There's a question in charles curtis linear algebra book which states:(adsbygoogle = window.adsbygoogle || []).push({});

Let ##f1, f2, f3## be functions in ##\mathscr{F}(R)##.

a. For a set of real numbers ##x_{1},x_{2},x_{3}##, let ##(f_{i}(x_{j}))## be the ##3-by-3## matrix

whose (i,j) entry is ##(f_{i}(x_{j}))##, for ##1\leq i,j \leq 3##. Prove that ##f_{1}, f_{2}, f_{3}## are linearly independent if the rows of the matrix ##(f_{i}(x_{j}))## are linearly independent.

Obviously if ##f_1,f_2,f_3## are in terms of basis vectors than they are linearly independent.

But can I say that if matrix ##A = (f_{i}(x_{j}))## is linearly independent, then they are in echelon form.

Therefore, I can row reduce the matrix to a diagonal matrix s.t. ## a_{i,i} \neq 0##.

Since the rows are linearly independent then ##f_{i} \neq 0 \quad 1 \leq i \leq 3##, therefore for

##\alpha_{i} f_{i} = 0## only if ##\alpha_{i} = 0##.

Is this a good proof for that question?

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# Show linear independence

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