I am having a difficult time with the following question. Any help will be much appreciated.

Let $A$ be an $n×n$ real matrix such that $A^T = A$. We call such matrices “symmetric.” Prove that the eigenvalues of a real symmetric matrix are real (i.e. if $\lambda$ is an eigenvalue of $A$, show that $\lambda = \overline{\lambda}$ )

Rodrigo de Azevedo
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    A real $n\times n$ matrix only can have real eigenvalues (every complex zero of the characteristic is no eigenvalue of the real matrix) – Dominic Michaelis Apr 07 '13 at 19:12
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    @Susan : see Dominic's answer. You will need to use the "complex inner product" $\langle \mathbf{x}, \mathbf{y} \rangle = \sum_{i=1}^n {\bar x_i}y_i$. Also see Lepidopterist's answer, where $C^*$ is the conjugate transpose of $C$, $C^* = {\bar {C^T}}$. – Stefan Smith Apr 07 '13 at 19:13
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    @DominicMichaelis : do you really mean that (a real square matrix can have only real eigenvalues)? I'm afraid you might confuse Susan. What about $[0, 1;-1, 0]$ with eigenvalues $\pm i$? (Sorry, I don't remember the $\LaTeX$ for writing a matrix – Stefan Smith Apr 07 '13 at 19:45
  • @StefanSmith if you consider it as a real matrix, it doesn't have any eigenvalues. – Dominic Michaelis Apr 07 '13 at 19:49
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    Going Dominic's way or Lepidopterist's way, you will easily see that the (a priori complex) eigenvalues must be real. Note that the exact same proofs show that the eigenvalues of a hermitian $A^*=A$ matrix are real in the general complex case. – Julien Apr 07 '13 at 19:59

9 Answers9


Let $(\lambda,v)$ be any eigenpair of $A$. Since $A=A^T=A^\ast$, $$\langle Av,Av\rangle=v^*A^*Av=v^\ast A^2v=v^*(A^2v)=\lambda^2||v||^2.$$

Therefore $\lambda^2=\frac{\langle Av,Av\rangle}{||v||^2}$ is a real nonnegative number. Hence $\lambda$ must be real.

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  • A quotient of non-negative real numbers? But the argument is very neat. – Chris Godsil Apr 07 '13 at 19:08
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    only because of $\lambda^2$ is real $\lambda$ is not real in general. and still you need non negative, because $Av=0$ will surely not be positive. and your equation is confusing – Dominic Michaelis Apr 07 '13 at 19:19
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    If $\lambda^2$ is positive and real, then $\lambda$ is real. What is confusing? And ok, we can assume $\lambda$ is not zero! Zero is real obviously. – Lepidopterist Apr 07 '13 at 19:23
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    You don't need to bother about $\lambda =0$. All you need is $\|v\|>0$, which is given as you take a (nonzero) eigenvector. So you do find $\lambda^2\geq 0$. Which is equivalent to $\lambda $ being real. – Julien Apr 07 '13 at 19:44
  • Nice concise answer, by the way, +1. – Julien Apr 07 '13 at 19:46
  • @Lepidopterist : I upvoted your answer, and I like it very much, but $\lambda^2$ is a quotient of _nonnegative_ real numbers, since $A\mathbf{v}$ might be $\mathbf{0}$. I'm being picky, but we don't want to confuse Susan. – Stefan Smith Apr 07 '13 at 19:48
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    Detail: but this is the quotient of a nonnegative number by a positive one, hence a nonnegative number, and not a positive one, to include the case of $0$ eigenvalue. Then you find $\lambda^2\geq 0$ which, in $\mathbb{C}$, is equivalent t0 $\lambda\in\mathbb{R}$. – Julien Apr 07 '13 at 19:49
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    I decided to edit your answer directly to avoid long chat discussions. I hope you don't mind. Just roll back to your version if you disagree and I will leave it alone. – Julien Apr 07 '13 at 19:51
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    I understood it as, $⟨Av,Av⟩=(Av)^∗(Av)=(λv)^∗(λv)= \bar λ λv^∗v=||λ||^2⟨v,v⟩ $. Since inner products have the property of Positive-definiteness, i.e. $⟨v,v⟩≥0$ we conclude $||λ||^2=(x+iy)(x−iy)=x^2−y^2≥0$ So this just means $x≥y$ right? How can we conclude $y=0$ and λ is actually real. – Aditya P Jul 13 '18 at 07:02
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    3 years late, but the answer to the question of @AdityaP is Lepidopterist did not convert $(Av)^*$ to $(\lambda v)^*$, but used a different calculation. Since $Av = \lambda v$, we have $A^2v = A(Av) = A(\lambda v) = \lambda Av = \lambda(\lambda v) = \lambda^2 v$. – Paul Sinclair Apr 08 '21 at 12:57
  • Let us call $\lambda = a+bi$, we have $\lambda^2=(a^2-b^2) + 2ab i $, since $\lambda^2$ is real, we have that $ab=0$. If, by contradiction, $b\not=0,$ necessarily $a=0$. Hence, $\lambda^2=(a^2-b^2) + 2ab i = -b^2<0,$ which contradicts with the fact that $\lambda^2$ is real and nonnegative. – R. W. Prado Aug 18 '21 at 23:06

Let $Ax=\lambda x$ with $x\ne 0$, with $\lambda\in\mathbb{R}$, then \begin{align} \lambda \bar x^T x &= \bar x^T(\lambda x)\\ &=\bar x^T A x \\ &=(A^T \bar{x})^T x \\ &=(A \bar x)^T x \\ &=(\bar A \bar x)^T x \\ &=(\bar\lambda\bar x)^T x\\ &=\bar \lambda \bar x^T x.\\ \end{align} Because $x\ne 0$, then $\bar{x}^T x\ne 0$ and $\lambda=\bar \lambda$.

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    @ThePointer How is this uninformative? It is a complete proof. – YiFan Sep 26 '19 at 23:24
  • Nice proof. The only step that isn't obvious to me is $\bar A \bar x = \bar \lambda \bar x$ – user1754322 Dec 28 '20 at 20:52
  • Actually, if that was true, the proof could be much simpler: $A x = \lambda x = \bar A x = \bar \lambda x \Rightarrow \lambda = \bar \lambda$ – user1754322 Dec 28 '20 at 21:03
  • @user1754322 the complex conjugation $\sigma(x): x\mapsto \bar{x}$ is stable under addition and multiplication. Once that it is verified it is relatively straightforward to see that the same holds for matrix multiplication (which just consists of indexwise multiplications and additions operations which can be pulled out of the complex conjugation). So you would go $\bar{A}\bar{x} = \sigma(Ax) = \sigma(\lambda x) = \bar{\lambda}\bar{x}$ – Felix B. Jun 28 '21 at 08:46
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    Good looking proof, except for one point. At the first line, there says "$\lambda\in\mathbb R$". This is what you're going to prove and actually is proved via "$\lambda=\bar\lambda$" So I think the first line should be like "Let $Ax=\lambda x$ with $x\neq0$, then" – govin Nov 02 '21 at 02:14

Hint: Prove that $$x^\ast A x=\langle x , A x\rangle = \langle Ax, x\rangle = x^\ast A^\ast x $$ Where $A^\ast=\overline{A}^T$

Dominic Michaelis
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    As $A^*=A$, you even get $x^*A^*x=x^*Ax$ directly without any intermediate step. ANd that's all you need for nonzero eigenvalues to be real. I have just seen you hint, by the way, +1. – Julien Apr 07 '13 at 19:56
  • It's homework so I thought i only give a hint. yeah the equation is pretty obvious but it's worth mentioning in homework i guess – Dominic Michaelis Apr 07 '13 at 19:59
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    I'm afraid you misunderstood my comment. It is great that you only gave a hint. I was just saying that the first two equalities are not necessary. But sure, it does not hurt either to include them. – Julien Apr 07 '13 at 20:01

If $\lambda$ is any eigenvalue of a Hermitian (in particular real symmetric) matrix $A$, then for some non-zero vector $x$, $$Ax = \lambda x \implies x^*Ax = \lambda x^*x \implies \lambda = \dfrac{x^*Ax}{x^*x}.$$

Now $$\lambda^* = \dfrac{x^* A^* x}{x^*x} = \dfrac{x^*Ax}{x^*x} = \lambda.$$ Therefore, $\lambda$ is real.

Note: $(AB)^* = B^*A^*$.

M. Vinay
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I found the proof in this document to be informative and educational.

The Spectral Theorem states that if $A$ is an $n \times n$ symmetric matrix with real entries, then it has $n$ orthogonal eigenvectors. The first step of the proof is to show that all the roots of the characteristic polynomial of $A$ (i.e. the eigenvalues of $A$) are real numbers.

Recall that if $z = a + bi$ is a complex number, its complex conjugate is defined by $\bar{z} = a − bi$. We have $z \bar{z} = (a + bi)(a − bi) = a^2 + b^2$, so $z\bar{z}$ is always a nonnegative real number (and equals $0$ only when $z = 0$). It is also true that if $w$, $z$ are complex numbers, then $\overline{wz} = \bar{w}\bar{z}$.

Let $\mathbf{v}$ be a vector whose entries are allowed to be complex. It is no longer true that $\mathbf{v} \cdot \mathbf{v} \ge 0$ with equality only when $\mathbf{v} = \mathbf{0}$. For example,

$$\begin{bmatrix} 1 \\ i \end{bmatrix} \cdot \begin{bmatrix} 1 \\ i \end{bmatrix} = 1 + i^2 = 0$$

However, if $\bar{\mathbf{v}}$ is the complex conjugate of $\mathbf{v}$, it is true that $\mathbf{v} \cdot \mathbf{v} \ge 0$ with equality only when $\mathbf{v} = 0$. Indeed,

$$\begin{bmatrix} a_1 - b_1 i \\ a_2 - b_2 i \\ \dots \\ a_n - b_n i \end{bmatrix} \cdot \begin{bmatrix} a_1 + b_1 i \\ a_2 + b_2 i \\ \dots \\ a_n + b_n i \end{bmatrix} = (a_1^2 + b_1^2) + (a_2^2 + b_2^2) + \dots + (a_n^2 + b_n^2)$$

which is always nonnegative and equals zero only when all the entries $a_i$ and $b_i$ are zero.

With this in mind, suppose that $\lambda$ is a (possibly complex) eigenvalue of the real symmetric matrix $A$. Thus there is a nonzero vector $\mathbf{v}$, also with complex entries, such that $A\mathbf{v} = \lambda \mathbf{v}$. By taking the complex conjugate of both sides, and noting that $A = A$ since $A$ has real entries, we get $\overline{A\mathbf{v}} = \overline{\lambda \mathbf{v}} \Rightarrow A \overline{\mathbf{v}} = \overline{\lambda} \overline{\mathbf{v}}$. Then, using that $A^T = A$,

$$\overline{\mathbf{v}}^T A \mathbf{v} = \overline{\mathbf{v}}^T(A \mathbf{v}) = \overline{\mathbf{v}}^T(\lambda \mathbf{v}) = \lambda(\overline{\mathbf{v}} \cdot \mathbf{v}),$$

$$\overline{\mathbf{v}}^T A \mathbf{v} = (A \overline{\mathbf{v}})^T \mathbf{v} = (\overline{\lambda} \overline{\mathbf{v}})^T \mathbf{v} = \overline{\lambda}(\overline{\mathbf{v}} \cdot \mathbf{v}).$$

Since $\mathbf{v} \not= \mathbf{0}$,we have $\overline{\mathbf{v}} \cdot \mathbf{v} \not= 0$. Thus $\lambda = \overline{\lambda}$, which means $\lambda \in \mathbb{R}$.

For further information on how the author gets from $\overline{\mathbf{v}}^T(\lambda \mathbf{v})$ to $\lambda(\overline{\mathbf{v}} \cdot \mathbf{v})$ and from $(\overline{\lambda} \overline{\mathbf{v}})^T \mathbf{v}$ to $\overline{\lambda}(\overline{\mathbf{v}} \cdot \mathbf{v})$, see this question.

The Pointer
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Since $A^* = A$, $(x^*Ax)^* = x^*Ax$. Therefore $x^*Ax$ is a real number for any $x$. If $x$ is an eigenvalue of $A$ with eigenvalue $\lambda$, we have $x^*Ax = x^*(\lambda x) = \lambda x^*x$. Since $x^*Ax$ and $x^*x$ are always real (and $x^*x$ is not zero for an eigenvector $x$), this means $\lambda$ must be real too.

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We are given that A is real symmetric, i.e \begin{align*} A = A^T \end{align*} If A were to have complex eigenvalues, then we can write \begin{align*} Ax = \lambda x \\ A\bar{x} = \bar{\lambda}\bar{x} \end{align*} Under complex conjugation, we can write \begin{align} \bar{x}^TAx = \bar{x}^T\lambda x = \lambda ||x||^2 \tag{i} \\ x^TA\bar{x} = x^T\bar{\lambda}x = \bar{\lambda}||x||^2 \tag{ii}.\\ \end{align} Since A is symmetric, $$\begin{align} \bar{x}^TAx = & (Ax)^{T} \bar{x} \\ = & x^{T} A^{T} \bar{x} \\ = & x^{T} A \bar{x}. \end{align}$$ Subtracting (i) from (ii), we get \begin{align*} \bar{\lambda}||x||^2 - \lambda ||x||^2 = 0\\ (\bar{\lambda}-\lambda)||x||^2 = 0 \end{align*} Only way this is possible for a non-zero z is if \begin{align*} \lambda = \bar{\lambda} \end{align*} Therefore, $\lambda$ is real.

R. W. Prado
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Hint: for every $n\times n$ matrix $M$ $$ \langle Mv , w\rangle ~=~ \langle v , M^H w\rangle $$ where $M^H$ is the conjugate transpose of $M$ and $\langle\,\cdot\,,\,\cdot\,\rangle$ is the complex inner product (i.e. $\langle v,w\rangle=v^Hw$).

  1. Think about how the eigenvalues of $M^H$ and those of $M$ are related
  2. Let $v$ be a $\lambda$-eigenvector of $M$ and try what happens choosing $v=w$ in the above equation
  3. Now, if $A$ is real valued and symmetric then $A^H=A$; try again the second point with $M=A$...
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Consider the real operator $$u := (x \mapsto Ax)$$ for all $x \in \mathbb{R}^{n}$ and the complex operator $$\tilde{u} := (x \mapsto Ax) $$ for all $x \in \mathbb{C}^{n}$. Both operators have the same characteristic polynomial, say $p(\lambda) = \det(A - \lambda I)$. Since $A$ is symmetric, $\tilde{u}$ is an hermitian operator. For the spectral theorem for hermitian operators all the eigenvalues (i.e. the roots of the $p(\lambda)$) of $\tilde{u}$ are real. Hence, all the eigenvalues (i.e. the roots of the $p(\lambda)$) of $u$ are real.

We have shown that the eigenvalues of a symmetric matrix are real numbers as a consequence of the fact that the eigenvalues of an Hermitian matrix are reals.

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