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I'm reading my linear algebra textbook and there are two sentences that make me confused.

(1) Symmetric matrix $A$ can be factored into $A=Q\lambda Q^{T}$ where $Q$ is orthogonal matrix : Diagonalizable
($Q$ has eigenvectors of $A$ in its columns, and $\lambda$ is diagonal matrix which has eigenvalues of $A$)

(2) Any symmetric matrix has a complete set of orthonormal eigenvectors
whether its eigenvalues are distinct or not.

It's a contradiction, right?
Diagonalizable means the matrix has n distinct eigenvectors (for $n$ by $n$ matrix).
If symmetric matrix can be factored into $A=Q\lambda Q^{T}$, it means that
symmetric matrix has n distinct eigenvalues.
Then why the phrase "whether its eigenvalues are distinct or not" is added in (2)?

After reading eigenvalue and eigenvector part of textbook, I conclude that every symmetric matrix is diagonalizable. Is that true?

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Diagonalizable doesn't mean it has distinct eigenvalues. Think about the identity matrix, it is diagonaliable (already diagonal, but same eigenvalues. But the converse is true, every matrix with distinct eigenvalues can be diagonalized.

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    $\begingroup$ Oh, in my textbook says "diagonalization can fail only if there are repaeated eigenvalues" and added "even then, it does not always fail, think about identity matrix" in the next page. I overlooked it. Thanks! $\endgroup$ – niagara Dec 10 '12 at 17:38
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It is definitively NOT true that a diagonalizable matrix has all distinct eigenvalues--take the identity matrix. This is sufficient, but not necessary. There is no contradiction here.

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  • $\begingroup$ Oh, in my textbook says "diagonalization can fail only if there are repaeated eigenvalues" and added "even then, it does not always fail, think about identity matrix" in the next page. I overlooked it. Thanks! $\endgroup$ – niagara Dec 10 '12 at 17:38
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There is a difference between algebraic multiplicity of eigenvalues (how many times an eigenvalue shows up in the characteristic polynomial) and their geometric multiplicity (nullity). I presume this is the confusing part!

For diagonalizability, it is required that geometric multiplicity to be equal to the algebraic multiplicity of eigenvalues.

See similarity transformation for more details.

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In symmetric matrix geometric multiplicity to be equal to the algebraic multiplicity of eigenvalues.Hence we are heaving complete set of the eigen vectors and Eigenvectors of the symmetric can always be made orthogonal by gram schmidt orthogonalisation. Second thing is that Diagonalizable does'nt mean n distinct eigenvalues.for example identity matrix has repeated eigenvalues still it is diagonalizable.

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Actually, symmetric matrix is a Hermitian matrix with specialty of having a equal conjugate. And all Hermitian matrix can find an unitary matrix to diagonalize no matter its eigenvalues are distinct or not.

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    $\begingroup$ 1) I agree that the OP was probably talking about real matrices but, since you introduced the complex field, I think you should also say that a symmetric matrix is hermitian if (and only if) it is real. 2) You are trying to explain the spectral theorem for real matrices by saying that it's a special case of the spectral theorem for complex matrices; which adds no information whatsoever. One could (and should) ask you why what you said holds for hermitian matrices. $\endgroup$ – user228113 May 26 '15 at 12:09

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