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$T$ is a linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$ such that $T(u) = u$, $T(v) = 2v$, $T(w) = 3w$, where $u,v,w$ are non-zero. Then which are these are necessarily true:

  1. $det(T) = 6$.
  2. ${u,v,w}$ is a basis of $\mathbb{R}^3$.
  3. $T$'s characteristic polynomial is $(x-1)(x-2)(x-3)$.

Or is there not enough information?

Attempt

If ${u,v,w}$ are the standard basis of $\mathbb{R}^3$, then all three results follow. But I am kinda stuck when considering ${u,v,w}$ that are not the standard basis.

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You can write down the matrix of a map with respect to any basis that you like. So if you knew 2, you would know 1 and 3 (for the same reason you know it when u, v, and w are the standard basis vectors). But how can eigenvectors with distinct eigenvalues be linearly dependent? –  user29743 Nov 4 '12 at 2:49
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1 Answer

up vote 5 down vote accepted

Assuming $u,v,w$ are nonzero,

on the hypotheses, 1, 2, and 3 are all eigenvalues of $T$. It's not hard to prove that eigenvectors belonging to distinct eigenvalues are linearly independent, so $u,v,w$ form a basis. The characteristic polynomial has the eigenvalues as roots, and the determinant is the product of the eigenvalues.

but if even one of $u,v,w$ is zero, 2 is false, and 1 and 3 might be false.

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Could the question be a trick question and maybe we're missing a hypothesis on the vectors? Because if one of them is the $0$ vector then certainly the statements would not necessarily hold, and statement 2 would be false. –  Adrián Barquero Nov 4 '12 at 2:48
    
@Adrián, you're absolutely right. I will edit. –  Gerry Myerson Nov 4 '12 at 3:00
    
Gosh I missed the key insight of seeing them as eigenvectors/eigenvalues! I'll edit in the fact that $u,v,w$ are non-zero. Thanks! –  Legendre Nov 4 '12 at 3:28
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