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.

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    $\begingroup$ 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. $\endgroup$
    – user12014
    Nov 15, 2011 at 21:19
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    $\begingroup$ 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. $\endgroup$ Nov 15, 2011 at 21:19
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    $\begingroup$ @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. $\endgroup$ Nov 15, 2011 at 21:32
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    $\begingroup$ @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. $\endgroup$ Nov 15, 2011 at 21:40
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    $\begingroup$ @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. $\endgroup$ Nov 15, 2011 at 21:55

6 Answers 6


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.

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    $\begingroup$ 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. $\endgroup$
    – level1807
    Oct 6, 2014 at 7:47
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    $\begingroup$ @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. $\endgroup$
    – user177692
    Jul 12, 2015 at 13:38
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    $\begingroup$ Notation question: $\langle\mathbf{a}, \mathbf{b}\rangle = \mathbf{a} \cdot \mathbf{b}$? $\endgroup$ Nov 22, 2015 at 18:53
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    $\begingroup$ What if two eigenvalues are the same? $\endgroup$
    – user377507
    Dec 7, 2016 at 8:07
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    $\begingroup$ @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. $\endgroup$ Apr 9, 2017 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 that 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, that is, all Jordan blocks are one-dimensional? Indeed, by referencing the theorem that any symmetric matrix is diagonalizable, Arturo effectively thew the baby out with the bathwater: showing that a matrix is diagonalizable is tautologically equivalent to showing that it has a full set of eigenvectors. Assuming this as a given dismisses half of the question: we were asked to show that $\Lambda$ is diagonal, and not just a generic Jordan form. Here I will untangle this bit of circular logic.

We prove by induction in the number of eigenvectors, namely it turns out that finding an eigenvector (and at least one exists for any matrix) of a symmetric matrix always allows us to generate another eigenvector. So we will run out of dimensions before we run out of eigenvectors, making the matrix diagonalizable.

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.


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.


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 $Q^T = Q^{-1}$

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    $\begingroup$ This solves the wrong direction of the problem. $\endgroup$
    – Phonon
    May 20, 2014 at 0:05
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    $\begingroup$ $A^t = A$ is related to eigenvectors how? $\endgroup$
    – Don Larynx
    Mar 4, 2015 at 22:07
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    $\begingroup$ 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. $\endgroup$ Jan 27, 2016 at 1:45
  • $\begingroup$ Thank you. Let $A$ be symmetric, then there exists a matrix $D$ such that $A=QDQ^T$ is theorem? Why you assumed that? $\endgroup$
    – Avv
    Feb 24, 2021 at 1:55
  • $\begingroup$ Your assumption that $A = QDQ^{\top}$, $D$ diagonal, is precisely what's being asked of you to prove. $\endgroup$
    – Joe Shmo
    Aug 14, 2021 at 13:41

Note a real symmetric matrix is a linear operator on Euclidean space with respect standard basis (orthonormal). So the fact that it equals to its conjugate transpose implies it is self-adjoint.

For two distinct eigenvalues $\lambda_1, \lambda_2$ and corresponding eigenvectors $v_2, v_2$, $$(\lambda_1-\lambda_2)<v_1, v_2>=<Tv_1,v_2>-<v_1,\bar{\lambda_2} v_2>=<Tv_1,v_2>-<v_1,T^* v_2>=0$$ where the 2nd last equality follows from properties of self-adjoint (thus normal) linear operator (Lemma below).

Lemma: Assume $T$ is normal. If $(\lambda, v)$ is eigenvalue and eigenvector of $T$, $(\bar{\lambda}, v)$ is eigenvalue and eigenvector of the adjoint $T^*$. (pf.) Trivial from definition of normality.


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