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If A is an nxn real symmetric matrix then A is diagonalisable. In other words, If A is a symmetric nxn matrix, then there exists an orthogonal matrix $P$ such that $P_{-1}AP=P_{T}AP=D$, a diagonal matrix. The eigenvalues of A lie on the main diagonal of D.

Proof of this statement must be considered in two cases.

It is clear that all eigenvalues of a real symmetric matrix are real and if they are all distinct, then eigenvectors $x_i$, i=1,2,..,n corresponding to $\lambda_i$ i=1,2,..,n are orthogonal. We can obtain an orthonormal set of eigenvectors eigenvectors $u_i$, i=1,2,..,n using these eigenvectors $x_i$, i=1,2,..,n Then we can construct matrix $P$ as $P=[u_1 u_2 ... u_n]$ such that $D=P_{-1}AP$.

I have some troubles in the second case that is not all eigenvalues are simple.
How can I prove that eigenvalue of multiplicity k of a real symmetric matrix A has exactly k linearly independent eigenvectors, i.e., dimension of solution space of $(A-\lambda I)$ is k?

Şeyma Kayan
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  • I typed $P_{-1}$ and $P_T$ wrong. The correct spelling is as follows $P^{-1}$ and $P^T$. – Şeyma Kayan Jun 28 '16 at 11:30
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    Do some search... http://math.stackexchange.com/questions/871454/why-is-algebraic-multiplicity-always-equal-to-the-geometric-multiplicity-of-dist http://math.stackexchange.com/questions/255622/symmetric-matrix-is-always-diagonalizable http://math.stackexchange.com/questions/482599/why-symmetric-matrices-are-diagonalizable – leonbloy Jun 28 '16 at 11:47
  • @user1551 I noticed my mistake an I added a comment right after. Thanks again. – Şeyma Kayan Jun 30 '16 at 08:57
  • @leonbloy I have done my research. None of the questions you share gives an explicit proof for repeated eigenvalue case. In the proof of Theorem 7.9 in Kolman, Hill, Elementary Linear Algebra stated that "It can be shown that if a symmetric matrix $A$ has an eigenvalue $\lambda$ of multiplicity k, then the solution space of the homogeneous system$ (A-\lambda I)x = 0 $ has dimension k. For a proof, see J. M. Ortega, Matrix Theory: A Secolld Course, New York: Plenum Press, 1987. This is about hermitian matrices but I could not find a proof in that book unfortunately. – Şeyma Kayan Jun 30 '16 at 09:20
  • I use the answer of this question to prove that ever real symmetric matrix is diagonalisable. – Şeyma Kayan Jun 30 '16 at 09:25
  • You're right, my bad. I added an answer. – leonbloy Jun 30 '16 at 11:47
  • Duplicate of https://math.stackexchange.com/questions/482599/why-are-real-symmetric-matrices-diagonalizable – Hans Lundmark Apr 19 '21 at 06:41
  • @HansLundmark I think you are voting to close as a duplicate incorrectly. You need to click the "duplicate" reason, rather than "other/add a comment". This is important because of review queues, where potential duplicates are treated in a specific way. – user1729 Apr 23 '21 at 09:29
  • @user1729: OK, fair enough, I wasn't aware of that. It's just that I find the boilerplate “Does this answer your question?” comment (that's auto-generated when voting for duplicates in the usual way) very inappropriate in many cases. (OP starts debating, then the comment is auto-removed when the question is closed, and everything is left in a confusing-looking state. To name but one reason.) – Hans Lundmark Apr 23 '21 at 09:36
  • @HansLundmark Sure - I guess clicking "duplicate", then deleting the auto-generated comment and adding your own would work then? Nicer comment, and easier for those coming later. – user1729 Apr 23 '21 at 09:39
  • @user1729: Thanks for the suggestion, that makes sense. A bit of a hassle, though... – Hans Lundmark Apr 23 '21 at 09:44
  • @HansLundmark Yes, but otherwise it is less likely that people will close the post as a duplicate (in the review queue I actually voted to leave open a question you had commented on, as your comment was hidden and the question looked fine). If you feel strongly about the wording, then you could bring it up on the meta site? – user1729 Apr 23 '21 at 09:50
  • It's already been discussed on Meta: https://math.meta.stackexchange.com/questions/31622/flagged-duplicate-does-this-answer-your-question-text-is-misleading, https://meta.stackoverflow.com/questions/391289/does-this-answer-your-question-duplicate-comment-text-is-silly-and-confusing, https://meta.stackoverflow.com/questions/390887/new-post-notices-closed-on-hold-etc-rolling-out-on-stack-overflow/390894#390894 – Hans Lundmark Apr 23 '21 at 10:03

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I'd say you have three ways:

  1. Prove first that a symmetric matrix is diagonalizable. Then see that your desired property is true for diagonal matrices and that a similarity transform preserves both algebraic and geometric multiplicities (eg: see prop 7.5 here). I think this is the most usual way.

  2. Continuity argument: in the space of symmetric matrices, we can go continuosly from different eigenvalues to repeated eigenvalues; because the starting point had orthogonal eigenvectors, these cannot degenerate in LD eigenvectors. This argument is easy to visualize but hard to formalize.

  3. Prove it directly. Here's an sketch (essentially from here)

Let ${\bf A}$ be a $n \times n$ symmetric matrix, let $\lambda_i$ be an eigenvalue with (algebraic) multiplicity $1<m \le n$. Then there exists some eigenvector ${\bf p}_{i1}$ with $|{\bf p}_{i1}|=1$.

Let ${\bf B}=({\bf p}_{i1} \, {\bf C})$ be an orthogonal matrix (orthonormal columns) with ${\bf p}_{i1}$ as first column (it can be constructed by Grand-Schmidt process). Then consider

$$ {\bf B}' {\bf A} {\bf B} = \begin{pmatrix} \lambda_i &0 \\ 0 & {\bf C}' {\bf A} {\bf C} \end{pmatrix}$$

By considering the characteristic polynomial, we see that (because the multiplicity of $\lambda_i>1$)$|{\bf C}' {\bf A} {\bf C} -\lambda_i I_{n-1}|=0$ Hence there exists some non null ${\bf q}$ with $({\bf C}' {\bf A} {\bf C} -\lambda_i I_{n-1})){\bf q}=0$. Next, see that ${\bf p}_{i2}={\bf C}{\bf q}$ is eigenvector of ${\bf A}$, and is orthogonal to ${\bf p}_{i1}$. We can repeat the procedure (if $m>2$), by rebuilding ${\bf B}$ with this eigenvector as second column, etc.

leonbloy
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