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I have a math problem I am struggling with:

If a linear transform $A: \mathbb{R}^n\to\mathbb{R}^n$ and we have a basis of $\mathbb{R}^n$ of eigenvectors of $A$, can't we just orthonormalize them and get a matrix $P$ such that $P^{-1}=P^\mathrm{T}$ and thus $P^\mathrm{T}AP$ is diagonal?

Solution (so far):

Let $B$ be the matrix of eigenvectors of $A$.

Now, I know that if $B$ is orthogonalizable, then the rest of the problem is true, and I also know that B is diagonalizable, because it is a basis of $\mathbb{R}^n$ and is therefore linearly independent. I also know (from prodding the professor) that the conjecture is false (i.e. we can't just orthonormalize $B$). I can't see how the process of normalizing the vectors in a matrix would cause a problem, so it must come down to whether the [linearly independent] vectors of $B$ are orthogonalizable.

Question:

Which matrix of linearly independent vectors is not orthogonalizable? (and why?) Thanks

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    http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process2012-11-09
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    A good advice is to try Latex code rather than -->. ^^2012-11-09
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    What do you mean by "orthogonalizable"2012-11-09
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    If you have a basis, you can obtain an orthornormal basis by a process called Gram-Schmidt process. Check this out at wikipedia, you'll see the details.2012-11-09
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    I know about gram-schmidt. But that doesn't tell you when a matrix isn't orthogonalizable, except if the norm is zero, which it can't be.2012-11-09
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    By orthogonalizable, I mean make the vectors orthogonal.2012-11-09

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