# Eigenvectors of real symmetric matrices are orthogonal

Can someone point me to a paper, or show here, why symmetric matrices have orthogonal eigenvectors? In particular, I'd like to see proof that for a symmetric matrix $A$ there exists decomposition $A = Q\Lambda Q^{-1} = Q\Lambda Q^{T}$ where $\Lambda$ is diagonal.

• If $A$ is symmetric, we have $AA^* = A^2 = A^*A$ so $A$ is normal. The assertion then follows directly from the spectral theorem. So just go read any proof of the spectral theorem, there are many copies available online. – user12014 Nov 15 '11 at 21:19
• The statement is imprecise: eigenvectors corresponding to distinct eigenvalues of a symmetric matrix must be orthogonal to each other. Eigenvectors corresponding to the same eigenvalue need not be orthogonal to each other. However, since every subspace has an orthonormal basis, you can find orthonormal bases for each eigenspace, so you can find an orthonormal basis of eigenvectors. – Arturo Magidin Nov 15 '11 at 21:19
• @Phonon: It's false otherwise, but you can find a basis for the eigenspace made up of orthogonal eigenvectors: just take any basis for the eigenspace, and apply Gram-Schmid. Once you have a basis of eigenvectors for all of $\mathbb{R}^n$, $Q$ is the matrix whose columns are the elements of the basis. – Arturo Magidin Nov 15 '11 at 21:32
• @Phonon: Might I add: if you already knew it was true for distinct eigenvalues, why not say so in your question? It would have saved me the trouble of writing it out, and then it would have been clear what your doubt was: you could have gotten a response that didn't re-tread stuff you already knew. – Arturo Magidin Nov 15 '11 at 21:40
• @Phonom. Two different ways: first, you can not compute $Q$ until after you have an orthonormal basis of eigenvectors. Second way: Work the other way: You have $\Lambda = Q^{-1}AQ$. Now orthonormalize the columns of $Q$ by multiplying on the right by elementary matrices, and adjust the inverse by multiplying by the inverse of the elementary matrices. So at each step you get $E^{-1}\Lambda E = E^{-1}Q^{-1}AQE$. But $E^{-1}\Lambda E$ is diagonal, so you get $\Lambda' = Q'AQ'^{-1}$. Lather, rinse, repeat until $Q^{-1}$ has orthonormal columns. – Arturo Magidin Nov 15 '11 at 21:55

## 5 Answers

For any real matrix $A$ and any vectors $\mathbf{x}$ and $\mathbf{y}$, we have $$\langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle.$$ Now assume that $A$ is symmetric, and $\mathbf{x}$ and $\mathbf{y}$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda$ and $\mu$. Then $$\lambda\langle\mathbf{x},\mathbf{y}\rangle = \langle\lambda\mathbf{x},\mathbf{y}\rangle = \langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle = \langle\mathbf{x},A\mathbf{y}\rangle = \langle\mathbf{x},\mu\mathbf{y}\rangle = \mu\langle\mathbf{x},\mathbf{y}\rangle.$$ Therefore, $(\lambda-\mu)\langle\mathbf{x},\mathbf{y}\rangle = 0$. Since $\lambda-\mu\neq 0$, then $\langle\mathbf{x},\mathbf{y}\rangle = 0$, i.e., $\mathbf{x}\perp\mathbf{y}$.

Now find an orthonormal basis for each eigenspace; since the eigenspaces are mutually orthogonal, these vectors together give an orthonormal subset of $\mathbb{R}^n$. Finally, since symmetric matrices are diagonalizable, this set will be a basis (just count dimensions). The result you want now follows.

• This answer, though intuitively satisfying, assumes that $A$ has the maximum number of eigenvectors, i. e. no generalized eigenvectors. For a more general proof see my answer. – level1807 Oct 6 '14 at 7:47
• @level1807 It doesn't "assume" anything: any square symmetric matrix is diagonalizable $\; \iff\;$ there is a basis of eigenvectors of the matrix for the space we're working on, and thus there is no need of worrying about generalized eigenvectors. – user177692 Jul 12 '15 at 13:38
• Notation question: $\langle\mathbf{a}, \mathbf{b}\rangle = \mathbf{a} \cdot \mathbf{b}$? – Gordon Gustafson Nov 22 '15 at 18:53
• What if two eigenvalues are the same? – Linfeng Li Dec 7 '16 at 8:07
• @AshkanRanjbar Nobody called anything "non-sequitur preference". You really need to work on your reading comprehension. Nobody asked you to accept anything, especially five years after the fact. Nor did I expect you to acknowledge your error of claiming the argument assumed there were more than one eigenvalue. So, frankly, my dear, I couldn't care less what you accept or don't accept. I don't need your approval. Once you get to 125 reputation, you are welcome to come back and downvote it and get 2 points off my 229,000+ reputation. It'll be the second downvote on this one. – Arturo Magidin Apr 9 '17 at 19:06

Since being symmetric is the property of an operator, not just its associated matrix, let me use $\mathcal{A}$ for the linear operator whose associated matrix in the standard basis is $A$. Arturo and Will proved that a real symmetric operator $\mathcal{A}$ has real eigenvalues (thus real eigenvectors) and eigenvectors corresponding to different eigenvalues are orthogonal. One question still stands: how do we know that there are no generalized eigenvectors of rank more than 1?

We prove by induction. Suppose $\lambda_1$ is an eigenvalue of $A$ and there exists at least one eigenvector $\boldsymbol{v}_1$ such that $A\boldsymbol{v}_1=\lambda_1 \boldsymbol{v}_1$. Choose an orthonormal basis $\boldsymbol{e}_i$ so that $\boldsymbol{e}_1=\boldsymbol{v}_1$. The change of basis is represented by an orthogonal matrix $V$. In this new basis the matrix associated with $\mathcal{A}$ is $$A_1=V^TAV.$$ It is easy to check that $\left(A_1\right)_{11}=\lambda_1$ and all the rest of the numbers $\left(A_1\right)_{1i}$ and $\left(A_1\right)_{i1}$ are zero. In other words, $A_1$ looks like this: $$\left( \begin{array}{c|ccc} \lambda_1 & \\ \hline & & \\ & & B_1 & \\ & & \end{array} \right)$$ Thus the operator $\mathcal{A}$ breaks down into a direct sum of two operators: $\lambda_1$ in the subspace $\mathcal{L}\left(\boldsymbol{v}_1\right)$ ($\mathcal{L}$ stands for linear span) and a symmetric operator $\mathcal{A}_1=\mathcal{A}\mid_{\mathcal{L}\left(\boldsymbol{v}_1\right)^{\bot}}$ whose associated $(n-1)\times (n-1)$ matrix is $B_1=\left(A_1\right)_{i > 1,j > 1}$. $B_1$ is symmetric thus it has an eigenvector $\boldsymbol{v}_2$ which has to be orthogonal to $\boldsymbol{v}_1$ and the same procedure applies: change the basis again so that $\boldsymbol{e}_1=\boldsymbol{v}_1$ and $\boldsymbol{e}_2=\boldsymbol{v}_2$ and consider $\mathcal{A}_2=\mathcal{A}\mid_{\mathcal{L}\left(\boldsymbol{v}_1,\boldsymbol{v}_2\right)^{\bot}}$, etc. After $n$ steps we will get a diagonal matrix $A_n$.

There is a slightly more elegant proof that does not involve the associated matrices: let $\boldsymbol{v}_1$ be an eigenvector of $\mathcal{A}$ and $\boldsymbol{v}$ be any vector such that $\boldsymbol{v}_1\bot \boldsymbol{v}$. Then $$\left(\mathcal{A}\boldsymbol{v},\boldsymbol{v}_1\right)=\left(\boldsymbol{v},\mathcal{A}\boldsymbol{v}_1\right)=\lambda_1\left(\boldsymbol{v},\boldsymbol{v}_1\right)=0.$$ This means that the restriction $\mathcal{A}_1=\mathcal{A}\mid_{\mathcal{L}\left(\boldsymbol{v}_1\right)^{\bot}}$ is an operator of rank $n-1$ which maps ${\mathcal{L}\left(\boldsymbol{v}_1\right)^{\bot}}$ into itself. $\mathcal{A}_1$ is symmetric for obvious reasons and thus has an eigenvector $\boldsymbol{v}_2$ which will be orthogonal to $\boldsymbol{v}_1$.

It would appear that you want to write vectors as rows, so your preferred multiplication will be on the left side, as in $v \mapsto v A.$

The ordinary dot product is then $v \cdot w = v w^T = w v^T = w \cdot v.$ Note that $v w^T$ is a number, or a 1 by 1 matrix, and is equal to its transpose.

In the same way, $v A \cdot w = v A w^T.$ However, $v A w^T$ is again a 1 by 1 matrix and is equal to its transpose, and $A^T = A,$ so we get $$v A \cdot w = v A w^T = (v A w^T)^T = (w^T)^T A^T v^T = w A v^T = w A \cdot v$$

First suppose $v,w$ are eigenvectors with distinct eigenvalues $\lambda, \mu.$ We have $$v A \cdot w = \lambda v \cdot w = w A \cdot v = \mu w \cdot v.$$ Or, $\lambda v \cdot w = \mu v \cdot w,$ finally $$(\lambda - \mu) v \cdot w = 0.$$ So, eigenvectors with distinct eigenvalues are orthogonal.

It is possible that an eigenvalue may have larger multiplicity. However, for a fixed eigenvalue $\lambda,$ the set of vectors $v$ for which $v A = \lambda v$ is a subspace, of full dimension (meaning the Jacobi form has no off-diagonal elements), and we may simply choose an orthonormal basis for this subspace. Choosing, in this way, all basis vectors to be length 1 and orthogonal, we get an orthonormal basis of eigenvalues of $A.$ Write those as rows of a matrix $P,$ we get $P A P^T = \Lambda.$

The only difficult aspect here is this: if an eigenvalue has algebraic multiplicity larger than one, that is the characteristic polynmial has a factor of $(x-\lambda)^k$ for some $k \geq 2,$ how can I be sure that the geometric multiplicity is also $k?$ That is, with $A$ symmetric, how do I know that $$v (A - \lambda I)^k = 0 \; \; \Rightarrow \; \; v (A - \lambda I) = 0?$$ Working on it. It appears that this is, at heart, induction on $k,$ and takes many pages. Give me some time.

Alright, this works. An induction on dimension shows that every matrix is orthogonal similar to an upper triangular matrix, with the eigenvalues on the diagonal (the precise statement is unitary similar). How do we know the eigenvalues are real? We have an eigenvalue $\lambda$ with an eigenvector $v,$ perhaps both with complex entries. As is traditional, for a vector or matrix define $v^\ast = \bar{v}^T$ and $A^\ast = \bar{A}^T.$ It is easy to see that $v v^\ast$ is a positive real number unless $v = 0.$ In any case $A^\ast = A.$ So, given $v A = \lambda v,$ $$( v A v^\ast)^\ast = (v^\ast)^\ast A^\ast v^\ast = v A v^\ast.$$ As a result, the complex number $v A v^\ast$ is actually a real number. At the same time, $v A v^\ast = \lambda v v^\ast,$ and since both $v A v^\ast$ and $v v^\ast$ are real numbers, the latter nonzero, it follows that $\lambda$ is real.

Put these together, we get that each real matrix with real characteristic values is orthogonal similar to an upper triangular real matrix. However, as $A$ is symmetric, this upper triangular matrix is actually diagonal.

How about Let $A$ be symmetric, then there exists a matrix $D$ such that $A=QDQ^T$, taking the transpose of $A$, namely

$$\left(A\right)^T = \left(QDQ^T\right)^T$$ $$A^T = \left(Q^T\right)^TD^TQ^T$$ $$A^T = QDQ^T$$

thus $A^T = A$ if and only if $A$ is symmetric.

It is noteworthy that $D^T = D$ since $D$ is diagonal and $Q$ is the matrix of normed eigenvectors of $A$, Thus $Q^T = Q^{-1}$

• This solves the wrong direction of the problem. – Phonon May 20 '14 at 0:05
• $A^t = A$ is related to eigenvectors how? – Don Larynx Mar 4 '15 at 22:07
• When you start with $A=A^T$ and the eigendecomposition is written as $A=QDQ^{-1}$, then the transpose of this yields $A^T=\left(Q^{-1}\right)^TDQ^T$, but has to be equal to the initial decomposition, which will only be the case if $Q^{-1}=Q^T$ which is the definition of an orthogonal matrix. – Kwin van der Veen Jan 27 '16 at 1:45

Let's assume that $$x$$ is an eigenvector of $$A$$ corresponding to the eigenvalue $$\lambda_1$$ and $$y$$ an eigenvector of $$A$$ corresponding to the eigenvalue $$\lambda_2$$, with $$\lambda_1 \neq \lambda_2$$.

$$Ax=\lambda_1x \\ Ay=\lambda_2y$$

After taking into account the fact that A is symmetric ($$A=A^*$$):

$$y^{\intercal}Ax=\lambda_1y^{\intercal}x \\ x^{\intercal}A^{\intercal}y=\lambda_2x^{\intercal}y$$

Now subtract the second equation from the first one and use the commutativity of the scalar product:

$$y^{\intercal}Ax-x^{\intercal}A^{\intercal}y=\lambda_1y^{\intercal}x - \lambda_2x^{\intercal}y \\ 0 = (\lambda_1 - \lambda_2)y^{\intercal}x$$

Hence $$x$$ and $$y$$ are orthogonal.

• Welcome to MSE. Your answer adds nothing new to the already existing answers. – José Carlos Santos Feb 8 at 14:02

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