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I am studying about symmetric matrices. I find out a lot of books state the theorem of existence of real eigenvector of symmetric matrices but didn't provide any solution. I am wander this problem.

Prove that all the eigenvectors of symmetric matrices are real.
Long Do
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  • Probably the most straightforward answer at the question to which HallaSurvivor just linked is the one by naturer. – Brian M. Scott Jul 10 '20 at 04:34
  • I can't for the life of me see why this question is a duplicate of the linked question, since this one asks to show a symmetric matrix has real *eigenvectors, whereas that question concerns real eigenvalues; naturer's answer clearly admits the possibility of complex eigenvectors, so doesn't show there have to be real* eigenvectors. Not to put too fine a point on it, but it seems to me closing this question as a duplicate of that question is erroneous. – Robert Lewis Jul 10 '20 at 05:17
  • I think that my question is NOT DUPLICATE, PLEASE READ THE QUESTION CAREFULLY! – Long Do Jul 10 '20 at 09:03

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It is not necessarily true that eigenvectors of real symmetric matrices are real; what is true, however, is that every real eigevalue has a real eigenvector, and since every eigenvalue of a symmetric matrix is real, a complete set real eigenvectors may be found. For if

$Ax = \lambda x, \tag 1$

then taking complex conjugates yields

$A \bar x = \lambda \bar x; \tag 2$

adding (1) and (2),

$A(x + \bar x) = \lambda(x + \bar x); \tag 3$

if

$x + \bar x = 0, \tag 4$

then

$\bar x = -x, \tag 5$

that is, $x$ is purely imaginary, hence $ix$ is real, and from (1),

$A(ix)= \lambda(ix), \tag 6$

so $\lambda$ has a real eigenvector; on the other hand, if

$x + \bar x \ne 0, \tag 7$

then $x + \bar x$ is a real eigenvector associated to $\lambda$.

Robert Lewis
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  • Coud you give an counter-example for your command: It is not necessarily true??? – Long Do Jul 10 '20 at 08:56
  • In the Linear Algebra's Lang, 
    definite scalar product. Let A: V ~ V be a symmetric linear map. Then
A has a nonzero eigenvector.

    After that he used this theorem to prove the well-known spectral theorem by indicating that there exists a non-zero eigenvector of A with A is A symmetric transformation over real vector space and he used this vector adterthat to be one of the basis vector. So it should be real.

     – Long Do Jul 10 '20 at 08:57