I know that different eigenvectors from different eigenspace are automatically orthogonal.

My question is:

Suppose we are doing spectrum decomposition to a 3x3 symmetric matrix and we have only two different eigenvalues(which means we have one repeated eigenvalue). Then we can choose two eigenvectors for that repeated eigenvalue and then do Grand-Schmit orthogonalization to these two vectors. But G-S orthogonalization requires linear independent vector here.

Does it mean:

1.We should always pick well selected linearly independent eigenvectors?

2.if 1 holds, can we always guarantee that we can find linearly independent eigenvectors for repeated eigenvalue?

  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax. $\endgroup$ – dantopa May 18 at 19:47
  • $\begingroup$ Are you dealing with a symmetric matrix? $\endgroup$ – Minus One-Twelfth May 18 at 20:03
  • $\begingroup$ @ Minus One-Twelfth Yes and I have edited my answer $\endgroup$ – Tian May 18 at 20:08
  • $\begingroup$ @Tian any symmetric matrix is diagonalizable. The eigenspace related to the repeated eigenvalue (double in your case) is 2-dimensional. Hence it has an orthogonal basis consisting of 2 vectors. Any vector from this eigenspace is orthogonal to any eigenvector related to the other eigenvalue. $\endgroup$ – user376343 May 18 at 20:20
  • $\begingroup$ (2) holds. Choose any eigenvector, the second chosen evec shouldd be orthogonal to it $\endgroup$ – user376343 May 18 at 20:23

When you computed the null space of $A-\lambda I$ for the repeated eigenvalue, you should’ve ended up with a basis for that space. Apply Gram-Schmidt process to these two vectors.

Alternatively, since you’re working in $\mathbb R^3$ you have a short-cut available to you. As user376343 noted, the two eigenspaces are orthogonal complements. Once you have an eigenvector $\mathbf v$ for the simple eigenvalue, then, choose any vector orthogonal to it. You can generate one via a simple manipulation of that vector’s components. This orthogonal vector is guaranteed to be an eigenvector of the repeated eigenvalue, and its cross product with $\mathbf v$ is another.


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