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I know that if it is positive definite, with the fact that it is diagonalizable and the eigenvalues are positive, the answer is simple and well-known.

But what if it is not positive definite ?

In this other question, it is stated that (in the general case of a real matrix), two conditions must be satisfied :

  1. the dimension differences between iterated kernels must not contain two successive occurrences of the same odd integer
  2. there must be an even number of Jordan blocks of each size for every negative eigenvalue.

My question is : can any (or both) of these conditions be reduced in the case of a symmetric real matrix ?

Johann
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    You surely mean a square root with real entries, but it would even be better if you write it down in your text. – Jean Marie Jul 12 '17 at 18:23
  • What is the matrix is diagonal? – kimchi lover Jul 12 '17 at 19:00
  • Real values, yes it was implied. I should have specified it. Well, the diagonal case needs even multiplicity for the negative eigenvalues, like in daw's example. I was "worried" about the iterated kernels but didn't really thought it through evidently ^^ Thanks to everyone for your help – Johann Jul 12 '17 at 20:28

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If the matrix is symmetric, then you do not need to worry about iterated kernels, as $\ker(A)=\ker(A^k)$ for all $k$. Since $A$ is symmetric, it can be diagonalized. Hence you only need the square root of diagonal matrices.

If the matrix has negative eigenvalues, then the multiplicity of those has to be even in order to obtain a real square root. The idea of the construction of the square root is $$ \pmatrix{0&1\\-1&0}^2 = \pmatrix{-1&0\\0&-1}. $$ If the multiplicity of a negative eigenvalue is odd, there is no real square root as the trivial example $A=\pmatrix{-1}$ shows.

daw
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