your functions.

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  • $\begingroup$ @ josecarlos santos,,,,,im not getting...as i have expressed c1f2 +c2f2+c3f3 = 0 where C1=c2=c3= 0,,,,,here it is clearly see that f1 ,f2 and f3 are linearly independent,,,,so they will forms basis,, $\endgroup$
    – jasmine
    Oct 28, 2017 at 9:14
  • $\begingroup$ @lomberlego How do you define basis? $\endgroup$
    – José Carlos Santos
    Oct 28, 2017 at 9:16
  • $\begingroup$ @ jose Carlos santos,,here i m getting three function,,,f1(t)=e^a1t , and f2(t) =e^a2t and f3(t) = e^a3(t)....these three are function are linearly independent,,,so they will form basis $\endgroup$
    – jasmine
    Oct 28, 2017 at 9:21
  • $\begingroup$ @lomberlego Please answer my question: how do you define basis? $\endgroup$
    – José Carlos Santos
    Oct 28, 2017 at 9:24
  • $\begingroup$ @ jose carlos santos,,,, collection B = { v 1, v 2, …, v r } of vectors from V is said to be a basis for V if B is linearly independent and spans V. ...... $\endgroup$
    – jasmine
    Oct 28, 2017 at 9:30

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