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 $M_n(\mathbf{F})$ can be useful sometimes for proving properties of the corresponding endomorphism.

Definition. The standard basis for $M_n(\mathbf{F})$ 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 $i \not= j$.

Since $\text{End}(V) \cong M_n(\mathbf{F})$ for a given basis $\Set{v_1, …, v_n}$ for $V$, the $n^2$ linear maps $\Set{\alpha_{ij}}$ in which their matrix representations with respect to $\Set{v_i}$ are $\Set{E_{ij}}$ form a basis for $\text{End}(V)$. In particular, $\alpha_{ij} \in \text{End}(V)$ defined by

\[\alpha_{ij}(v_j) = v_i \quad \alpha_{ij}(v_k) = 0, k \not= j\]

form a basis for $\text{End}(V)$.

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 $\text{GL}_n(\mathbf{F})$ denotes all the invertible matrices so $\text{GL}_n(\mathbf{F})$ acts on $M_n(\mathbf{F})$ 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 $\text{tr}: M_n(\mathbf{F}) \to \mathbf{F}$ such that $\text{tr}\,A = \sum A_{ii}$.

Proposition. Suppose that $A \in M_{n,m}(\mathbf{F})$ and $B \in M_{m,n}(\mathbf{F})$. 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 $f \in M_n(\mathbf{F})^\ast$ such that $f(AB) = f(BA)$ for all $A, B \in M_n(\mathbf{F})$. 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 $\text{tr}\,A^\intercal = \text{tr}\,A$.

Minimal Polynomials

Since $\dim \text{End}(V) = n^2$, for a fixed linear map $\alpha \in \text{End}(V)$, the $n^2 + 1$ powers of $\alpha$, i.e. $\Set{1, \alpha, \alpha^2, …, \alpha^{n^2}}$ are linearly dependent. That means there exists a nonzero polynomial

\[f(x) = a_0 + a_1 x + ... + a_{n^2} x^{n^2} \in \mathbf{F}[x]\]

such that $f(\alpha) = 0$.

Proposition. There exists a uniquely determined integer $r \le n^2$ such that $\Set{1, \alpha, …, \alpha^{r-1}}$ are linearly independent but not $\Set{1, \alpha, …, \alpha^{r}}$. Then

\[\alpha^r = a_0 \iota + a_1 \alpha + ... + a_{r-1} \alpha^{r-1}\]

Let $m(x) = x^r - a_{r-1} x^{r-1} - … - a_0 \iota \in \mathbf{F}[x]$. Then

  • $m(x)$ is nonzero and $m(\alpha) = 0$;

  • If $f(x)$ is any polynomial such that $f(\alpha) = 0$, then $m(x) \mid f(x)$.

Proof.

Base on the above we can see that $m(x)$ exists.

Let $f(x)$ be any polynomial such that $f(\alpha) = 0$. By division theorem, $f(x) = m(x) Q(x) + R(x)$ with $\deg R < r$ and $R(\alpha) = f(\alpha) - m(\alpha) Q(\alpha) = 0$. Since $\Set{1, \alpha, …, \alpha^{r-1}}$ are linearly independent, there doesn’t exist a nonzero polynomial $R(x)$ with $\deg R < r$ such that $R(\alpha) = 0$. Hence, $R = 0$ and $m(x) \mid f(x)$.

Definition. The minimal polynomial of a linear map $\alpha \in \text{End}(V)$ is the nonzero monic polynomial of least degree such that $m(\alpha) = 0$ and it is well-defined.

The uniqueness of minimal polynomial is the result from the fact that if both $m(x)$ and $m’(x)$ are minimal polynomials then $m \mid m’$ and $m’ \mid m$ and hence $m = m’$. Also, the bijective correspondence between linear maps and matrices allows us to define same for matrices.

Proposition. The minimal polynomials of a linear map $\alpha$ and its matrix representation $A$ with respect any basis of $V$ are the same, i.e. if matrices $A$ and $B$ are similar, then they have the same minimal polynomial.

Proof.

$A$ and $B$ are similar means $B = P^{-1}AP$. Let $m_A(x)$ and $m_B(x)$ be the minimal polynomial of $A$, then

\[m_A(B) = m_A(P^{-1}AP) = P^{-1} m_A(A) P = 0\]

since $a_k(P^{-1}AP)^k = P^{-1}(a_kA^k)P$ and $a_0 I = P^{-1} (a_0 I) P$. Hence, $m_B(x) \mid m_A(x)$ and similarily $m_A(x) \mid m_B(x)$ so $m_A(x) = m_B(x)$.

Invariant Subspaces

Definition. A subspace $U$ of $V$ is called an invariant subspace relative to $\alpha$ (or simply $\alpha$-invariant subspace or $\alpha$-subspace) if $\alpha(u) \in U$ for all $u \in U$.

Lemma. For any polynomial $f(x) \in \mathbf{F}[x]$, the kernel of $f(\alpha)$, i.e. the set of vectors such that $f(\alpha)(v) = 0$, is a $\alpha$-invariant subspace.

Proof.

Since $f(\alpha) \in \text{End}(V)$, the nullspace of $f(\alpha)$ is a subspace of $V$. For any $v \in \ker f(\alpha)$, we have to show that $\alpha(v) \in \ker f(\alpha)$ as well which means $f(\alpha)(\alpha(v)) = 0$.

\[f(\alpha)(\alpha(v)) = (f(\alpha)\alpha)(v) = (\alpha f(\alpha))(v) = \alpha(f(\alpha)(v)) = 0\]

Lemma. Suppose that there are nonzero linear maps $\Set{\beta_1, …, \beta_r}$ in $\text{End}(V)$ such that

  • $\beta_1 + … + \beta_r = \iota$;

  • $\beta_i \beta_j = \beta_j \beta_i = 0$ for $i \not= j$.

Then

  • $\beta_i^2 = \beta_i$;

  • $V = \beta_1 V \oplus \cdots \oplus \beta_r V$ and each subspace $\beta_i V$ is different from zero.

Proof.

$\beta_i = \beta_i \iota = \beta_i(\beta_1 + … + \beta_r) = \beta_i^2$.

$\beta_i V$ is nonzero subspace since $\beta_i$ is nonzero linear map.

For any $v \in V$, $v = \iota(v) = \beta_1(v) + \cdots + \beta_r(v)$ so $V = \beta_1 V + \cdots + \beta_r V$.

Let $v_i \in \beta_i V$. Since $\beta_i \beta_j = 0$ and $v_j \in \beta_j V$, $\beta_i v_j = 0$ for $i \not= j$. On the other hand, $\beta_i v_i = \beta_i^2 v = \beta_i v = v_i$. Suppose that $v_1 + \cdots + v_r = 0$. Then for each $1 \le i \le r$,

\[\beta_i (v_1 + \cdots + v_r) = \beta_i v_i = v_i = 0\]

so $v_1 + \cdots + v_r = 0$ implies $v_1 = \cdots = v_r = 0$. Hence, $V = \beta_1 V \oplus \cdots \oplus \beta_r V$.

Definition. A linear map $\alpha \in \text{Env}(V)$ is idempotent if $\alpha^2 = \alpha \not= 0$.

We are now ready to connect the minimal polynomial of a linear map with its underlying vector space, which is often called the primary decomposition theorem.

Theorem. Let $m(x) = p_1(x)^{e_1} \cdots p_r(x)^{e_r}$ be the minimal polynomial of $\alpha \in \text{End}(V)$ factored into powers of distinct primes. Then the subspaces $\ker p_i(\alpha)^{e_i}$ are $\alpha$-invariant subspaces and

\[V = \bigoplus_{i=1}^r \ker p_i(\alpha)^{e_i}\]

Proof.

By showing that there exists polynomials $\Set{f_1, …, f_r}$ such that the linear maps $\beta_i = f_i(\alpha)$ satisfy

  • $\beta_i \not= 0$;

  • $\beta_1 + … + \beta_r = \iota$;

  • $\beta_i \beta_j = \beta_j \beta_i = 0$ for $i \not= j$;

  • $\beta_i V = \ker p_i(\alpha)^{e_i}$,

we can prove the statement using the above lemmas.

Let

\[q_i(x) = {m(x) \over p_i(x)^{e_i}}\]

Since all the $\Set{q_i(x)}$ are relatively prime, there exists polynomials $a_i(x)$ such that

\[q_1(x) a_1(x) + \cdots + q_r(x) a_r(x) = 1\]

Let $f_i(x) = q_i(x) a_i(x)$ and $\beta_i = f_i(\alpha) = q_i(\alpha) a_i(\alpha)$, we have

\[\beta_1 + \cdots + \beta_r = \iota\]

and similar to the above lemma, $V = \beta_1 V + \cdots + \beta_r V$.

Next, since $m(x) \mid q_i(x) q_j(x)$ for $i \not= j$, we have

\[\beta_i \beta_j = \beta_j \beta_i = q_i(\alpha) a_i(\alpha) q_j(\alpha) a_j(\alpha) = 0\]

Furthermore, if $\beta_i V = 0$ for some $i$, then $V = \sum_{j \not= i} \beta_j V$ and

\[q_i(\alpha) V = q_i(\alpha) \sum_{j \not= i} \beta_j V = \sum_{j \not= i} q_i(\alpha) q_j(\alpha) a_j(\alpha) V = 0\]

since $m(x) \mid q_i(x) q_j(x)$. Thus, $q_i(\alpha) = 0$ contradicting the assumption that $m(x)$ is the minimal polynomial and therefore $\beta_i V \not= 0$ and $\beta_i \not= 0$.

By the above lemma, we can conclude that

\[V = \beta_1 V \oplus \cdots \oplus \beta_r V\]

Moreover,

\[\alpha(\beta_i V) = (\alpha f_i(\alpha)) V = (f_i(\alpha) \alpha) V = \beta_i (\alpha V) \subseteq \beta_i V\]

so $\beta_i V$ is a $\alpha$-invariant subspace.

Finally, for all $v \in V$, we have

\[p_i(\alpha)^{e_i}(\beta_i(v)) = (p_i(\alpha)^{e_i} q_i(\alpha) a_i(\alpha))(v) = (m(\alpha)a_i(\alpha))(v) = 0\]

so $\beta_i V \subseteq \ker p_i(\alpha)^{e_i}$.

On the other hand, let $v \in \ker p_i(\alpha)^{e_i}$ and express $v$ in the form $v = \beta_1 v_1 + \cdots + \beta_r v_r$, then

\[p_i(\alpha)^{e_i}(v) = p_i(\alpha)^{e_i} \beta_1 v_1 + \cdots + p_i(\alpha)^{e_i} \beta_r v_r = 0\]

Since $V$ is the direct sum of the subspaces $\beta_i V$, we have

\[p_i(\alpha)^{e_i} \beta_1 v_1 = \cdots = p_i(\alpha)^{e_i} \beta_r v_r = 0\]

For $j \not= i$, since $p_i(x)$ and $p_j(x)$ are distinct primes, there exists $a_i(x)$ and $a_j(x)$ such that

\[p_i(x)^{e_i} a_i(x) + p_j(x)^{e_j} a_j(x) = 1\]

and by substituting $\alpha$ and applying both sides to $\beta_j v_j$, we have

\[\beta_j v_j = a_i(\alpha)(p_i(\alpha)^{e_i} \beta_j v_j) + a_j(\alpha)(p_j(\alpha)^{e_j} \beta_j v_j) = 0\]

which means every $\beta_j v_j$ can be expressed as a sum of the above two terms in which the second term is always zero and first term being zero is concluded earlier. Hence,

\[v = \beta_1 v_1 + \cdots + \beta_r v_r = \beta_i v_i \in \beta_i V\]

and $\ker p_i(\alpha)^{e_i} \subseteq \beta_i V$. Hence, $\beta_i V = \ker p_i(\alpha)^{e_i}$.

Definition. $\ker (\alpha - \lambda \iota)^e$ is called the generalized $\lambda$-eigenspace of $\alpha$.

References