Eigenvectors and orthogonal basis

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SUMMARY

This discussion centers on the impossibility of forming an orthogonal basis of eigenvectors for a specific linear transformation in ##\mathbb{R}^3##. The eigenvectors identified are ##(1,-3,0), (1,0,3), (\frac{1}{2}, \frac{1}{2},1)##. The user explores methods to demonstrate that these eigenvectors are not mutually orthogonal, considering techniques such as the Gram-Schmidt process and Schur decomposition. The implications of having an orthogonal basis include geometric advantages and numeric stability in computations.

PREREQUISITES
  • Understanding of linear transformations in ##\mathbb{R}^3##
  • Familiarity with eigenvectors and eigenvalues
  • Knowledge of the Euclidean inner product
  • Basic concepts of orthogonal and unitary matrices
NEXT STEPS
  • Learn how to compute the dot product of vectors in ##\mathbb{R}^3##
  • Study the Gram-Schmidt process for orthonormalization
  • Explore Schur decomposition and its applications in linear algebra
  • Investigate the geometric implications of orthogonal matrices in vector spaces
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, as well as anyone interested in the properties of eigenvectors and their applications in computational methods.

0kelvin
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Homework Statement


I have a linear transformation ##\mathbb{R}^3 \rightarrow \mathbb{R}^3##. The part that asks for a basis of eigenvectors I've already solved it. The possible eigenvectors are ##(1,-3,0), (1,0,3), (\frac{1}{2}, \frac{1}{2},1) ##. Now the exercise wants me to show that there is no orthogonal basis of eigenvectors for this particular linear transformation.

How do I show it?

The exercise doesn't ask this, but what's the implication of eigenvectors forming an orthogonal basis?

Homework Equations

The Attempt at a Solution


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With the euclidean inner product I can clearly see that the eigenvectors are not orthogonal to each other. But I'm not sure if calculating many pairs of dot products is the way to show it.

I thought about Gram-Schmidt but doing that would make the vectors not be eigenvectors anymore.
 
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If you have got three eigenvectors, then either they are mutually orthogonal or they are not.
 
0kelvin said:
Now the exercise wants me to show that there is no orthogonal basis of eigenvectors for this particular linear transformation.

How do I show it?

I suppose you could do a Schur decomposition, and show that you get a matrix ##\mathbf{UTU^*}## where ##\mathbf{T}## has eigenvalues along the diagonal but it is upper triangular only (i.e. it is not both upper triangular and lower triangular -- i.e. not diagonal). Out of curiosity, what are the eigenvalues here?

More importantly:

0kelvin said:
The exercise doesn't ask this, but what's the implication of eigenvectors forming an orthogonal basis?

Orthogonal (and unitary if complex) matrices are extremely pleasant to work with. From a geometric standpoint, they are length preserving. Consider some real valued vector ##\mathbf{x}## and an orhogonal matrix ##mathbf{U}##: ##||\mathbf{x}||_2^2 = \mathbf{x^Tx} = ||\mathbf{Ux}||_2^2 =\mathbf{x^TU^TUx} = \mathbf{x^TIx} = \mathbf{x^T x}##. Having mutually orthonormal eigenvectors is immensely useful in manipulating things like quadratic forms -- e.g. maximizing or minimizing ##\mathbf{x^T A x}##, which comes up all the time (e.g. Hessian matrix of second derivatives in calc).

On top of all this, orthogonal matrices are very nice for preserving numeric stability if you're doing serious computational work.
 

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