es
There are two reasons to use the sum of two vector spaces. One of them is the way to build new vector spaces from old ones. Another reason is to decompose the known vector space into sum of two (smaller) spaces. Since we consider linear transformations between vector spaces, these sums lead to representations of these linear maps and corresponding matrices into forms that reflect these sums. In many very important situations, we start with a vector space V and can identify subspaces “internally” from which the whole space V can be built up using the construction of sums. However, the most fruitful results we obtain for a special sum, called the direct sum.

Sums of Subspaces

Let A and B be nonempty subsets of a vector space V. The sum of A and B, denoted A + B, is the set of all possible sums of elements from both subsets: \( A+B = \left\{ a+b \, : \, a\in A, \ b\in B \right\} . \)

Theorem 1: Let W₁, W₂, … , Wn be subspaces of a vector space V over field 𝔽, then their sum W₁ + W₂ + ⋯ + Wn = {w₁ + w₂ + ⋯ + wn | w₁ ∈ W₁, w₂ ∈ W₂, … , wn ∈ Wn } is a sub- space of V and it is the smallest subspace of V containing all subspaces W₁, W₂, … , Wn.
Since W₁, W₂, … , Wn are subspaces of V and zero vector belongs to each subspace, then \[ 0 = 0 + 0 + \cdots + 0 \in W_1 + W_2 + \cdots + W_n . \] Now let v, w ∈ W₁ + W₂ + ⋯ + Wn and r ∈ 𝔽. Then v = v₁ + v₂ + ċ + vn and w = w₁ + w₂ + ċ + wn, where vi, wi ∈ Wi for all i = 1, 2, … , n. As each Wi is a subspace of V, each partial sum vi + wi belongs to Wi. Hence, their sum is \[ \mathbf{v} + \mathbf{w} = \sum_{i=1}^n \mathbf{v}_i + \mathbf{w}_i \in W_1 + \cdots + W_n . \] Similarly, rvi ∈ Wi for all i = 1, 2, … , n. So, scalar multiple becomes \[ r\,\mathbf{v} = \sum_{i=1}^n r\,\mathbf{v}_i \in W_1 + \cdots + W_n . \] Therefore, W₁ + W₂ + ⋯ + Wn is a subspace of V.

Now to prove that W₁ + W₂ + ⋯ + Wn is the smallest subspace containing W₁, W₂, … , Wn, we will show that any subspace of V containing W₁, W₂, … , Wn contains the sum W₁ + W₂ + ⋯ + Wn.

Let W be any subspace containing W₁, W₂, … , Wn. Let w = w₁ + w₂ + ċ + wn ∈ WW₁ + W₂ + ⋯ + Wn, where wi ∈ Wi for all i = 1, 2, … , n. Since W is a subspace of V and W contains the sum W₁ + W₂ + ⋯ + Wn, w ∈ W.

   
Example 1: Let V = ℝ². Let us consider W₁ = { (x, y) : x = 2y, x, y ∈ ℝ } and W₂ = { (x, y) : x = −2x, x, y ∈ ℝ }. Then W₁, W₂ are subspaces of V.
RJB Fig 2.6 on page 68 of A Course in LinAlg
Subspace W₁
     
Subspace W₂

Any vector . (x, y) ∈ ℝ² can be written as a linear combination of elements of W₁ and . W₂ as follows: \[ \mathbb{R}^2 \ni (x, y) = \left( \frac{x+ 2\,y}{2} , \ \frac{x + 2\,y}{4} \right) + \left( \frac{x - 2\,y}{2} , \ \frac{2\,y -x}{4} \right) \in W_1 + W_2 . \] As W₁ + W₂ is a two-dimensional subspace of ℝ², this implies that W₁ + W₂ = Ropf;². Also observe that the representation of any vector as the sum of elements in W₁ and W₂ is unique here.    ■
End of Example 1
   
Example 2: Let V = ℝ2×2 be the vector space of 2-by-2 matrices with real entries. Let us consider two its subspaces \[ W_1 = \left\{ \begin{bmatrix} a_{1,1} & a_{1,2} \\ 0 & a_{2,2} \end{bmatrix} \ : \quad a_{1,1} , a_{1,2}, a_{2.2} \in \mathbb{R} \right\} \] and \[ W_2 = \left\{ \begin{bmatrix} a_{1,1} & 0 \\ a_{2,1} & a_{2,2} \end{bmatrix} \ : \quad a_{1,1}, a_{2,1}, a_{2,2} \in \mathbb{R} \right\} . \] Then W₁ and W₂ are subspaces of V. Indeed, W₁ contains only upper triangular matrices and W₂ contains lower triangular matrices. Any sum of upper (or lower) triangular matrices is again upper (lower) triangular matrix. Also multiplication by a scalar does not ulter elements from i>W₁ and W₂, respectively.

Also any 2-by-2 matrix from V can be expressed as a sum of elements in W₁ and W₂. However, this expression is not unique. For example, \[ \begin{bmatrix} 3&4 \\ 1& 2 \end{bmatrix} = \begin{bmatrix} 3&4 \\ 0& 2 \end{bmatrix} + \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} \] and \[ \begin{bmatrix} 3&4 \\ 1& 2 \end{bmatrix} = \begin{bmatrix} 0&4 \\ 0&0 \end{bmatrix} + \begin{bmatrix} 3&0 \\ 1&2 \end{bmatrix} . \]    ■

End of Example 2

Direct Sums

Now we come to a particular but very important case of sums of subspaces when every vector from a vector space V can be uniquely decomposed into sum of vectors from given subspaces.

A vector space V is called the direct sum of V₁ and V₂ if V₁ and V₂ are subspaces of V such that \( V_1 \cap V_2 = \{ 0 \} \) and \( V_1 + V_2 = V. \) This means that every vector v from V is uniquely represented via sum of two vectors \( {\bf v} = {\bf v}_1 + {\bf v}_2 , \quad {\bf v}_1 \in V_1 , \ {\bf v}_2 \in V_2 . \) We denote that V is the direct sum of V1 and V2 by writing \( V = V_1 \oplus V_2 . \)

A complement to a subspace of a vector space is another subspace which forms a direct sum. Two such spaces are mutually complementary. , that is: Equivalently, every element of V can be expressed uniquely as a sum of an element of U and an element of W.
Suppose that an n-dimensional vector space V is the direct sum of two subspaces \( V = U\oplus W . \) Let \( {\bf e}_1 , {\bf e}_2 , \ldots , {\bf e}_k \) be a basis of the linear subspace U and let \( {\bf e}_{k+1} , {\bf e}_{k+2} , \ldots , {\bf e}_n \) be a basis of the linear subspace W. Then \( {\bf e}_1 , {\bf e}_2 , \ldots , {\bf e}_n \) form a basis of the whole linear space V. Any linear transformation written in this basis has a matrix representation:

\[ {\bf A} = \begin{bmatrix} {\bf A}_{k \times k} & {\bf 0}_{k \times (n-k)} \\ {\bf 0}_{k \times (n-k)} & {\bf A}_{(n-k)\times (n-k)} \end{bmatrix} . \]
Therefore, the block diagonal matrix A is the direct sum of two matrices of lower sizes.
Example 3: Consider the Cartesian plane ℝ² when every element is represented by an ordered pair v = (x, y). This vector has a unique decomposition \( {\bf v} = (x,y) = {\bf v}_1 + {\bf v}_2 = (x,0) + (0,y) , \) where vectors (x, 0) and (0, y) can be identified with a one-dimensional space ℝ¹ = ℝ.

If we choose two arbitrary not parallel vectors u and v on the plane, then spans of these vectors generate two vectors spaces that we denote by U and V, respectively. Therefore, U and V are two lines containing vectores u and v, respectively. Their sum, \( U + V = \left\{ {\bf u} +{\bf v} \,: \ {\bf u} \in U, \ {\bf v} \in V \right\} \) is the whole plane \( \mathbb{R}^2 . \)

g1 = Graphics[{Blue, Thickness[0.01], Arrow[{{0, 0}, {3, 1}}]}]
g2 = Graphics[{Blue, Thickness[0.01], Arrow[{{0, 0}, {1, 3}}]}]
g3 = Graphics[{Green, Thickness[0.01], Arrow[{{0, 0}, {4, 4}}]}]
g4 = Graphics[{Cyan, Thickness[0.005], Line[{{1, 3}, {4, 4}}]}]
g5 = Graphics[ Text[Style[
ToExpression["u + v \\notin U\\cup V", TeXForm, HoldForm],
FontSize -> 14, FontWeight -> "Bold", FontColor -> Black], {3.6, 3.5}, {0, 1}, {1, 1}]]
g6 = Graphics[
Text[StyleForm["u", FontSize -> 14, FontWeight -> "Bold",
FontColor -> Blue], {2.6, 1.2}, {0, 0.8}, {2.8, 1}]]
g7 = Graphics[
Text[StyleForm["v", FontSize -> 14, FontWeight -> "Bold",
FontColor -> Blue], {1.1, 2.8}]]
Show[g1, g2, g3, g4, g5, g6, g7]
The union \( U \cup V \) of two subspaces is not necessarily a subspace.
End of Example 3

Suppose we have a vector space V over field 𝔽 and subspaces W₁, W₂, … , Wn of V. Then it is not easy to check whether every element in V has a unique representation as the sum of elements of W₁, W₂, … , Wn. The following theorem provides a solution for this.

Theorem 2: Let V be a vector space over field 𝔽 and W₁, W₂, … , Wn be subspaces of V. Then V = = W₁ ⊕ W₂ ⊕ ⋯ ⊕ Wn if and only if the following conditions are satisfied:

  1. V = W₁ + W₂ + ⋯ + Wn;
  2. zero vector has only the trivial representation.
Let V = W₁ ⊕ W₂ ⊕⋯ ⊕ Wn. Then by the definition of direct sum both conditions (i) and (ii) hold. Conversely, suppose that both (i) and (ii) hold. Let v ∈ V have two representations, namely \[ \mathbf{v} = v_1 + v_2 + \cdots + v_n \] and \[ \mathbf{v} = u_1 + u_2 + \cdots + u_n , \] where vi, ui ∈ Wi for all i = 1, 2, … , n. Then subtracting these equations gives \[ 0 = \left( v_1 - u_1 \right) + \left( v_2 - u_2 \right) + \cdots + \left( v_n - u_n \right) \] and as zero has trivial representation only, vi − ui = 0 for all i = 1, 2, … , n which implies vi = ui for all i = 1, 2, … , n. That is, every vector has a unique representation. Therefore. V = W₁ ⊕ W₂ ⊕⋯ ⊕ Wn.
   
Example 4: Let E denote the set of all polynomials of even powers: \( E = \left\{ a_n t^{2n} + a_{n-1} t^{2n-2} + \cdots + a_0 \right\} , \) and O be the set of all polynomails of odd powers: \( O = \left\{ a_n t^{2n+1} + a_{n-1} t^{2n-1} + \cdots + a_0 t \right\} . \) Then the set of all polynomials P is the direct sum of these sets: \( P = O\oplus E . \)

It is easy to see that any polynomial (or function) can be ubiquely decomposed into direct sum of even and odd counterparts:

\[ p(t) = \frac{p(t) + p(-t)}{2} + \frac{p(t) - p(-t)}{2} . \]
End of Example 2

Before formulating the Primary Decomposition Theorem, we need to recall some definitions and facts that were explained in other sections. We remind that the minimal polynomial of a square matrix A (or corresponding lineat transformation) is the (unique) monic polynomial ψ(λ) of least degree that annihilates the matrix A, that is ψ(A) = 0. The minimal polynomial \( \psi_u (\lambda ) \) of a vector \( {\bf u} \in V \ \mbox{ or } \ {\bf u} \in \mathbb{R}^n \) relative to A is the monic polynomial of least degree such that \( \psi_u ({\bf A}) {\bf u} = {\bf 0} . \) It follows that \( \psi_u (\lambda ) \) divides the minimal polynomial ψ(λ) of the matrix A. There exists a vector \( {\bf u} \in V (\mathbb{R}^n ) \) such that \( \psi_u (\lambda ) = \psi (\lambda ) . \) This result can be proved by representing the minimal polynomial as the product of simple terms to each of which corresponds a subspace. Then the original vector space (or \( \mathbb{R}^n \) ) is the direct sum of these subspaces.

A subspace U of a vector space V is said to be T-cyclic with respect to a linear transformation T : V ⇾ V if there exists a vector \( {\bf u} \in U \) and a nonnegative integer r such that \( {\bf u}, T\,{\bf u} , \ldots , T^r {\bf u} \) form a basis for U. Thus, for the vector u if the degree of the minimal polynomial \( \psi_u (\lambda ) \) is k, then \( {\bf u}, T\,{\bf u} , \ldots , T^{k-1} {\bf u} \) are linearly independent and the space U is spanned by these k vectors is T-cyclic.

Theorem (Primary Decomposition Theorem): Let V be an n-dimensional vector space (n is finite) and T is a linear transformation on V. Then V is the direct sum of T-cyclic subspaces. ■

Let k be the degree of the minimal polynomial ψ(λ) of transformation T (or corresponding matrix written is specified basis), and let u be a vector in V with \( \psi_u (\lambda ) = \psi (\lambda ) . \) Then the space U spanned by \( {\bf u}, T{\bf u} , \ldots , T^{k-1} {\bf u} \) is T-cyclic. We shall prove that if \( U \ne V \quad (k \ne n), \) then there exists a T-invariant subspace W such that \( V = U\oplus W . \) Clearly, by induction on the dimension, W will then be the direct sum of T-cyclic subspaces and the proof is complete.

To show the existence of W enlarge the basis \( {\bf e}_1 = {\bf u}, {\bf e}_2 = T{\bf u} , \ldots , {\bf e}_k = T^{k-1} {\bf u} \) of U to a basis \( {\bf e}_1 , {\bf e}_2 , \ldots , {\bf e}_k , \ldots , {\bf e}_n \) of V and let \( {\bf e}_1^{\ast} , {\bf e}_2^{\ast} , \ldots , {\bf e}_k^{\ast} , \ldots , {\bf e}_n^{\ast} \) be the dual basis in the dual space. Recall that the dual space consists of all linear forms on V or, equivalently, of all functionals on V. To simplify notation, let z = ek*. Then

\[ \langle {\bf z} , {\bf e}_i \rangle =0 \quad\mbox{for } 1 \le i \le k-1 \quad\mbox{and}\quad \langle {\bf z} , {\bf e}_k \rangle =1 . \]
Consider the dual space U* spanned by \( {\bf z}, T^{\ast} {\bf z} , \ldots , T^{\ast\, k-1} {\bf z} . \) Since ψ(λ) is also the minimal polynomial of T*, the space U* is T*-invariant. Now observe that if \( U^{\ast} \cap U^{\perp} = \{ 0 \} \) and dim U* = k, then \( V^{\ast} = U^{\ast} \oplus U^{\perp} , \) where U* and \( U^{\perp} \) are T*-invariant (since dim \( U^{\perp} \) = n-k). This in turn implies the desired decomposition \( V = U^{\perp\perp} \oplus U^{\ast\perp} = U \oplus W , \) where \( U^{\perp\perp} = U \) and \( U^{\ast\perp} = W \) are T-invariant.

Finally, we shall prove that \( U^{\ast} \cap U^{\perp} = \{ 0 \} \) and dim U* = k simultaneously as follows. Suppose that \( a_0 {\bf z} + a_1 T^{\ast} {\bf z} + \cdots + a_s T^{\ast s} {\bf z} \in U^{\perp} , \) where \( a_s \ne 0 \) and \( 0 \le s \le k-1 . \) Then

\[ T^{\ast k-1-s} \left( a_0 {\bf z} + a_1 T^{\ast} {\bf z} + \cdots + a_s T^{\ast s} {\bf z} \right) = a_0 T^{\ast k-1-s} {\bf z} + a_1 T^{\ast k-s} {\bf z} + \cdots + a_s T^{\ast k-1} {\bf z} \]
is in \( U^{\perp} \) since \( U^{\perp} \) is T*-invariant. Therefore,

\[ \left\langle T^{\ast k-1-s} \left( a_0 {\bf z} + a_1 T^{\ast} {\bf z} + \cdots + a_s T^{\ast s} {\bf z} \right) , {\bf u} \right\rangle =0 . \]
This implies
\[ \left\langle {\bf z} , \left( a_0 T^{k-1-s} + a_1 T^{k-s} + \cdots + a_s T^{k-1} \right) {\bf u} \right\rangle = a_0 \left\langle {\bf z} , {\bf e}_{k-s} \right\rangle + a_1 \left\langle {\bf z} , {\bf e}_{k-s-1} \right\rangle + \cdots + a_s \left\langle {\bf z} , {\bf e}_k \right\rangle = a_s = 0 , \]
which is a contradiction. ■

Example 5: Consider the defective matrix (not diagonalizable)
\[ {\bf A} = \begin{bmatrix} 15&-6&2 \\ 35&-14&5 \\ 7&-3&2 \end{bmatrix} . \]
This matrix has the characteristic polynomial \( \chi (\lambda ) = \left( \lambda -1 \right)^3 \) while its minimal polynomial is \( \psi (\lambda ) = \left( \lambda -1 \right)^2 . \) The matrix A has two linearly independent eigenvectors
\[ {\bf u} = \left[ -1, 0, 7 \right]^{\mathrm T} \qquad\mbox{and} \qquad {\bf v} = \left[ 3, 7, 0 \right]^{\mathrm T} . \]
Let U and V be one-dimensional subspaces generated by spans of vectors u and v, respectively. The minimal polynomials of these vectors are the same: \( \psi_u (\lambda ) = \psi_v (\lambda ) = \lambda -1 \) because
\[ {\bf A}\,{\bf u} = {\bf u} \qquad\mbox{and} \qquad {\bf A}\,{\bf v} = {\bf v} . \]
Each of these one-dimensional subspaces U and V are A-cyclic and they cannot form the direct sum of \( \mathbb{R}^3 . \) We choose a vector \( {\bf z} = \left[ 7, -3, 1 \right]^{\mathrm T} , \) which is perpendicular to each u and v. matrix A transfers z into the vector \( {\bf A}\,{\bf z} = \left[ 125, 192, 60 \right]^{\mathrm T} , \) which is perpendicular to neither u nor v. Next application of A yields
\[ {\bf A}^2 \,{\bf z} = \begin{bmatrix} 243 \\ 587 \\ 119 \end{bmatrix} \qquad\Longrightarrow \qquad \det \begin{bmatrix} 7&125&243 \\ -3&192&587 \\ 1&60&119 \end{bmatrix} \ne 0 . \]
Hence, vectors z, Az, and A2z are linearly independent and \( \mathbb{R}^3 \) is the direct sum of A-cyclic. ■
End of Example 4

Annihilators and Direct Sums

Consider a direct sum decomposition of a vector space over field 𝔽:
\begin{equation} \label{EqDirect.1} V = S \oplus T . \end{equation}
Then any linear functional φT ∈ T* can be extended to a linear functional φ on V by setting φ(S) = 0. Let us call this extension by 0. Clearly, φ ∈ S⁰, the annihilator of S. Therefore, the mapping φT ⇾ φ is an iso,orphism from T* to S⁰, whose inverse is the restriction to T.

Theorem 4: Let V = S ⊕ T be a direct decomposition of a vector space V. The extension by 0 map is an isomorphism from T* to S⁰, and so \[ T^{\ast} \cong S^0 . \] If V is finite-dimensional, then \[ \dim\left( S^0 \right) = \mbox{codim}\left( S \right) = \dim \left( V/S \right) = \dim V - \dim S . \]

Example 6: Let V be the vector space over ℤ₂ with a countably infinite ordered basis ε = (e₁, e₂, e₃, …). We consider two its subspaces that are spanned on two distinct sets of basis elements: S = span{e₁} and T = span{e₂, e₃, …}. The annihilator of S is congruent to S⁰ ≌ T* ≌ V*.
End of Example 6
The annihilator provides a way to describe the dual space of a direct sum.

Theorem 5: A linear functional on the direct sum V = S ⊕ T can be written as a sum of a linear functional that annihilates S and a linear functional that annihilates T, that is, \[ \left( S \oplus T \right)^{\ast} = S^0 \oplus T^0 . \]

Clearly S⁰ ∩ T⁰ = {0} because any functional that annihilates both S and T must annihilate S ⊕ T = V. Hence, the sum S⁰ + T⁰ is direct. We have \[ V^{\ast} = \{ 0\}^0 = \left( S \cap T \right)^0 = S^0 + T^0 = S^0 \oplus T^0 . \] Alternatively, since φS + φT is the identity map, if φ ∈ V*, then we can write \[ \phi = \phi \circ \left( \phi_S + \phi_T \right) = \left( \phi \circ \phi_S \right) + \left( \phi \circ \phi_T \right) \in S^0 \oplus T^0 , \] so V ≌ S⁰ ⊕ T⁰.
   
Example 7: Let ℭ[−π, π] denote the class of all real-valued continuous functions on closed interval [−π, π]. This is a vector space (with respect to regular addition and scalar multiplication), but infinite dimensional one. We introduce two subspaces: \[ W_e = \left\{ f(x) \ : \ f(x) = f(-x) \right\} \] and \[ W_o = \left\{ f(x) \ : \ f(x) = -f(-x) \right\} \] We and Wo are respectively the collection of all odd functions and even functions. They are subspaces of ℭ[−π, π]. For any f ∈ ℭ[−π, π], consider \[ f_1 (x) = \frac{f(x) - f(-x)}{2} , \qquad f_2 (x) = \frac{f(x) + f(-x)}{2} . \] Then \[ f_1 (-x) = \frac{f(-x) - f(x)}{2} = - f_1 (x) , \] and \[ f_2 (-x) = \frac{f(-x) + f(x)}{2} = f_2 (x) . \] Thus, f₁ ∈ Wo and f₂ ∈ We. Clearly, f = f ₁ + f ₂ and hence ℭ[−π, π] = We + Wo. Also observe that We ∩ Wo = {0}. For if f ∈ We ∩ Wo, then f(−x) = −f(x) = f(x) for all x ∈ [−π, π]. This gives . f(x) ≡= 0 for all x ∈ [−π, π]. Thus, we can conclude that ℭ[−π, π] = We ⊕ Wo.    ■
End of Example 7

 

 


  1. Axier, S., Linear Algebra Done Right. Undergraduate Texts in Mathematics (3rd ed.). Springer. 2015, ISBN 978-3-319-11079-0.
  2. Beezer, R.A., A First Course in Linear Algebra, 2017.
  4. Dillon, M., Linear Algebra, Vector Spaces, and Linear Transformations, American Mathematical Society, Providence, RI, 2023.
  5. Halmos, Paul Richard (1974) [1958]. Finite-Dimensional Vector Spaces. Undergraduate Texts in Mathematics (2nd ed.). Springer. ISBN 0-387-90093-4.
  6. Roman, Steven (2005). Advanced Linear Algebra. Undergraduate Texts in Mathematics (2nd ed.). Springer. ISBN 0-387-24766-1.