Linear Algebra: Eigenvectors and Orthonormal Bases

Niles
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Homework Statement


Consider a symmetric (and hence diagonalizable) n x n matrix A. The eigenvectors of A are all linearly independant, and hence they span the eigenspace Rn.

Since the matrix A is symmetric, there exists an orthonormal basis consisting of eigenvectors.

My questions are:

1) Will this orthonormal basis of eigenvectors also span the same space Rn?

2) If two vectors are linearly independant, will they also be orthogornal?
 
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For 1): when does a set of vectors span the vector space? Do the eigenvectors satisfy these conditions? [Actually, you already gave the answer yourself... do you see where? ]

For 2): Consider (1, 0) and (1, 1) in R2.
 
1) They ar elinearly dependant, so yes - I guess I answered my own question there!

2) Great, a counter-example, so no. Thanks!
 
1) Yep, it follows from the fact that "the eigenvectors are linearly independent" and that there are n of them. That is, they form a basis, as you said in the question. And of course a basis always spans the space (even a non-orthogonal and/or not normalized one).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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