First, some clarifications. We are considering linear maps from a finite-dimensional vector space to itself (also called an operator L(V)), as opposed to linear maps from one vector space to another vector space.

Suppose V is a finite-dimensional vector space, T ∈ L(V), and λ ∈ F.

Some definitions to know:

• Invariant subspace: U is invariant under T if T|_U is an operator on U.

• λ ∈ F is an eigenvalue of T if ∃ v ∈ V s.t. v ≠ 0 and Tv = λv.

• The eigenspace of T corresponding to λ, E(λ, T) = null(T - λI) (the set of all eigenvectors of T corresponding to λ, along with the 0 vector.)

• Concepts from a previous post

Basics

• A linear map is invertible if and only if it is injective and surjective.

• For operators on a finite-dimensional vector space, however, either injectivity or surjectivity alone implies the other condition, and thus implies invertibility. (proved using Rank theorem)

• If λ is an eigenvalue of T, then by definition v ∈ ker(T - λI) so ker(T - λI) ≠ 0. Hence, T - λI is not injective (or surjective). That is, T - λI is not invertible, det(T - λI) = 0.

• Eigenvectors corresponding to distinct eigenvalues are linearly independent.

• Each operator on V has at most dim V distinct eigenvalues.

• If A is an n ×n matrix, then the sum of the n eigenvalues of A is the trace of A and the product of the n eigenvalues is the determinant of A.

• If λ is an eigenvalue of the T, λ² is an eigenvalue of T².

• n × n matrix A and its transpose A^T have the same eigenvalues.

• Suppose A and B are similar matrices. Then A and B have the same characteristic polynomial and hence the same eigenvalue.

• If λ is an eigenvalue of A, then the dimension of E_λ is at most the multiplicity of λ.

• If λ^* is an eigenvalue of A, then the multiplicity of λ^* is at least the dimension of the eigenspace E_λ^* .

Intermediate

• Any two polynomials of an operator commute.

• Every operator on a finite-dimensional, nonzero, complex vector space has an eigenvalue.

• Over C, every operator has an upper-triangular matrix

• Suppose T ∈ L(V) has an upper-triangular matrix with respect to some basis of V, then T is invertible if and only if all the entries on the diagonal of that upper-triangular matrix are nonzero;

• Suppose T ∈ L(V) has an upper-triangular matrix with respect to some basis of V, then the eigenvalues of T are precisely the entries on the diagonal of that upper-triangular matrix.

• Let λ₁, …, λ_m denote the distinct eigenvalues of T. Then, T is diagonalizable

         <=> V has a basis consisting of eigenvectors of T

         <=> ∃ 1-dimensional subspaces U₁, …, U_n of V, each invariant under T, such that V          = U₁ ⊕ … ⊕ U_n

         <=> V = E(λ₁, T) ⊕ … ⊕ E(λ_m, T)

         <=> dim V = dim E(λ₁, T) + … + dim E(λ_m, T)

• If T has dim V (enough) distinct eigenvalues, then T is diagonalizable. (The converse is not true, as the diagonalizable identity operator only has only 1 eigenvalue λ = 1)