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Properties of Eigenvalues 📂Matrix Algebra

Properties of Eigenvalues

Definition

$n \times n$ matrix $A$ and a $n$-dimensional vector $\mathbf{x} \ne \mathbf{0}$, a scalar $\lambda$ that satisfies the following equation is called an eigenvalue of $A$.

$$ A \mathbf{x} = \lambda \mathbf{x} $$

Also, in this case $\mathbf{x}$ is called an eigenvector corresponding to $\lambda$.

Properties

(a) Every matrix has at least one eigenvalue.

(b) For a positive integer $k$, if $\lambda$ is an eigenvalue of the matrix $A$ and $\mathbf{x}$ is an eigenvector corresponding to $\lambda$, then $\lambda ^{k}$ is an eigenvalue of $A^{k}$ and $\mathbf{x}$ is the eigenvector corresponding to $\lambda ^{k}$.

(c) For a real matrix $A \in M_{n \times n}(\mathbb{R})$, if $\lambda$ is an eigenvalue of $A$, then $\lambda$ is also an eigenvalue of $A^{\mathsf{T}}$. Here $A^{\mathsf{T}}$ is the transpose of $A$.

(c’) For a complex matrix $A \in M_{n \times n}(\mathbb{C})$, if $\lambda$ is an eigenvalue of $A$, then $\overline{\lambda}$ is an eigenvalue of $A^{\ast}$. Here $\overline{\lambda}$ is the complex conjugate of $\lambda$, and $A^{\ast}$ is the conjugate transpose of $A$.

(c’’) If $\lambda_{i}$ is an eigenvalue of $A$, then the eigenvalue of $A^{\mathsf{T}}A$ is $\lambda_{i}^{2}$ and the eigenvalue of $A^{\ast}A$ is $|\lambda_{i}|^{2}$.

Explanation

In (c), the sets of eigenvalues of $A$ and $A^{\mathsf{T}}$ are the same, but in general the eigenvectors corresponding to each eigenvalue are not the same.

Proof

(a)

The characteristic equation of the matrix $n \times n$ is a polynomial of degree $n$, which therefore has at least one root; hence every matrix has at least one eigenvalue.

(b)

Trivial.

(c)

An eigenvalue of $A$ is a root of the characteristic equation $\det(A - \lambda I) = 0$. Since the value of a determinant does not depend on transposition (see $\det A = \det A^{\mathsf{T}}$), we have

$$ \det(A - \lambda I) = \det(A^{\mathsf{T}} - \lambda I^{\mathsf{T}}) = \det (A^{\mathsf{T}} - \lambda I) $$

Therefore $A$ and $A^{\mathsf{T}}$ have the same eigenvalues. By the same reasoning, (c’) also holds.