next up previous
Next: Spherical harmonics Up: Useful mathematics Previous: Elliptic expansions

Matrix eigenvalue theory

Suppose that $ {\bf A}$ is a real symmetric square matrix of dimension $ n$ . If follows that $ {\bf A}^\ast = {\bf A}$ and $ {\bf A}^T = {\bf A}$ , where $ ~^\ast$ denotes a complex conjugate, and $ ~^T$ denotes a transpose. Consider the matrix equation

$\displaystyle {\bf A} \,{\bf x} = \lambda\,{\bf x}.$ (A.144)

Any column vector $ {\bf x}$ which satisfies this equation is called an eigenvector of $ {\bf A}$ . Likewise, the associated number $ \lambda $ is called an eigenvalue of $ {\bf A}$ (Gradshteyn and Ryzhik 1980c). Let us investigate the properties of the eigenvectors and eigenvalues of a real symmetric matrix.

Equation (A.144) can be rearranged to give

$\displaystyle ({\bf A} - \lambda\,{\bf 1})\,{\bf x} = {\bf0},$ (A.145)

where $ {\bf 1}$ is the unit matrix. The preceding matrix equation is essentially a set of $ n$ homogeneous simultaneous algebraic equations for the $ n$ components of $ {\bf x}$ . A well-known property of such a set of equations is that it only has a nontrivial solution when the determinant of the associated matrix is set to zero (Gradshteyn and Ryzhik 1980c). Hence, a necessary condition for the preceding set of equations to have a nontrivial solution is that

$\displaystyle \vert{\bf A} - \lambda\,{\bf 1}\vert = 0,$ (A.146)

where $ \vert\cdots\vert$ denotes a determinant. This formula is essentially an $ n$ th-order polynomial equation for $ \lambda $ . We know that such an equation has $ n$ (possibly complex) roots. Hence, we conclude that there are $ n$ eigenvalues, and $ n$ associated eigenvectors, of the $ n$ -dimensional matrix $ {\bf A}$ .

Let us now demonstrate that the $ n$ eigenvalues and eigenvectors of the real symmetric matrix $ {\bf A}$ are all real. We have

$\displaystyle {\bf A}\,{\bf x}_i = \lambda_i\,{\bf x}_i,$ (A.147)

and, taking the transpose and complex conjugate,

$\displaystyle {\bf x}_i^{\ast\,T}\,{\bf A} = \lambda_i^{\,\ast}\,{\bf x}_i^{\ast\,T},$ (A.148)

where $ {\bf x}_i$ and $ \lambda_i$ are the $ i$ th eigenvector and eigenvalue of $ {\bf A}$ , respectively. Left multiplying Equation (A.147) by $ {\bf x}_i^{\ast\,T}$ , we obtain

$\displaystyle {\bf x}_i^{\ast\,T} {\bf A}\,{\bf x}_i = \lambda_i\,{\bf x}_i^{\ast\,T}{\bf x}_i.$ (A.149)

Likewise, right multiplying Equation (A.148) by $ {\bf x}_i$ , we get

$\displaystyle {\bf x}_i^{\ast\,T}\,{\bf A}\,{\bf x}_i = \lambda_i^{\,\ast}\,{\bf x}_i^{\ast\,T}{\bf x}_i.$ (A.150)

The difference of the previous two equations yields

$\displaystyle (\lambda_i - \lambda_i^{\,\ast})\,{\bf x}_i^{\ast\,T} {\bf x}_i = 0.$ (A.151)

It follows that $ \lambda_i=\lambda_i^{\,\ast}$ , because $ {\bf x}_i^{\ast\,T}{\bf x}_i$ (which is $ {\bf x}_i^{\,\ast}\cdot{\bf x}_i$ in vector notation) is real and positive definite. Hence, $ \lambda_i$ is real. It immediately follows that $ {\bf x}_i$ is real.

Next, let us show that two eigenvectors corresponding to two different eigenvalues are mutually orthogonal. Let

    $\displaystyle {\bf A}\,{\bf x}_i$ $\displaystyle = \lambda_i\,{\bf x}_i,$ (A.152)
and   $\displaystyle {\bf A}\,{\bf x}_j$ $\displaystyle = \lambda_j\,{\bf x}_j,$     (A.153)

where $ \lambda_i\neq \lambda_j$ . Taking the transpose of the first equation and right multiplying by $ {\bf x}_j$ , and left multiplying the second equation by $ {\bf x}_i^T$ , we obtain

    $\displaystyle {\bf x}_i^T\,{\bf A}\,{\bf x}_j$ $\displaystyle = \lambda_i\,{\bf x}_i^T{\bf x}_j,$ (A.154)
and   $\displaystyle {\bf x}_i^T\,{\bf A}\,{\bf x}_j$ $\displaystyle = \lambda_j\,{\bf x}_i^T{\bf x}_j.$     (A.155)

Taking the difference of the preceding two equations, we get

$\displaystyle (\lambda_i-\lambda_j)\,{\bf x}_i^T{\bf x}_j = 0.$ (A.156)

Because, by hypothesis, $ \lambda_i\neq \lambda_j$ , it follows that $ {\bf x}_i^T{\bf x}_j = 0$ . In vector notation, this is the same as $ {\bf x}_i \cdot{\bf x}_j=0$ . Hence, the eigenvectors $ {\bf x}_i$ and $ {\bf x}_j$ are mutually orthogonal.

Suppose that $ \lambda_i=\lambda_j=\lambda$ . In this case, we cannot conclude that $ {\bf x}_i^T{\bf x}_j = 0$ by the preceding argument. However, it is easily seen that any linear combination of $ {\bf x}_i$ and $ {\bf x}_j$ is an eigenvector of $ {\bf A}$ with eigenvalue $ \lambda $ . Hence, it is possible to define two new eigenvectors of $ {\bf A}$ , with the eigenvalue $ \lambda $ , which are mutually orthogonal. For instance,

    $\displaystyle {\bf x}_i'$ $\displaystyle = {\bf x}_i,$ (A.157)
and   $\displaystyle {\bf x}_j'$ $\displaystyle = {\bf x}_j - \left(\frac{{\bf x}_i^T{\bf x}_j}{{\bf x}_i^T{\bf x}_i}\right) {\bf x}_i.$     (A.158)

It should be clear that this argument can be generalized to deal with any number of eigenvalues that take the same value.

In conclusion, a real symmetric $ n$ -dimensional matrix possesses $ n$ real eigenvalues, with $ n$ associated real eigenvectors, which are, or can be chosen to be, mutually orthogonal.

next up previous
Next: Spherical harmonics Up: Useful mathematics Previous: Elliptic expansions
Richard Fitzpatrick 2016-03-31