If $\ A$ is a real symmetric matrix of order n then $\ A$ has n linearly independent eigenvectors.

Question

How do you prove the question above? This is a sub-part of a proof that real symmetric matrices of order n is diagonalizable.

See the answer to your previous question: http://math.stackexchange.com/questions/1317936/why-does-a-symmetric-matrix-have-a-complete-set-of-eigenvectors-and-eigenvalues — muaddib, Jun 11 '15 at 02:50

score 0 · Answer 1 · edited Apr 13 '17 at 12:20

0

Eigenvectors of real symmetric matrices are orthogonal

Look at this one. It has your answer. Try looking around the forum to see if the question you have asked has already been dealt with, in some other post.

edited Apr 13 '17 at 12:20

Community

1

answered Jun 11 '15 at 03:18

HumbleStudent

1,625

If $\ A$ is a *real symmetric matrix of order n* then $\ A$ has *n* linearly independent eigenvectors.

1 Answers1

If $\ A$ is a real symmetric matrix of order n then $\ A$ has n linearly independent eigenvectors.