Let \(A\in\mathbb{F}^{\,n\times n}\) be a square matrix and \(\underline{v} \in \mathbb{F}^n\) be a non-zero vector. \(\underline{v}\) is an eigenvector of \(A\) if and only if

\[A\underline{v}=\lambda\underline{v}\]

where \(\lambda\in\mathbb{F}\) is called an eigenvalue of \(A\) if \(\underline{v}\) exists.

The linear transformation \(A\underline{v}\) where \(\underline{v}\) is an eigenvector of \(A\) is equivalent to the scalar transformation \(\lambda\underline{v}.\)

Let \(\lambda_1,\ldots,\lambda_n \in \mathbb{C}\) be the eigenvalues of \(A,\) repeated according to their algebraic multiplicity (defined below). The eigenvalues satisfy the following properties:

\[\sum_{i=1}^n \lambda_i = \mathrm{tr}(A)\] \[\prod_{i=1}^n \lambda_i = \lvert A\rvert\]

The expression \(\lvert A-\lambda I_n\rvert\) is a polynomial of degree \(n\) in \(\lambda\) called the characteristic polynomial of \(A.\) The roots of \(\lvert A-\lambda I_n\rvert=0\) are the eigenvalues of \(A.\)

The set of all distinct eigenvalues of \(A\) is called the eigen-spectrum \(\sigma_A\) of \(A.\)

\[\sigma_A=\{\lambda : \lvert A-\lambda I_n\rvert=0\}\]

The characteristic polynomial of \(A\) will admit \(m\) distinct complex roots where \(1\leq m\leq n.\) Therefore, taking eigenvalues over \(\mathbb{C},\) the cardinality of the eigen-spectrum \(\sigma_A\) is \(m.\) A real matrix can have complex eigenvalues: the rotation matrix \(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\) has eigenvalues \(\pm i\) and no real eigenvalues. The results below that count \(n\) eigenvalues assume that all roots of the characteristic polynomial lie in \(\mathbb{F},\) which always holds for \(\mathbb{F}=\mathbb{C}.\)

For each eigenvalue \(\lambda_i\in\sigma_A,\) any non-zero solution to

\[\left(A-\lambda_i I_n\right)\underline{v_i}=\underline{0}\]

is an eigenvector \(\underline{v_i}\) of \(A\) corresponding to \(\lambda_i.\)

The eigen-space \(\mathcal{E}_{\lambda}\) of \(A\) with respect to eigenvalue \(\lambda\) is the vector space that is spanned by all eigenvectors of \(A\) corresponding to \(\lambda.\) Equivalently, it is the kernel of \(A-\lambda I_n,\) which contains these eigenvectors together with the zero vector:

\[\mathcal{E}_{\lambda}=\{\underline{v} \in \mathbb{F}^n : \left(A-\lambda I_n\right)\underline{v}=\underline{0}\}\]

The algebraic multiplicity \(\mu_A\left(\lambda\right)\) of \(\lambda\) with respect to the matrix \(A\) is the number of times \(\lambda\) appears as a root of the characteristic polynomial of \(A.\) The sum of the algebraic multiplicities of all eigenvalues of \(A\) is \(n.\)

The geometric multiplicity \(\gamma_A\left(\lambda\right)\) of \(\lambda\) with respect to the matrix \(A\) is the maximum number of linearly independent eigenvectors of \(A\) corresponding to \(\lambda\) and is equivalent to the dimension of the eigen-space of \(A\) with respect to \(\lambda.\)

The algebraic and geometric multiplicities are related by:

\[1\leq\gamma_A\left(\lambda\right)\leq\mu_A\left(\lambda\right)\leq n\,\,\forall\,\lambda\in\sigma_A\]

Eigenvectors corresponding to distinct eigenvalues are always linearly independent, but this inequality shows that a repeated eigenvalue does not necessarily correspond to multiple linearly independent eigenvectors.

\(A\) will admit \(n\) linearly independent eigenvectors if and only if:

\[\gamma_A\left(\lambda_i\right)=\mu_A\left(\lambda_i\right)\,\,\forall\,\lambda_i\in\sigma_A.\]

\(A\) will admit fewer than \(n\) linearly independent eigenvectors if and only if:

\[\gamma_A\left(\lambda_i\right)<\mu_A\left(\lambda_i\right)\,\,\mathrm{for}\,\mathrm{some}\,\,\lambda_i\in\sigma_A\]

and \(A\) is referred to as defective.

For a non-defective matrix \(A,\) the \(n\) linearly independent eigenvectors will form a basis for \(\mathbb{F}^n\) and is called an eigen-basis.

Let \(A\) be a non-defective matrix with \(n\) linearly independent eigenvectors \(\underline{v}_1,\ldots,\underline{v}_n\) and corresponding eigenvalues \(\lambda_1,\ldots,\lambda_n.\) The following matrices can be constructed:

\[P=\begin{bmatrix} \uparrow & & \uparrow \\ \underline{v}_1 & \ldots & \underline{v}_n \\ \downarrow & & \downarrow \end{bmatrix}\] \[\Lambda=\begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & \lambda_n \end{bmatrix}\]

where \(P\) is the eigenvectors in columnar form stacked horizontally to form a square matrix and \(\Lambda\) is a diagonal matrix with eigenvalues along the diagonal.

The diagonalisation of \(A\) is the decomposition:

\[A=P\Lambda P^{-1}\]