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An introduction to the concept of eigenvalues and eigenvectors of a symmetric real matrix in the context of spectral graph theory. It explains how to find eigenvalues and eigenvectors, their relationship with the characteristic polynomial, and the orthogonal properties of eigenvectors. The document also includes an example of finding eigenvalues and eigenvectors for a specific matrix.
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Let A be an n × n-matrix. If there exists a non-zero column vector α and a number λ satisfying Aα = λα, we say λ is an eigenvalue of A and α is an eigenvector of A corresponding to the eigenvalue λ. For any c 6 = 0, cα is also an eigenvector of A. We can normalize it so that α becomes a unit vector. Note that α and λ is NOT necessary to be real even if A is a real matrix. All eigenvalues λ 1 , λ 2 ,... , λn are the root of the characteristic polynomial
det(λI − A) =
∏^ n i=
(λ − λi).
By comparing the coefficient, we have
∏^ n i=
λi = det(A) ∑^ n i=
λi = tr(A).
Here tr(A) = ∑ i=1 aii is called the trace of a matrix. Let f (x) be any polynomial of x. It is easy to check that f (A)α = f (λ)α.
In conclusion, f (A) has eigenvalues f (λ 1 ), f (λ 2 ),... , f (λn). In particular, we have (^) ∑n
i=
λki = tr(Ak). From now on, we assume A is a symmetric real matrix (i.e., A′^ = A). Then all eigenvalues are real and all eigenvectors can be made into an orthogonal base. Two eigenvectors corresponding to the distinct eigenvalues are always or- thogonal to each other. Two eigenvectors corresponding to the same eigenvalue is not always orthogonal to each other. A matrix O is orthogonal if O′^ = O−^1. There exists an orthogonal matrix O satisfying AO = OΛ. Here Λ is a diagonal matrix with diagonal entries λ 1 , λ 2 ,... , λn. Let αi be the i-th column of O. Then αi is the eigenvector corresponding to λi. I.e.,
Aαi = λiαi.
Example 1: Find all eigenvalues and eigenvectors of
A =
Solution:
det(λI − A) = det
λ − 1 0 − 1 λ − 1 0 − 1 λ
= λ^3 − 2 λ = λ(λ −
2)(λ +
Thus, A has three distinct eigenvalues √ 2 , 0 , −√2. To find an eigenvector for λ 1 = √2, we solve
x y z
We have y = √ 2 x and y = √ 2 z. This system of linear equations has infinitely many solutions of form
c
Up to a sign, we get a unit eigenvector
α 1 =
√^12 22 (^12)
To find an eigenvector for λ 1 = 0, we solve
x y z
We have y = 0 and x + z = 0. This system of linear equations has infinitely many solutions of form
c
Up to a sign, we get a unit eigenvector
α 2 =
√ 2 (^20) − √ 2 2