Let \(A\) be an \(n \times n\) matrix. The scalar \(\lambda\) is called an eigenvalue of \(A\) when there is a nonzero vector \(\mathbf{x}\) such that
\begin{equation*}
A \mathbf{x}=\lambda \mathbf{x}.
\end{equation*}
The vector \(\mathbf{x}\) is an eigenvector of \(A\) corresponding to \(\lambda\text{.}\)
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, \((A-\lambda I_n) \mathbf{x}=\mathbf{0}\text{.}\)
The matrix equation \(A\mathbf{x}=\mathbf{0}\) has nontrivial solution if and only if \(\underline{\operatorname{det}(A)= 0}\)
Thus the eigenvalue of a matrix should satisfy \(\underline{\hspace{5cm}}\text{.}\) The corresponding eigenvectors should satisfy \(\underline{\hspace{5cm}}\text{.}\)
Example7.1.1.
Find the eigenvalues and eigenvectors of the matrix
The number \(k_{i}\) is called the algebriac multiplicity of the eigenvalue \(\lambda_i\text{.}\)
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 \(\{\mathbf{x}\in \mathbb{R}^{n}|(A-\lambda_{i}I_{n})\mathbf{x}=\mathbf{0}\}.\)
Note that the sum of algebraic multiplicities of all eigenvalues is the size of the matrix \(A\text{.}\) Thus if for every eigenvalue \(\lambda_{i}\text{,}\) the geometric multiplicity is equal to the algebraic multiplicity, then we will have \(n\) linear independent eigenvectors.
Theorem7.1.3.
If for every eigenvalue \(\lambda\) of \(A\text{,}\) the algebraic multiplicity equals the geometric multiplicity, then the matrix \(A\) is diagonalizable.