Uncountable Infinite Basis

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The question and my partial solution is in the link. Please help me finish it off. Thanks.
 
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matt grime

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Please post your work in a non-proprietary format that those of us not blessed with MS word can read easily. Freely available programs that open Word documents are usually terrible for those that contain anything other than the plainest text (i.e. any maths content will be completely buggered).

Better yet, just type it in to the forum directly.
 
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When I tried to copy the contents of my Word file onto here, the symbols did not show. Ok, so I took a gif picture of my Word file. The links are below.
 
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AKG

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You got confused with the notation. At one moment, the hint is using the subscript on fr to determine the function, i.e. fr = (1, r, r2, ...). Next, they are using the subscripts to index, so when they talk about f1, ..., fn, they just mean any n functions (from T), i.e. not that fn = (1, n, n2, ...). To be more clear, let's instead think about

[tex]f_{r_1},\, f_{r_2},\, \dots ,\, f_{r_n}[/tex]

If these are linearly dependent, then there exists a set of reals, {a1, ..., an} not all zero such that:

[tex]\sum _{i = 1} ^n a_if_{r_i} = 0[/tex]

Thus, for ANY vector v in V, we get:

[tex]\sum _{i = 1} ^n a_if_{r_i}(v) = 0(v) = 0[/tex]

In particular, this holds (simultaneously) for v0, v1, ..., vn-1, so we get:

[tex]\sum _{i = 1} ^n a_if_{r_i}(v_0) = \sum _{i=1} ^n a_i = 0[/tex]

[tex]\sum _{i = 1} ^n a_if_{r_i}(v_0) = \sum _{i=1} ^n a_ir_i = 0[/tex]

.
.
.

[tex]\sum _{i = 1} ^n a_if_{r_i}(v_0) = \sum _{i=1} ^n a_ir_i^{n-1} = 0[/tex]

So:

[tex]\left(
\begin{array}{cccc}
1 & 1 & \dots & 1\\
r_1 & r_2 & \dots & r_n \\
\vdots & \vdots & \ddots & \vdots \\
r_1^{n-1} & r_2^{n-1} & \dots & r_n^{n-1}
\end{array}
\right)
\left(
\begin{array}{c}
a_1\\
a_2\\
\vdots \\
a_n
\end{array}
\right)
=
\left(
\begin{array}{c}
0\\
0\\
\vdots \\
0
\end{array}
\right)[/tex]

You should be able to prove that the ai are all zero from here. Prove it by contradiction, and think about how many roots a k-degree polynomial can have. The fact that T is linearly independent then follows immediately. I don't think you need to show that T spans your space, because showing that T is a basis is overkill, is it not? If T is uncountable and linearly independent, then surely any (not the) basis for your space is uncountable.
 
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Thank you very much AKG. So the problem was that the hint was vaguely written. Had I known what they meant exactly, I could have found that eventually.

The columns of the nxn matrix in your last line are linearly independent and so the matrix is invertible. Left multiplying both sides of the equation by the inverse matrix results in the ai all being zero.
 
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AKG

Science Advisor
Homework Helper
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Thank you very much AKG. So the problem was that the hint was vaguely written. Had I known what they meant exactly, I could have found that eventually.

The columns of the nxn matrix in your last line are linearly independent and so the matrix is invertible. Left multiplying both sides of the equation by the inverse matrix results in the ai all being zero.
No, that's not it. You are trying to prove that the columns are linearly independent.
 
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I've already proven that the columns are linearly independent. I was just explaining the remaining proof after that.
 

AKG

Science Advisor
Homework Helper
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Oh, I see. I went about it a different way:

If there is a non-trivial choice of the ai that makes the linear combination of the fri 0, then there is a non-zero column vector in the kernel of that nxn matrix, which means that the columns are linearly dependent, which means that there is an (n-1)-degree polynomial with n distinct roots, contradiction. Thus there is no non-trivial choice for the ai.
 
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Ok, the problem is solved. AKG, please check your private mail. I believe your specialty is linear algebra, abstract algebra, topology, and vector calculus. The exercises that I will be doing for the next many months will be differential topology, differential geometry, tensor analysis, riemannian geometry.
 

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