To solve the matrix equation above, the key is to notice that \(\lambda \mathbf{x}=(\lambda I_n)\mathbf{x} \text{.}\) Thus \(A \mathbf{x}=\lambda \mathbf{x}\) is equivalent to \(A \mathbf{x}-(\lambda I_n) \mathbf{x}=\mathbf{0}\) that is, \((A-\lambda I_n) \mathbf{x}=\mathbf{0}\text{.}\)
Thus the eigenvalue of a matrix should satisfy \(\underline{\hspace{5cm}}\text{.}\) The corresponding eigenvectors should satisfy \(\underline{\hspace{5cm}}\text{.}\)
Let \(A\) be a square matrix. The polynomial \(\det(A-\lambda I_n)\) is called the characteristic polynomial of \(A\text{.}\) The eigenvalues of \(A\) are the roots of the characteristic equation \(\det(A-\lambda I_n)=0\) .
The Nullity(\(A-\lambda_{i}I_{n}\)) is called the geometric multiplicity of the eigenvalue \(\lambda_{i}\text{.}\) Note that the Nullity is always \(\geq 1\text{.}\) Recall that the nullity is the dimension of the subspace \(\{\mathbf{x}\in \mathbb{R}^{n}|(A-\lambda_{i}I_{n})\mathbf{x}=\mathbf{0}\}.\)
Subsection7.1.4The eigenvalue and eigenvectors of \(f(A)\)
Let \(A\) be an \(n\times n\) matrix with eigenpairs \((\lambda_i,\mathbf{v}_i)\text{,}\)\(i=1,\dots,k\text{.}\) For a polynomial function \(f\) we define
Let matrix \(A\) be an \(n \times n\) square matrix. Let \(\mathbf{v}\) be an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\text{.}\) Prove that: