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Could any one tell me how to prove:

$\lambda$ be an eigen value of a symmetric matrix $A$ then how to show that the geometric multiplicity and algebraic multiplicity are equal?

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  • $\begingroup$ What facts do you know so far? If you know the fact that all symmetric matrices are orthogonally diagonalizable then this is easy. If not then it gets a bit more complicated. $\endgroup$ – EuYu May 16 '13 at 4:57
  • $\begingroup$ see the thing is was reading that proof from my local writer book, and he says "let A be orthogonally diagonalizable..." I understand that that then $A$ is symmetric, but for the converse part he uses this lemma which he has not proved in his books so I asked. $\endgroup$ – Marso May 16 '13 at 5:02
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    $\begingroup$ I think the standard way to prove that symmetric matrices are orthogonally diagonalizable uses the fact that matrices with real eigenvalues are orthogonally triangularizable. This is sometimes referred to as Schur's theorem. Have you heard of that before? $\endgroup$ – EuYu May 16 '13 at 5:05
  • $\begingroup$ No Dear Sir EUYU $\endgroup$ – Marso May 16 '13 at 6:14
  • $\begingroup$ @Marso The last theorem may help quandt.com/papers/basicmatrixtheorems.pdf $\endgroup$ – Shatabdi Sinha Dec 19 '19 at 6:22
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Let's review some basic matrix theorems first:

  1. Geometric Multiplicity $n$ is the dimension of $ker(A- \lambda I)$; Algebraic Multiplicity $k$ is the sum of the sizes of all Jordan Blocks corresponding to an eigenvalue $\lambda$; and $n\leq k$.
  2. Suppose $A$ is symmetric, we can diagonalize it as $A=Q\Lambda Q^{T},\ Q^{T}=Q^{-1}$.
    $\Lambda$ is a diagonal matrix with eigenvalues of matrix $A$ on the main diagonal line; $Q$ is an Orthonormal matrix with column space as span of Eigenspaces. We can always construct an Eigenspace for each $\lambda$ with size of Algebraic Multiplicity k.

For a specific eigenvalue $\lambda$, if Geometric Multiplicity $n$ is equal to Algebraic Multiplicity $k$, this means the largest Jordan Block should be size 1 and there are k blocks for $\lambda$. In another words, for any eigenvaule $\lambda$ and its eigevector $v$, we have generalized vector form with $k=1$: $$(A-\lambda I)^{k}v=0\iff (A-\lambda I)v=0$$

Prove: $$(A-\lambda I)^{k}v=(Q\Lambda Q^{T}-\lambda I)^{k}v=[Q(\Lambda-\lambda I)Q^{T}]^{k}v=[Q(\Lambda-\lambda I)^{k}][Q^{T}v]=0$$Since column vectors of $Q$ are constructed as orthonormal basis, we get the equation above. Take an example for clarification, assume Algebraic Multiplicity of $\lambda_{i}$ is $2$ and set $\lambda=\lambda_{i}$, $v = v_{i_{1}}$ as eigenvector while another base vector $v_{i_{2}} \bot v_{i_{1}}$, which span eigenspace of $\lambda_{i}$. The last step can be viewing as 2 parts: $$ \left( \begin{array}{cccccc} |&\cdots&|&|&\cdots&| \\ (\lambda_{1}-\lambda)^{2}v_{1}&\cdots&(\lambda_{i}-\lambda)^{2}v_{{i}_{1}}&(\lambda_{i}-\lambda)^{2}v_{i_{2}}&\cdots&(\lambda_{n}-\lambda)^{2}v_{n} \\ |&\cdots&|&|&\cdots&| \end{array} \right) \left( \begin{array}{ccc} 0\\ \vdots\\ 1\\ 0\\ \vdots\\ 0 \end{array} \right) = (\lambda_{i} -\ \lambda)^{2}v=0 $$ $$hence\ for\ i\in \mathbb{Z}^{+},\ (A-\lambda I)^{i}v=Q(\Lambda-\lambda I)^{i}Q^{T}v=0$$ $$(A-\lambda I)v=0$$
vice versa.

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  • $\begingroup$ BTW, I'm confused about the voted answer. e.g., 'it has a full set of orthogonal eigenvectors and is conjugate to a diagonal matrix', what this mean? 'Symmetric matrices have no Jordan block in their spectral decomposition', is that so? @DVD $\endgroup$ – Kuo Nov 11 at 19:08
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Every symmetric matrix is diagonalizable (this can be proved by small perturbation argument), that is: it has a full set of orthogonal eigenvectors and is conjugate to a diagonal matrix. So, you only need to prove the statement for diagonal matrix. Symmetric matrices have no Jordan block in their spectral decomposition, that cause discrepancy in the geometric and algebraic multiplicities of eigenvalues.

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  • $\begingroup$ Out of curiosity, I've never heard of proving a symmetric matrix to be diagonalizable using perturbation. Can you sketch a proof? $\endgroup$ – EuYu May 16 '13 at 5:14
  • $\begingroup$ Well, if a matrix has all eigenvalues of multiplicity 1 then it is obviously diagonalizable. Suppose, it has an eigenvalue of higher multiplicity (belongs to a thin set). We perturbe the matrix, keeping it symmetric and making eigenvalues simple. Perturbed matrices have full orthogonal sets of unit eigenvectors, which cannot collapse to parallel vectors in the limit, because they are orthogonal and unit. Hope it helps:) $\endgroup$ – DVD May 16 '13 at 5:30
  • $\begingroup$ Very clear, thanks :) $\endgroup$ – EuYu May 16 '13 at 5:35
  • $\begingroup$ Readers may be interested in this related thread with a slightly more detailed argument being used for a related purpose. $\endgroup$ – Silverfish Jan 3 '15 at 0:01
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If $A$ is real symmetric then it is diagonalizable. That is, there is some orthogonal $P$ such that $AP=PD$, where $D$ is diagonal. The columns of $P$ are each eigenvectors, and form a basis. Hence all geometric multiplicities equal algebraic multiplicities.

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