Ngô Quốc Anh

August 16, 2009

The eigenvalues of a common tridiagonal matrix

Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 3:36

The eigenvalues of then N \times N tridiagonal matrix

\left( {\begin{array}{*{20}{c}}    a & b & {} & {} & {} & {} & {}  \\    c & a & b & {} &...

are

{\lambda _s} = a + 2\sqrt {bc} \cos \frac{{s\pi }} {{N + 1}}

and its conresponding eigenvector is

{v_s}^T = \left( {{{\left( {\frac{c} {b}} \right)}^{\frac{1} {2}}}\sin \frac{{s\pi }} {{N + 1}},{{\left( {\frac{c} {b}} \righ...

Proof. Let \lambda present an eigenvalue of the given matrix (denoted by A) and v the corresponding eigenvector with components v_1,...,v_N. Then

\left( {\begin{array}{*{20}{c}}    a & b & {} & {} & {} & {} & {}  \\    c & a & b & {} &...

If we defined v_0 = v_{N+1}=0, then we have

c{v_{i - 1}} + \left( {a - \lambda } \right){v_i} + b{v_{i + 1}} = 0,i = \overline {1,N} .

The solution of the above equation is of the form

{v_i} = Bm_1^i + Cm_2^i

where B, C are constants and m_1, m_2 are the roots of the equation

c+(a-\lambda)m + bm^2=0.

Since v_0 = v_{N+1}=0, then 0=B+C and 0 = Bm_1^{N + 1} + Cm_2^{N + 1}. Hence,

{\left( {\frac{{{m_1}}} {{{m_2}}}} \right)^{N + 1}} = 1 = {e^{\sqrt { - 1} i2\pi }}

or

\frac{{{m_1}}} {{{m_2}}} = {e^{\frac{{\sqrt { - 1} i2\pi }} {{N + 1}}}}.

We also have

{m_1}{m_2} = \frac{c} {b}, \quad {m_1} + {m_2} = \frac{{\lambda  - a}} {b}.

Finally,

\lambda  = a + b\sqrt {\frac{c} {b}} \left( {{e^{\frac{{\sqrt { - 1} i2\pi }} {{N + 1}}}} + {e^{ - \frac{{\sqrt { - 1} i2\pi ...

The j-component of the eigenvector is

{v_j} = Bm_1^j + Cm_2^j = B\sqrt {{{\left( {\frac{c} {b}} \right)}^j}} \left( {{e^{\frac{{\sqrt { - 1} i2\pi }} {{N + 1}}}} -...

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