Endomorphisms

Definition. An endomorphism of $V$ is a linear map $\alpha: V \to V$. The vector space of endomorphisms of $V$ is denoted by $\text{End}(V)$ and the identity endomorphism is denoted by $\iota$.

The key difference while considering endomorphisms as matrices is that the same basis will be used for both the domain and the range which leads to some other properties.

Basis

The standard basis of $\text{Mat}_n(\mathbb{F})$ can be useful sometimes for proving properties of the corresponding endomorphism.

Definition. The standard basis for $\text{Mat}_n(\mathbb{F})$ consists of $n^2$ matrices $\Set{E_{ij}}$ in which

\[(E_{ij})_{kl} = \begin{cases} 1 & i = k, j = l \\ 0 & \text{otherwise} \end{cases}\]

Proposition. $E_{ij} E_{kl} = \delta_{jk} E_{il}$.

Proof.

\[(E_{ij} E_{kl})_{pq} = \sum_{r = 1}^n (E_{ij})_{pr} (E_{kl})_{rq}\]

$(E_{ij})_{pr} = 1$ if $p = i$ and $r = j$ and $(E_{kl})_{rq} = 1$ if $q = l$ and $r = k$ so only $(E_{ij} E_{kl})_{il}$ can be non-zero provided that $j = k$.

Definition. The commutator of two elements $a$ and $b$ of a ring is the element $[a, b] = ab - ba$. The elements $a$ and $b$ are said to commute if the commutator is zero, i.e. $ab = ba$.

Proposition. Every off-diagonal basis matrices are commutators, specifically $E_{ij} = E_{ik}E_{kj} - E_{kj}E_{ik}$ for $i \not= j$.

Proof.

$E_{ik}E_{kj} = E_{ij}$ and $E_{kj}E_{ik} = 0$ when $i \not= j$.

Invariants

We will first address properties of matrices that are invariant up to similarity.

Proposition. Suppose that $(e_1, …, e_n)$ and $(f_1, …, f_n)$ are bases for $V$ such that $f_i = \sum P_{ki} e_k$. Let $\alpha \in \text{End}(V)$, $A$ be the matrix representing $\alpha$ with respect to $(e_i)$ and $B$ the matrix representing $\alpha$ with respect to $(f_i)$. Then $B = P^{-1}AP$.

Proof.

$P$ is the identity map from basis $(f_i)$ to $(e_i)$. The relation is just a special case of the change of basis formula with $Q = P$.

Definition. Matrices $A$ and $B$ are similar (or conjugate) if $B = P^{-1}AP$ for some invertible matrix $P$.

In terms of group theory, recall $\text{GL}_n(\mathbb{F})$ denotes all the invertible matrices so $\text{GL}_n(\mathbb{F})$ acts on $\text{Mat}_n(\mathbb{F})$ by conjugation and two matrices are similar if they lie in the same orbit, and similarity is an equivalence relation.

Definition. The trace of a matrix $A$ is defined by $\text{tr}: \text{Mat}_n(\mathbb{F}) \to \mathbb{F}$ such that $\text{tr}\,A = \sum A_{ii}$.

Proposition. Suppose that $A \in \text{Mat}_{n,m}(\mathbb{F})$ and $B \in \text{Mat}_{m,n}(\mathbb{F})$. Then $\text{tr}\,AB = \text{tr}\,BA$.

Proof.

\[\text{tr}\,AB = \sum_{i=1}^n \left( \sum_{j=1}^m A_{ij} B_{ji} \right) = \sum_{j=1}^m \left( \sum_{i=1}^n B_{ji} A_{ij} \right) = \text{tr}\,BA\]

Proposition. Suppose that $f \in \text{Mat}_n(\mathbb{F})^\ast$ such that $f(AB) = f(BA)$ for all $A, B \in \text{Mat}_n(\mathbb{F})$. Then $f$ is a scalar multiple of the trace function.

Proof.

Since every off-diagonal basis matrices are commutators, for $i \not= j$, $f(E_{ij}) = f(E_{ik}E_{kj} - E_{kj}E_{ik}) = 0$. For on-diagonal basis matrices, $f(E_{ii}) = f(E_{ik}E_{ki} - E_{ki}E_{ik}) = 0$ so $f(E_{ii}) = f(E_{kk})$ and all of them have the same scalar value $\lambda$. Hence, for any matrix $A$, $f(A) = \sum A_{ii} f(E_{ii}) = \lambda \text{tr}(A)$.

Proposition. If matrices $A$ and $B$ are similar. Then

  • $\text{tr}\,A = \text{tr}\,B$;

  • $\det A = \det B$.

Proof.

  • $\text{tr}\,B = \text{tr}\,P^{-1}AP = \text{tr}\,AP^{-1}P = \text{tr}\,A$.

  • $\det B = \det P^{-1}AP = \det A \det P^{-1} \det P = \det A$.

Since the trace and determinant are independent of the choice of basis, we can define:

Definition. The trace and determinant of an endomorphism $\alpha$, i.e. $\text{tr}\,\alpha$ and $\det \alpha$, are defined to be the trace and determinant of any matrix representing $\alpha$ respectively.

Proposition. Let $\alpha^\ast \in \text{End}(V^\ast)$ be the dual of $\alpha \in \text{End}(V)$. Then $\det \alpha^\ast = \det \alpha$ and $\text{tr}\,\alpha^\ast = \text{tr}\,\alpha$.

Proof.

Since $\det A^\intercal = \det A$ and $\text{tr}\,A^\intercal = \text{tr}\,A$.

References