Tag Archives: direct sum

Facts about power series of matrices

Let G(x) = \sum_{k \in \mathbb{N}} \alpha_k x^k be a power series over \mathbb{C} with radius of convergence R. Let A be an n \times n matrix over \mathbb{C}, and let P be a nonsingular matrix. Prove the following.

  1. If G(A) converges, then G(P^{-1}AP) converges, and G(P^{-1}AP) = P^{-1}G(A)P.
  2. If A = B \oplus C and G(A) converges, then G(B) and G(C) converge and G(B \oplus C) = G(B) \oplus G(C).
  3. If D is a diagonal matrix with diagonal entries d_i, then G(D) converges, G(d_i) converges for each d_i, and G(D) is diagonal with diagonal entries G(d_i).

Suppose G(A) converges. Then (by definition) the sequence of matrices G_N(A) converges entrywise. Let G_N(A) = [a_{i,j}^N], P = [p_{i,j}], and P^{-1} = [q_{i,j}]. Now G_N(P^{-1}AP) = P^{-1}G_N(A)P = [\sum_\ell \sum_k q_{i,k} a_{k,\ell}^Np_{\ell,j}]. That is, the (i,j) entry of G_N(P^{-1}AP) is \sum_\ell \sum_k q_{i,k} a_{k,\ell}^Np_{\ell,j}. Since each sequence a_{k,\ell}^N converges, this sum converges as well. In particular, G(P^{-1}AP) converges (again by definition). Now since G_N(P^{-1}AP) = P^{-1}G_N(A)P for each N, the corresponding sequences for each (i,j) are equal for each term, and so have the same limit. Thus G(P^{-1}AP) = P^{-1}G(A)P.

Now suppose A = B \oplus C. We have G_N(B \oplus C) = \sum_{k=0}^N \alpha_k (B \oplus C)^k = \sum_{k=0}^N \alpha_k B^k \oplus C^k = (\sum_{k = 0}^N \alpha_k B^k) \oplus (\sum_{k=0}^N \alpha_k C^k) = G_N(B) \oplus G_N(C). Since G_N(A) converges in each entry, each of G_N(B) and G_N(C) converge in each entry. So G(B) and G(C) converge. Again, because for each (i,j) the corresponding sequences G_N(A)_{i,j} and (G_N(B) \oplus G_N(C))_{i,j} are the same, they converge to the same limit, and thus G(B \oplus C) = G(B) \oplus G(C).

Finally, suppose D is diagonal. Then in fact we have D = \bigoplus_{t=1}^n [d_t], and so by the previous part, G(D) = \bigoplus_{t=1}^n G(d_t). In particular, G(d_t) converges, and G(D) is diagonal with diagonal entries G(d_t) as desired.

The minimal polynomial of a direct sum is the least common multiple of minimal polynomials

Let M = A \oplus B = \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix} be a direct sum of square matrices A and B. Prove that the minimal polynomial of M is the least common multiple of the minimal polynomials of A and B.


Given a linear transformation T on V, we let \mathsf{Ann}_T(V) denote the annihilator in F[x] of V under the action induced by x \cdot v = T(v).

Let p(x) \in \mathsf{Ann}_M(V). If p(x) = \sum r_ix^i, we have \sum a_iM^i = 0 as a linear transformation. Note that M^k = \begin{bmatrix} A^k & 0 \\ 0 & B^k \end{bmatrix}. So we have \begin{bmatrix} \sum r_i A^i & 0 \\ 0 & \sum r_i B^i \end{bmatrix} = 0, and thus \sum r_i A^i = 0 and \sum r_i B^i = 0. So p(x) \in \mathsf{Ann}_A(W_1) \cap \mathsf{Ann}_B(W_2), where V = W_1 \oplus W_2. Conversely, if p(x) \in \mathsf{Ann}_A(W_1) \cap \mathsf{Ann}_B(W_2), then p(A) = 0 and p(B) = 0 as linear transformations. Then p(M) = \sum r_i M^i = \begin{bmatrix} \sum r_iA^i & 0 \\ 0 & \sum r_iB^i \end{bmatrix} = 0, so that p(x) \in \mathsf{Ann}_M(V). So we have \mathsf{Ann}_M(V) = \mathsf{Ann}_A(W_1) \cap \mathsf{Ann}_B(W_2).

That is, (m_M) = (m_A) \cap (m_B), where m_T is the minimal polynomial of T.

Lemma: In a principal ideal domain R, if (a) \cap (b) = (c), then c is the least common multiple of a and b. Proof: Certainly c \in (a) and c \in (b), so that c is a multiple of both a and b. If d is a multiple of a and of b, then d \in (a) \cap (b) = (c), so that latex d$ is a multiple of c. \square

The result then follows.

Every torsion module over a principal ideal domain is the direct sum of its p-primary components

Let R be a principal ideal domain and let N be a torsion R-module. Prove that the p-primary component of N is a submodule for every prime p \in R. Prove that N is the direct sum of its p-primary components.


We proved this in a previous exercise.

Over an integral domain, the rank of a direct sum is the sum of the ranks

Let R be an integral domain and let A and B be (left, unital) R-modules. Prove that \mathsf{rank}(A \oplus B) = \mathsf{rank}(A) + \mathsf{rank}(B), where the rank of a module is the largest possible cardinality of a linearly independent subset.


Suppose A has rank n and B has rank m. By the previous exercise, there exist free submodules A_1 \subseteq A and B_1 \subseteq B having free ranks n and m, respectively, such that the quotients A/A_1 and B/B_1 are torsion. Note that A_1 \oplus B_1 \subseteq A \oplus B is free. By this previous exercise, we have (A \oplus B)/(A_1 \oplus B_1) \cong_R (A/A_1) \oplus (B/B_1). Note that since R is an integral domain, finite direct sums of torsion modules are torsion. Thus (A \oplus B)/(A_1 \oplus B_1) is torsion. Since A_1 \oplus B_1 is free and has free rank n+m, by this previous exercise, A \oplus B has rank n+m.

Show that a given stable subspace does not have a stable complement

Let F = \mathbb{R} and V = \mathbb{R}^2. Let v_1 = (1,0) and v_2 = (0,1) be the standard basis B for V. Let \varphi : V \rightarrow V be the linear transformation whose matrix with respect to this basis (on both sides) is A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}. Prove that the subspace W spanned by v_1 is stable under \varphi. Prove that there is no subspace W^\prime, also stable under \varphi, such that V = W \oplus W^\prime.


Note that \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} = 2v_1. Thus \varphi[W] \subseteq W, and so W is stable under \varphi.

If there does exist a subspace W^\prime, stable under \varphi, such that V = W \oplus W^\prime, then by this previous exercise, there is a basis E \subseteq V with respect to which the matrix realization of \varphi is (block) diagonal. That is, there exists an invertible matrix P = \begin{bmatrix} a & b \\ c & d \end{bmatrix} such that PAP^{-1} = D is diagonal. Evidently, we have PAP^{-1} = \begin{bmatrix} 2 - \frac{ac}{ad-bc} & \frac{a^2}{ad-bc} \\ \frac{-c^2}{ad-bc} & 2 + \frac{ac}{ad-bc} \end{bmatrix}. If this matrix is diagonal, then a = c = 0. This is a contradiction, however, since then P is not invertible. So no such subspace W^\prime exists.

A linear transformation on a finite dimensional vector space which has a stable subspace decomposes as a direct sum

Let V be a finite dimensional vector space over a field F and let \varphi : V \rightarrow V be a linear transformation. A subspace W \subseteq V is called \varphi-stable if \varphi[W] \subseteq W. Prove that if \varphi has a stable subspace W, then \varphi decomposes as a direct sum of linear transformations. Moreover, show that if each summand is nonsingular, then \varphi is nonsingular.

Conversely, prove that if \alpha \oplus \beta is nonsingular over a finite dimensional vector space, then \alpha and \beta are nonsingular. Prove that this statement is not true over an infinite dimensional vector space.


Suppose \varphi[W] \subseteq W. Letting \pi : V \rightarrow V/W denote the natural projection, note that \pi \circ \varphi : V \rightarrow V/W. Moreover, since W is \varphi-stable, we have W \subseteq \mathsf{ker}\ \pi \circ \varphi. By the generalized first isomorphism theorem, we have an induced linear transformation \overline{\varphi} : V/W \rightarrow V/W given by \overline{\varphi}(v+W) = \varphi(v) + W. It is clear that the restriction \varphi|_W : W \rightarrow W is a linear transformation. Recall that V \cong W \oplus V/W via the isomorphism \theta : (w, v+W) = v+w. Evidently, \theta \circ (\varphi|_W \oplus \overline{\varphi}) = \varphi \circ \theta. Thus \varphi decomposes as a direct sum: \theta = \alpha \oplus \beta, where V = A \oplus B.

Suppose \alpha : A \rightarrow A and \beta : B \rightarrow B are nonsingular linear transformations. That is, \mathsf{ker}\ \alpha = \mathsf{ker}\ \beta = 0. Suppose (a,b) \in \mathsf{ker}\ \alpha \oplus \beta; then a \in \mathsf{ker}\ \alpha and b \in \mathsf{ker}\ \beta, and we have (a,b) = 0. Thus \alpha \oplus \beta is nonsingular.

Now suppose V is finite dimensional, W \subseteq V is \varphi-stable for some linear transformation \varphi, and let \varphi|_W : W \rightarrow W and \overline{\varphi} : V/W \rightarrow V/W be the induced maps discussed above. Suppose \varphi is nonsingular. Since V is finite dimensional, in fact \varphi is an isomorphism. This induces the following short exact sequence of vector spaces.

a diagram of vector spaces

Note that \mathsf{ker}\ \varphi|_W \subseteq \mathsf{ker}\ \varphi, so that \varphi|_W is injective (I.e. nonsingular). Again, since W is finite dimensional, \varphi|_W is surjective. Using part (a) to this previous exercise, \overline{\varphi} is injective- that is, nonsingular.

Note that this proof strategy depends essentially on the fact that, on finite dimensional vector spaces, injectivity and surjectivity are equivalent. If we are going to find an infinite dimensional counterexample, then the subspace W must also be infinite dimensional and the induced mapping \varphi|_W must be injective but not surjective.

Consider the vector space V = \bigoplus_\mathbb{N} F. Let \varphi : V \rightarrow V be the “right shift operator” given by \varphi(a)_i = 0 if i = 0 and a_{i-1} otherwise. Let W = 0 \oplus \bigoplus_{\mathbb{N}^+} F; that is, all tuples in V whose first coordinate is 0. Certainly W is a subspace of V, and moreover is stable under \varphi. We also have that V/W \cong F \cong F \oplus \bigoplus_{\mathbb{N}^+} 0. Now \varphi|_W is injective but not surjective. Suppose v+W \in V/W; then \overline{\varphi}(v+W) = \varphi(v)+W. Since \varphi(v) \in W, in fact \overline{\varphi} = 0. So \overline{\varphi} is singular.

As an F-vector space, an infinite direct sum of F has strictly smaller dimension than an infinite direct power of F over the same index set

Let F be a field. Prove that a vector space V over F having basis B (regardless of the cardinality of B) is isomorphic as a vector space to \bigoplus_B F. Prove that \prod_B F is also an F-vector space which has strictly larger dimension than that of \bigoplus_B F.


(So a free module on any set is isomorphic to a direct sum. We’ve never gotten around to proving this in the best possible generality, though, so we’ll just prove it for vector spaces here.)

Note that, by the universal property of free modules, the natural injection B \rightarrow \bigoplus_B F which sends b to the tuple which has 1 in the bth component and 0 elsewhere induces a vector space homomorphism \varphi : V \rightarrow \bigoplus_B F. This mapping is clearly surjective, and also clearly injective. So V \cong_F \bigoplus_B F.

Certainly \prod_B F is an F-vector space which contains \bigoplus_B F, so that \mathsf{dim}\ \bigoplus_B F \leq \mathsf{dim}\ \prod_B F. Suppose these two dimensions are in fact equal. Identify B with the usual basis of \bigoplus_B F. By this previous exercise, there is a basis D of \prod_B F which contains B, and as argued above, \prod_B F \cong_F \bigoplus_D F. By our hypothesis, in fact B and D have the same cardinality, and so there exists a bijection \theta : B \rightarrow D. Now \theta induces a vector space isomorphism \Theta : \bigoplus_B F \rightarrow \prod_B F.

However, note that |\prod_B F| = |F|^{|B|}, while |\bigoplus_B F| = |\bigcup_{T \subseteq \mathcal{P}(X), T\ \mathrm{finite}} \prod_T F| = \sum_{|B|} |F|^{|T|} = \sum_{|B|} |F| = |B| \cdot |F|. Since |B| \cdot |F| < |F|^{|B|}, we have a contradiction. Thus the dimension of \bigoplus_B F is strictly smaller than that of \prod_B F.

The interaction of Hom with direct sums and direct products

Let R be a ring with 1. Let A be a left unital R-module, and let \{B_i\}_I be a nonempty family of left unital R-modules. Prove that, as abelian groups, \mathsf{Hom}_R(\bigoplus_I B_i, A) \cong \prod_I \mathsf{Hom}_R(B_i,A) and \mathsf{Hom}_R(A, \prod_I B_i) \cong \prod_I \mathsf{Hom}_R(A,B_i). Prove also that if R is commutative, these pairs are R-module isomorphic.


First we show that \mathsf{Hom}_R(\bigoplus_I B_i, A) \cong \prod_I \mathsf{Hom}_R(B_i,A) as abelian groups. Recall that for each i, we have the canonical injection \iota_i : B_i \rightarrow \bigoplus_I B_i. Define for each i \in I the map \varphi_i : \mathsf{Hom}_R( \bigoplus_I B_i, A) \rightarrow \mathsf{Hom}_R(B_i,A) by \varphi_i(\alpha) = \alpha \circ \iota_i; certainly each \varphi_i is well defined. By the universal property of direct products of abelian groups, there exists a unique group homomorphism \Phi : \mathsf{Hom}_R(\bigoplus_I B_i, A) \rightarrow \prod_I \mathsf{Hom}_R(B_i, A) such that \pi_i \circ \Phi = \varphi_i, where \pi_i denotes the ith natural projection from a direct product. We claim that this \Phi is a group isomorphism.

Suppose \alpha \in \mathsf{ker}\ \Phi, so \Phi(\alpha) = 0. Then (\pi_i \circ \Phi)(\alpha) = 0 for each i, so that \varphi_i(\alpha) = 0 for each i. Thus \alpha \circ \iota_i = 0 for all i. That is, \alpha applied to the ith component of any element in \bigoplus_I B_i is zero, for all i. So \alpha = 0, and thus \mathsf{ker}\ \Phi = 0. So \Phi is injective.

Now suppose \psi = (\psi_i) \in \prod_I \mathsf{Hom}_R(B_i,A). Define \alpha_\psi : \bigoplus_I B_i \rightarrow A by \alpha_\psi(b_i) = \sum \psi_i(b_i); this map is well defined since only finitely many terms of (b_i) are nonzero. Moreover, it is clear that \alpha_\psi is an R-module homomorphism, so in fact \alpha_\psi \in \mathsf{Hom}_R(\bigoplus_I B_i, A). Note that for all i \in I, \Phi(\alpha_\psi)_i(b) = \varphi_i(\alpha_\psi)(b) = (\alpha_\psi \circ \iota_i)(b) = \alpha_\psi(\iota_i(b)) = \psi_i(b). Thus \Phi(\alpha_\psi)_i = \psi_i for all i, and so \Phi(\alpha_\psi) = \psi. Thus \Phi is surjective.

So \Phi is an isomorphism of abelian groups. Suppose now that R is commutative, so that both \mathsf{Hom}_R(\bigoplus_I B_i,A) and \prod_I \mathsf{Hom}_R(B_i,A) are naturally left R-modules. For all r \in R and all module homomorphisms \alpha : \bigoplus_I B_i \rightarrow A, we have \Phi(ra) = ((r \alpha) \circ \iota_i) = (r(\alpha \circ \iota_i)) = r(\alpha \circ \iota_i) = r\Phi(\alpha). Thus \Phi is an isomorphism of left R-modules.

Next we show that \mathsf{Hom}_R(A, \prod_I B_i) \cong \prod_I \mathsf{Hom}_R(A,B_i) as abelian groups. For each i \in I, define \varphi_i : \mathsf{Hom}_R(A, \prod_I B_i) \rightarrow \mathsf{Hom}_R(A,B_i) by \varphi_i(\alpha) = \pi_i \circ \alpha, where \pi_i denotes the ith canonical projection from a direct product. By the universal property of direct products, we have a unique group homomorphism \Phi : \mathsf{Hom}_R(A,\prod_I B_i) \rightarrow \prod_I \mathsf{Hom}_R(A, B_i) such that \pi_i \circ \Phi = \varphi_i for all i. We claim that this \Phi is an isomorphism.

Suppose \alpha \in \mathsf{ker}\ \Phi. So \Phi(\alpha) = 0, and thus (\pi_i \circ \Phi)(\alpha) = 0 for all i. Thus \varphi_i(\alpha) = 0 for all i, so \pi_i \circ \alpha = 0 for all i. That is, every coordinate of any element in the image of \alpha is 0. So \alpha = 0. Thus \mathsf{ker}\ \Phi = 0, and so \Phi is injective.

Now let \psi = (\psi_i) \in \prod_I \mathsf{Hom}_R(A,B_i). Define \alpha_\psi : A \rightarrow \prod_I B_i by \alpha_\psi(a)_i = \psi_i(a). Certainly \alpha_\psi is a module homomorphism. Note that, for all i, \Phi(\alpha_\psi)_i(a) = \varphi_i(\alpha_\psi)(a) = (\pi_i \circ \alpha_\psi)(a) = \psi_i(a). So \Phi(\alpha_\psi)_i = \psi_i for all i, and we have \Phi(\alpha_\psi) = \psi. So \Phi is surjective, and thus an isomorphism of groups.

Finally, suppose R is commutative. If r \in R and \alpha : A \rightarrow \prod_I B_i is a module homomorphism, then \Phi(r\alpha) = (\pi_i \circ r\alpha) = (r(\pi_i \circ \alpha)) = r(\pi_i \circ \alpha) = r\Phi(\alpha). Thus in this case \Phi is an isomorphism of R-modules.

An arbitrary direct sum of modules is flat if and only if each direct summand is flat

Let R be a ring with 1 and let \{A_i\}_I be a family of (right, unital) R-modules. Prove that \bigoplus_I A_i is flat if and only if each A_i is flat.


We begin with a lemma.

Lemma: Let \{\varphi_i : A_i \rightarrow B_i\}_I be a family of (unital) R-module homomorphisms. Then \Phi : \bigoplus_I A_i \rightarrow \bigoplus_I B_i given by \Phi(a_i) = (\varphi_i(a_i)) is injective if and only if each \varphi_i is injective. Proof: Suppose \Phi is injective. If \varphi_k(a) = \varphi_k(b), and letting \iota_k denote the inclusion A_k \rightarrow \bigoplus_I A_i or B_k \rightarrow \bigoplus_I B_i, then \Phi(\iota_k(a)) = \Phi(\iota_k(b)), so that a = b. Thus each \varphi_k is injective. Conversely, suppose each \varphi_k is injective; then if \Phi(a_i) = \Phi(b_i), we have \varphi_i(a_i) = \varphi_i(b_i) for each i, so that a_i = b_i for each i. So (a_i) = (b_i), and thus \Phi is injective. \square

Suppose each A_i is flat. Now let L and M be left unital R-modules and let \theta : L \rightarrow M be a module homomorphism. Since each A_i is flat, 1_i \otimes \theta : A_i \otimes_R L \rightarrow A_i \otimes_R M is injective. Using the lemma, \Theta = \bigoplus_I (1_i \otimes \theta) : \bigoplus_I (A_i \otimes L) \rightarrow \bigoplus_I (A_i \otimes M) is injective. In this previous exercise, we constructed group isomorphisms \Phi : (\bigoplus_I A_i) \otimes_R L \rightarrow \bigoplus_I (A_i \otimes_R L) and \Psi : \bigoplus_I (A_i \otimes_R L) \rightarrow (\bigoplus_I A_i) \otimes_R M. We claim that \Psi \circ \Theta \circ \Phi = 1 \otimes \theta. To that end, note that (\Psi \circ \Theta \circ \Phi)((a_i) \otimes \ell) = (\Psi \circ \Theta)((a_i \otimes \ell)) = \Psi((a_i \otimes \theta(\ell))) = (\sum a_i) \otimes \theta(\ell) = (1 \otimes \theta)((a_i) \otimes \ell), as desired. Thus 1 \otimes \theta : (\bigoplus_I A_i) \otimes_R L \rightarrow (\bigoplus_I A_i) \otimes_R M is injective, and so \bigoplus_I A_i is flat.

Conversely, suppose \bigoplus_I A_i is flat. Let L and R be (unital, left) R-modules and \theta : L \rightarrow M a module homomorphism. Note that \Theta = 1 \otimes \theta : (\bigoplus_I A_i) \otimes_R L \rightarrow (\bigoplus_I A_i) \otimes_R M is injective. Recall the homomorphisms \Phi and \Psi from the previous paragraph; \Psi^{-1} \circ \Theta \circ \Phi^{-1} is an injective module homomorphism mapping \bigoplus_I (A_i \otimes_R L) to \bigoplus_I (A_i \otimes_R M), and in fact (\Psi^{-1} \circ \Theta \circ \Phi^{-1})((a_i \otimes \ell_i)) = (\Psi^{-1} \circ \Theta)(\sum \iota_i(a_i) \otimes \ell_i) = \Psi^{-1}(\sum \iota_i(a_i) \otimes \theta(\ell_i)) = (a_i \otimes \theta(\ell_i)) = ((1_i \otimes \theta)(a_i \otimes \ell_i)). By the lemma, each 1_i \otimes \theta is injective, so that each A_i is flat.

The direct sum of two modules is injective if and only if each direct summand is injective

Let R be a ring with 1 and let Q_1 and Q_2 be (left, unital) R-modules. Prove that Q_1 \oplus Q_2 is injective if and only if Q_1 and Q_2 are injective.


Recall Baer’s Criterion: an R-module Q is injective if and only if for every left ideal I \subseteq R and every R-module homomorphism \varphi : I \rightarrow Q, there exists an R-module homomorphism \Phi : R \rightarrow Q such that \Phi|_I = \varphi.

Suppose Q_1 and Q_2 are injective. Let I \subseteq R be a left ideal and let \varphi : I \rightarrow Q_1 \oplus Q_2 be an R-module homomorphism. Letting \pi_1 and \pi_2 denote the first and second coordinate projections, \pi_1 \circ \varphi : I \rightarrow Q_1 and \pi_2 \circ \varphi : I \rightarrow Q_2 are R-module homomorphisms. By Baer’s Criterion, there exist R-module homomorphisms \Phi_1 : R \rightarrow Q_1 and \Phi_2 : R \rightarrow Q_2 such that \Phi_1|_I = \pi_1 \circ \varphi and \Phi_2|_I = \pi_2 \circ \varphi. Define \Phi : R \rightarrow Q_1 \oplus Q_2 by \Phi(r) = (\Phi_1(r), \Phi_2(r)). Certainly \Phi is an R-module homomorphism. Moreover, \Phi_I =  \varphi. Thus Q_1 \oplus Q_2 is injective.

Suppose Q_1 \oplus Q_2 is injective. Let I \subseteq R be a left ideal and let \varphi_1 : I \rightarrow Q_1 and \varphi_2 : I \rightarrow Q_2 be R-module homomorphisms. Define \varphi : I \rightarrow Q_1 \oplus Q_2 by \varphi(i) = (\varphi_1(i), \varphi_2(i)). Certainly \varphi is an R-module homomorphism. By Baer’s Criterion, there exists an R-module homomorphism \Phi : R \rightarrow Q_1 \oplus Q_2 extending \varphi. Now let \Phi_1 = \pi_1 \circ \Phi and \Phi_2 = \pi_2 \circ \Phi; certainly \Phi_1 : R \rightarrow Q_1 and \Phi_2 : R \rightarrow Q_2 are R-module homomorphisms, and if a \in I then (\Phi_1(a), \Phi_2(a)) = \Phi(a) = \varphi(a) = (\varphi_1(a), \varphi_2(a)). Thus Q_1 and Q_2 are injective.