# Are the real eigenvalues of real symmetric matrices continuous?

Given an $$n \times n$$ symmetric matrix with real coefficients it has $$n$$ eigenvalues. I was wondering are the eigenvalues continuous with respect to the coefficients of the matrices? I have seen somewhere that the eigenvalues of matrices are continuous but it was for complex matrices.

Does this holds for real symmetric matrices? I would guess that it does, but I was not sure how to show it. Any reference or comments would be appreciated. Thank you.

Added after comment: For the complex case I understand as the characteristic polynomial is a polynomial in the coefficients of the matrix, and the roots (in $$\mathbb{C}$$) of a polynomial over $$\mathbb{C}$$ varies continuously.

However, over $$\mathbb{R}$$ I wasn't sure if I could still say the same, as for example, polynomial like $$x^2 + a$$ no longer has root after $$a > 0$$. So I wasn't sure if I could still say the same about real eigenvalues of real matrices. And so, I wasn't sure about the real eigenvlaues of symmetric real matrices either...

I know that the reals are a subset of the complexes, but I wasn't sure if that was enough to obtain the statement for the real eigenvalues of symmetric real matrices... Any clarification would be appreciated. Thank you.

• Aren't real numbers complex numbers with imaginary part equal to zero? – broncoAbierto Jun 26 at 10:59
• There must be something I'm missing here, I explained more in the question. Thank you – Johnny T. Jul 7 at 9:02
• If I understand your problem correctly, you should note that the eigenvalues of real symmetric matrices are necessarily real. Hence, there may be situtations in which variation of the matrix coefficients leads to different eigenvalues "fusing", but it can never happen that they all "disappear". – Mars Plastic Jul 7 at 11:08
• I guess you can approximate a solution to a polynomial equation by for example Newton's method, so when the elements of a matrix are perturbed a little bit the solution wont change much? But I am not sure if there would be chaos phenomenon. – wilsonw Jul 8 at 1:36
• math.stackexchange.com/questions/63196/… – leonbloy Jul 9 at 2:05

## 2 Answers

The set of real numbers is a subset of the set of complex numbers, if we consider that real numbers are complex numbers with imaginary part equal to zero. Therefore, whatever holds for all complex numbers holds for real numbers.

As you observe, a polynomial might not have real roots. However, all the eigenvalues of a symmetric real matrix are real. By definition, they are the roots of the characteristic polynomial, so you can be sure that an example like the one you proposed will not arise.

For order $$2$$ matrix $$A=\begin{pmatrix}a&b\\b&c\\\end{pmatrix}$$, eigenvalues are $$\alpha$$ and $$\beta$$ s.t. $$\alpha+\beta=a+c$$ and $$\alpha\beta=ac-b^2$$. Suppose $$\epsilon>0$$ is the change in eigenvalues of $$A$$, then there always exist a $$\delta>0$$ ,where $$\delta=\epsilon$$ is the change in the diagonal of $$A$$ s.t.

$$A+\delta I=\begin{pmatrix}a+\epsilon &b\\b&c+\epsilon\\\end{pmatrix}$$has characteristic equation $$t^2-(a+2\epsilon+c)t+[{ac-b^2+(a+c)\epsilon+\epsilon^2}]=0$$ which clearly suggests that the new roots are $$\alpha+\epsilon$$ and $$\beta+\epsilon$$.