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# 6.1: Basic Facts about Eigenvalue Problems

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Even if a matrix $$\mathbf{A}$$ is real, its eigenvectors and eigenvalues can be complex. For example,

$\begin{bmatrix}1&1\\-1&1\end{bmatrix} \begin{bmatrix}1\\i\end{bmatrix} = (1+i) \begin{bmatrix}1\\i\end{bmatrix}.$

Eigenvectors are not uniquely defined. Given an eigenvector $$\vec{x}$$, any nonzero complex multiple of that vector is also an eigenvector of the same matrix, with the same eigenvalue. We can reduce this ambiguity by normalizing eigenvectors to a fixed unit length:

$\sum_{n=0}^{N-1} |x_n|^2 = 1.$

Note, however, that even after normalization, there is still an inherent ambiguity in the overall complex phase. Multiplying a normalized eigenvector by any phase factor $$e^{i\phi}$$ gives another normalized eigenvector with the same eigenvalue.

## 6.1.1 Matrix Diagonalization

Most matrices are diagonalizable, meaning that their eigenvectors span the $$N$$-dimensional complex space (where $$N$$ is the matrix size). Matrices which are not diagonalizable are called defective. Many classes of matrices that are relevant to physics (such as Hermitian matrices) are always diagonalizable; i.e., never defective.

The reason for the term "diagonalizable" is as follows. A diagonalizable $$N\times N$$ matrix $$\mathbf{A}$$ has eigenvectors that span the $$N$$-dimensional space, meaning that we can choose $$N$$ linearly independent eigenvectors, $$\{\vec{x}_0, \vec{x}_1, \cdots \vec{x}_{N-1}\}$$, with eigenvalues $$\{\lambda_0, \lambda_1, \cdots \lambda_{N-1}\}$$. We refer to such a set of $$N$$ eigenvalues as the "eigenvalues of $$\mathbf{A}$$". If we group the eigenvectors into an $$N\times N$$ matrix

$\mathbf{Q} = [\vec{x}_0, \vec{x}_1, \cdots \vec{x}_{N-1}],$

then, since the eigenvectors are linearly independent, $$\mathbf{Q}$$ is guaranteed to be invertible. Using the eigenvalue equation, we can then show that

$\mathbf{Q}^{-1} \,\mathbf{A} \, \mathbf{Q} = \begin{bmatrix}\lambda_0 & 0& \cdots & 0 \\ 0 & \lambda_1 & \cdots & 0 \\ \vdots& \vdots & \ddots & \vdots \\ 0&0&\cdots&\lambda_{N-1}\end{bmatrix}.$

In other words, there exists a similarity transformation which converts $$\mathbf{A}$$ into a diagonal matrix. The $$N$$ numbers along the diagonal are precisely the eigenvalues of $$\mathbf{A}$$.

## 6.1.2 The Characteristic Polynomial

One of the most important consequences of diagonalizability is that the determinant of a diagonalizable matrix $$\mathbf{A}$$ is the product of its eigenvalues:

$\det(\mathrm{A}) = \prod_{n=0}^{N-1} \lambda_n$

This can be proven by taking the determinant of the similarity transformation equation, and using (i) the property of the determinant that $$\det(\mathbf{U}\mathbf{V}) = \det(\mathbf{U})\det(\mathbf{V})$$, and (ii) the fact that the determinant of a diagonal matrix is the product of the elements along the diagonal.

In particular, the determinant of $$\mathbf{A}$$ is zero if one of its eigenvalues is zero. This fact can be further applied to the following re-arrangement of the eigenvalue equation:

$\Big(\mathbf{A} - \lambda\mathbf{I}\Big) \, \vec{x} = 0,$

where $$\mathbf{I}$$ is the $$N\times N$$ identity matrix. This says that the matrix $$\mathbf{A}-\lambda\mathbf{I}$$ has an eigenvalue of zero, meaning that for any eigenvalue $$\lambda$$,

$\det\left(\mathbf{A} - \lambda\mathbf{I}\right) = 0.$

The left-hand side of the above equation is a polynomial in the variable $$\lambda$$, of degree $$N$$. This is called the characteristic polynomial of the matrix $$\mathbf{A}$$. Its roots are eigenvalues of $$\mathbf{A}$$, and vice versa.

For $$2\times 2$$ matrices, the standard way of calculating the eigenvalues is to find the roots of the characteristic polynomial. However, this is not a reliable method for finding the eigenvalues of larger matrices. There is a well-known and important result in mathematics, known as Abel's impossibility theorem, which states that polynomials of degree $$5$$ and higher have no general algebraic solution. (By comparison, degree-2 polynomials have a general algebraic solution, which is the familiar quadratic formula, and similar formulas exist for degree-3 and degree-4 polynomials.) A matrix of size $$N \ge 5$$ has a characteristic polynomial of degree $$N \ge 5$$, and Abel's impossibility theorem tells us that we can't calculate the roots of that characteristic polynomial by ordinary arithmetic.

In fact, Abel's impossibility theorem leads to an even stronger conclusion: there is no general algebraic method for finding the eigenvalues of a matrix of size $$N \ge 5$$, whether using the characteristic polynomial or any other method. For suppose we had such a method for finding the eigenvalues of a matrix. Then, for any polynomial equation of degree $$N \ge 5$$, of the form

$a_0 + a_1 \lambda + \cdots + a_{N-1} \lambda^{N-1} + \lambda^N = 0,$

we can construct an $$N\times N$$ "companion matrix" of the form

$\mathbf{A} = \begin{bmatrix}0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1& \cdots & 0 \\ \vdots&\vdots&\ddots&\ddots& \vdots \\ 0&0&0&\ddots&1\\-a_0& -a_1& -a_2 & \cdots & -a_{N-1} \end{bmatrix}.$

As you can check for yourself, each root $$\lambda$$ of the polynomial is also an eigenvalue of the companion matrix, with corresponding eigenvector

$\vec{x} = \begin{bmatrix}1\\\lambda\\ \vdots \\ \lambda^{N-1}\end{bmatrix}.$

Hence, if there exists a general algebraic method for finding the eigenvalues of a large matrix, that would allow us to find solve polynomial equations of high degree. Abel's impossibility theorem tells us that no such solution method can exist.

This might seem like a terrible problem, but in fact there's a way around it, as we'll shortly see.

## 6.1.3 Hermitian Matrices

A Hermitian matrix $$\mathbf{H}$$ is a matrix which has the property

$\mathbf{H}^\dagger = \mathbf{H},$

where $$\mathbf{H}^\dagger$$ denotes the "Hermitian conjugate", which is matrix transposition accompanied by complex conjugation:

$\mathbf{H}^\dagger \equiv \left(\mathbf{H}^T\right)^*, \quad \mathrm{i.e.}\;\;\left(H^\dagger\right)_{ij} = H_{ji}^*.$

Hermitian matrices have the nice property that all their eigenvalues are real. This can be easily proven using index notation:

\begin{align}\sum_j H_{ij} x_j = \lambda x_i \;\;&\Rightarrow\;\; \sum_j x_j^* H_{ji} = \lambda^* x_i^*\\ &\Rightarrow \sum_{ij} x^*_i H_{ij} x_j = \lambda \sum_i |x_i|^2 = \lambda^* \sum_j |x_j|^2 \\ &\Rightarrow \lambda = \lambda^*.\end{align}

In quantum mechanics, Hermitian matrices play a special role: they represent measurement operators, and their eigenvalues (which are restricted to the real numbers) are the set of possible measurement outcomes.

6.1: Basic Facts about Eigenvalue Problems is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Y. D. Chong via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.