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I am reading http://www.deeplearningbook.org/contents/linear_algebra.html Chapter $2$, page $44$ ($3$rd paragraph) of this book and got confused. Can any body help me to understand this paragraph? Thanks in advance.

While any real symmetric matrix $A$ is guaranteed to have an eigendecomposition, the eigendecomposition may not be unique. If any two or more eigenvectors share the same eigenvalue, then any set of orthogonal vectors lying in their span are also eigenvectors with that eigenvalue, and we could equivalently choose a $Q$ using those eigenvectors instead.

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I think it simply says: if $\;u,v\;$ two linearly independent eigenvectors corresponding to one same eigenvalue $\;\lambda\;$ , then any linear combination $\;\alpha u+\beta v\;$ is also an eigenvector for the same eigenvalue, and we can thus choose on of these lin. comb.'s instead of $\;u\;$, or of $\;v\;$ , say.

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  • $\begingroup$ Why linear combination again give eigenvector? $\endgroup$ Mar 5, 2017 at 15:54
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    $\begingroup$ @Arun, do it... $\endgroup$
    – DonAntonio
    Mar 5, 2017 at 16:11
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I'll start by assuming that you understand that if an $n \times n$ symmetric matrix has $n$ distinct eigenvalues then the eigenspaces are all one dimensional and any basis of eigenvectors is essentially unique: it contains one nonzero vector from each eigenspace.

One way to understand the nonuniqueness when an eigenvalue is repeated is to think about the $2 \times 2$ identity matrix. The two eigenvalues are $1$ and $1$. Every basis is a basis of eigenvectors!

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  • $\begingroup$ I think your first parraph is very confusing and even wrong: there are infinite basis of eigenvectors as if $\;u\;$is one of these then also $\;av\;$ is an eigenvector, for any scalar $\;a\;$ . What you probably meant is that each eigenspace is spanned by one single eigenvector, yet such an eigenspace contains infinite eigenvectors. $\endgroup$
    – DonAntonio
    Mar 5, 2017 at 15:51
  • $\begingroup$ @DonAntonio That's why I said "essentially" unique, and clarified what I meant. Each eigenspace is spanned by any one of its nonzero vectors. $\endgroup$ Mar 5, 2017 at 15:59
  • $\begingroup$ @E Still, that "essentially" doesn't seem to be correct at all. It's like saying that $\;\Bbb R^2\;$ has "essentially one unique" basis with eigenvectors of the Identity map (matrix), not to mention that saying "symmetric matrix" can be misleading, too: that much is true for any $\;n\times n\;$ matrix for which its characteristic polynomial factors as the product of $\;n\;$ different linear factors. No need of symmetry. I think this kind of things can be confusing for someone asking a question as elementary as the OP. $\endgroup$
    – DonAntonio
    Mar 5, 2017 at 16:15
  • $\begingroup$ @DonAntonio I think you misread my answer. The first paragraph is about matrices with distinct eigenvalues. The second uses the identity as an example with repeated eigenvalues. I specified symmetric matrices because that's what the OP asked about. Your answer and mine say the same thing differently so the OP has two answers. No need to continue our dialog. $\endgroup$ Mar 5, 2017 at 16:27

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