# For a symmetric matrix, the geometric and algebraic multiplicities are equal

Could anyone tell me how to prove the following proposition?

Let $$\lambda$$ be an eigenvalue of a symmetric matrix. Then, its geometric multiplicity equals its algebraic multiplicity.

• 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.
– EuYu
May 16, 2013 at 4:57
• 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. May 16, 2013 at 5:02
• 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?
– EuYu
May 16, 2013 at 5:05
• No Dear Sir EUYU May 16, 2013 at 6:14
• @Marso The last theorem may help quandt.com/papers/basicmatrixtheorems.pdf
– S.S
Dec 19, 2019 at 6:22

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.

• Out of curiosity, I've never heard of proving a symmetric matrix to be diagonalizable using perturbation. Can you sketch a proof?
– EuYu
May 16, 2013 at 5:14
• 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:)
– DVD
May 16, 2013 at 5:30
• Very clear, thanks :)
– EuYu
May 16, 2013 at 5:35
• Readers may be interested in this related thread with a slightly more detailed argument being used for a related purpose. Jan 3, 2015 at 0:01

I saw this question in Anton's Elementary Linear Algebra (Exercise 7.2.31). At there, this question was used to prove the spectral theorem. So I was supposed to unable to use the theorem.

Anton had given an outline of the proof. I just give the detail.

• Let $$B_0=\{\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k\}$$ be a basis for the eigenspace $$E_{\lambda}$$. Extend this basis to an orthonormal basis $$B=\{\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k, \mathbf{u}_{k+1}, ..., \mathbf{u}_n\}$$ for $$\mathbb{R}^n$$.

• Let $$P=[\mathbf{u}_1|\mathbf{u}_2|\cdots |\mathbf{u}_k |\mathbf{u}_{k+1} |\cdots |\mathbf{u}_n]$$. Then $$AP=P\begin{bmatrix}\lambda I_k & X \\ O & Y \end{bmatrix}$$ and $$P^{-1}=P^T$$ and $$P^T AP=\begin{bmatrix}\lambda I_k & X \\ O & Y \end{bmatrix}$$.

• Since $$A$$ is symmetric, we have $$\begin{bmatrix}\lambda I_k & X \\ O & Y \end{bmatrix} =P^T AP=P^T A^T P=(P^T AP)^T =\begin{bmatrix}\lambda I_k & X \\ O & Y \end{bmatrix}^T =\begin{bmatrix}\lambda I_k & O \\ X^T & Y^T \end{bmatrix}.$$ This follows that $$X=O$$. Hence, $$AP=P\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}$$ and $$P^T AP=\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}$$.

• Note that $$A$$ is similar to $$\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}$$. So they have the same characteristic polynomial. The characteristic polynomial of $$\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}$$ is $$\det{(\lambda I_k-xI_k)}\det{(Y-xI_{n-k})}$$. We will show that $$\lambda$$ cannot be a root of $$\det{(Y-xI_{n-k})}$$. Then the algebraic multiplicity of $$\lambda$$ is $$k$$, equals to geometric multiplicity of $$\lambda$$.

• Suppose that $$\lambda$$ is a root of $$\det{(Y-xI_{n-k})}$$. Let $$\mathbf{y}$$ be an eigenvector of $$Y$$ with eigenvalue $$\lambda$$. Then $$A\left(P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}\right) =\left(P\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}P^T\right)\left(P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}\right) =\lambda\left(P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}\right).$$ That is, $$P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda$$. Which implies that $$P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}\in E_{\lambda}=\text{span}(\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k)$$.

• On the other hand, note that $$P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix} =[\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k, \mathbf{u}_{k+1}, ..., \mathbf{u}_n]\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix} \in \text{span}(\mathbf{u}_{k+1}, \mathbf{u}_{k+2}, ..., \mathbf{u}_{n}).$$ This is impossible because $$\text{span}(\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k) \cap \text{span}(\mathbf{u}_{k+1}, \mathbf{u}_{k+2}, ..., \mathbf{u}_{n})=\{0\}$$.

• I think $P$ should be $B$ since step 2. $P^{-1}$ can not be its transpose considering its shape
– Kuo
Sep 8, 2022 at 12:10
• @Kuo Aha! You are right. Thanks for pointing it out. I have already edited it. Thanks. Sep 9, 2022 at 5:59

This is not a proof but something like paraphrase to help understand the question. For a detailed proof, I vote for @bfhaha's.

1. Geometric Multiplicity $$\gamma(\lambda)$$ is the dimension of $$ker(A- \lambda I)$$, and the number of Jordan Blocks for an eigenvalue $$\lambda$$; Algebraic Multiplicity $$\mu(\lambda)$$ is the sum of the sizes of all Jordan Blocks for an eigenvalue $$\lambda$$; and $$\gamma\leq \mu$$.
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 $$\mu(\lambda)$$.

For a specific eigenvalue $$\lambda$$, if Geometric Multiplicity $$\gamma(\lambda)$$ is equal to Algebraic Multiplicity $$\mu(\lambda)$$, this means the size of the largest Jordan Block should be 1 and there are $$\mu=\gamma$$ blocks for $$\lambda$$.

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.