The Sum and the Subspace Question
Combining vectors from two subspaces creates a new set: take one vector from the first subspace, one from the second, and add them. The central question is whether the resulting set still satisfies the defining conditions for a subspace. The answer follows from how addition and scalar multiplication distribute over the two parts of a sum.
Let \(V\) be a vector space over a field \(\mathbb{F}\), and let \(U\) and \(W\) be subspaces of \(V\). In particular, both are subsets of the same ambient space and use its addition and scalar multiplication. The zero vector of \(V\) is denoted by \(\mathbf{0}\).
To show that \(U+W\) is a subspace, we verify the subspace conditions: it contains the zero vector, is closed under addition, and is closed under scalar multiplication. Those checks must apply to arbitrary elements of the sum, not merely to selected examples. The subspace assumptions on \(U\) and \(W\) provide exactly the closure facts needed.
The Subspace Proof
Proof. First, since \(U\) and \(W\) are subspaces, both contain \(\mathbf{0}\). Therefore \(\mathbf{0}=\mathbf{0}+\mathbf{0}\) belongs to \(U+W\). Now take any \(\mathbf{x},\mathbf{y}\in U+W\). By the definition of the sum, there are \(\mathbf{u}_1,\mathbf{u}_2\in U\) and \(\mathbf{w}_1,\mathbf{w}_2\in W\) such that \(\mathbf{x}=\mathbf{u}_1+\mathbf{w}_1\) and \(\mathbf{y}=\mathbf{u}_2+\mathbf{w}_2\). Then $$ \mathbf{x}+\mathbf{y} =(\mathbf{u}_1+\mathbf{u}_2)+(\mathbf{w}_1+\mathbf{w}_2). $$ Because \(U\) and \(W\) are closed under addition, \(\mathbf{u}_1+\mathbf{u}_2\in U\) and \(\mathbf{w}_1+\mathbf{w}_2\in W\); hence \(\mathbf{x}+\mathbf{y}\in U+W\). Finally, take any scalar \(c\in\mathbb{F}\) and any \(\mathbf{x}\in U+W\), writing \(\mathbf{x}=\mathbf{u}+\mathbf{w}\) with \(\mathbf{u}\in U\) and \(\mathbf{w}\in W\). Then $$ c\mathbf{x}=c(\mathbf{u}+\mathbf{w})=c\mathbf{u}+c\mathbf{w}. $$ Closure under scalar multiplication gives \(c\mathbf{u}\in U\) and \(c\mathbf{w}\in W\), so \(c\mathbf{x}\in U+W\). The set contains zero and is closed under addition and scalar multiplication, so it is a subspace of \(V\). \(\square\)
The proof uses the same pattern in both closure checks: write each element of the sum in its two parts, perform the vector operation, and regroup the result into a part from \(U\) and a part from \(W\). For scalar multiplication, the scalar must be in the field over which \(V\), \(U\), and \(W\) are defined. The argument covers every such scalar, including \(c=0\).
Proof. Take any \(\mathbf{u}\in U\). Since \(\mathbf{0}\in W\), we can write \(\mathbf{u}=\mathbf{u}+\mathbf{0}\), so \(\mathbf{u}\in U+W\). Thus \(U\subseteq U+W\). Next, take any \(\mathbf{w}\in W\). Since \(\mathbf{0}\in U\), we can write \(\mathbf{w}=\mathbf{0}+\mathbf{w}\), so \(\mathbf{w}\in U+W\). Thus \(W\subseteq U+W\). \(\square\)
Coordinate Subspaces in \(\mathbb{R}^3\)
Worked Example: Two Coordinate Planes
Let \(U=\{(x,y,0):x,y\in\mathbb{R}\}\) and \(W=\{(0,0,z):z\in\mathbb{R}\}\), both subsets of \(\mathbb{R}^3\). These are subspaces: each contains \(\mathbf{0}\), and adding vectors or multiplying by a real scalar preserves the defining zero coordinates. Determine \(U+W\).
A vector in the sum has the form $$ (x,y,0)+(0,0,z)=(x,y,z), $$ so every vector in \(U+W\) belongs to \(\mathbb{R}^3\). Conversely, take any \((a,b,c)\in\mathbb{R}^3\). Its decomposition is $$ (a,b,c)=(a,b,0)+(0,0,c), $$ where \((a,b,0)\in U\) and \((0,0,c)\in W\). Therefore every vector of \(\mathbb{R}^3\) lies in \(U+W\), and \(U+W=\mathbb{R}^3\).
Check: For example, \((2,-1,4)=(2,-1,0)+(0,0,4)\). Adding the two displayed vectors gives \((2,-1,4)\), with the first summand in \(U\) and the second in \(W\).
Two Lines Whose Sum Is a Plane
Worked Example: Decomposing a Vector in the Plane
In \(\mathbb{R}^3\), let \(U=\{\alpha(1,1,0):\alpha\in\mathbb{R}\}\) and \(W=\{\beta(1,-1,0):\beta\in\mathbb{R}\}\). Each is a subspace: it consists of scalar multiples of a fixed vector, so it contains zero and is closed under addition and scalar multiplication. Find their sum.
Any vector in \(U+W\) has the form $$ \alpha(1,1,0)+\beta(1,-1,0) =(\alpha+\beta,\alpha-\beta,0). $$ Its third coordinate is zero, so \(U+W\) is contained in the \(xy\)-plane. To show that every vector in that plane is in the sum, take an arbitrary \((p,q,0)\). Choose \(\alpha=(p+q)/2\) and \(\beta=(p-q)/2\). Then $$ \alpha+\beta=\frac{p+q+p-q}{2}=p,\qquad \alpha-\beta=\frac{p+q-p+q}{2}=q. $$ Thus $$ (p,q,0)=\frac{p+q}{2}(1,1,0)+\frac{p-q}{2}(1,-1,0), $$ so \(U+W=\{(p,q,0):p,q\in\mathbb{R}\}\).
Check: For \((3,1,0)\), the coefficients are \(\alpha=(3+1)/2=2\) and \(\beta=(3-1)/2=1\). Hence $$ 2(1,1,0)+1(1,-1,0)=(2,2,0)+(1,-1,0)=(3,1,0). $$ The computed vector is in the stated sum.
A Sum of Matrix Subspaces
Worked Example: Symmetric and Skew-Symmetric Matrices
Work in the vector space \(\mathbb{R}^{2\times 2}\). Let \(U\) be the set of symmetric matrices, meaning matrices \(\mathbf{S}\) with \(\mathbf{S}^{\top}=\mathbf{S}\), and let \(W\) be the set of skew-symmetric matrices, meaning matrices \(\mathbf{K}\) with \(\mathbf{K}^{\top}=-\mathbf{K}\). Both are subspaces: the zero matrix satisfies each defining equation, and transposition respects addition and scalar multiplication, so each defining equation is preserved by those operations. Show that their sum is all of \(\mathbb{R}^{2\times 2}\).
Take an arbitrary matrix \(\mathbf{A}=\begin{bmatrix}p&q\\r&s\end{bmatrix}\). Define $$ \mathbf{S}=\begin{bmatrix}p&\frac{q+r}{2}\\\frac{q+r}{2}&s\end{bmatrix}, \qquad \mathbf{K}=\begin{bmatrix}0&\frac{q-r}{2}\\\frac{r-q}{2}&0\end{bmatrix}. $$ The off-diagonal entries of \(\mathbf{S}\) agree, so \(\mathbf{S}^{\top}=\mathbf{S}\) and \(\mathbf{S}\in U\). The off-diagonal entries of \(\mathbf{K}\) are negatives of each other and its diagonal entries are zero, so \(\mathbf{K}^{\top}=-\mathbf{K}\) and \(\mathbf{K}\in W\). Adding gives $$ \mathbf{S}+\mathbf{K} =\begin{bmatrix}p&\frac{q+r+q-r}{2}\\\frac{q+r+r-q}{2}&s\end{bmatrix} =\begin{bmatrix}p&q\\r&s\end{bmatrix} =\mathbf{A}. $$ Every matrix is therefore in \(U+W\), and since \(U+W\subseteq\mathbb{R}^{2\times2}\) by definition, \(U+W=\mathbb{R}^{2\times2}\).
Check: For \(\mathbf{A}=\begin{bmatrix}2&5\\1&-3\end{bmatrix}\), the formulas give \(\mathbf{S}=\begin{bmatrix}2&3\\3&-3\end{bmatrix}\) and \(\mathbf{K}=\begin{bmatrix}0&2\\-2&0\end{bmatrix}\). Their sum is \(\begin{bmatrix}2&3+2\\3-2&-3\end{bmatrix}=\begin{bmatrix}2&5\\1&-3\end{bmatrix}\), as required.
Why the Ambient Space and Both Parts Matter
The requirement that \(U\) and \(W\) be subspaces of the same \(V\) is not cosmetic. The expression \(\mathbf{u}+\mathbf{w}\) uses the addition in one vector space, so its two vectors must belong to a common ambient space. Also, the proof depends on closure in each component: when it forms \(\mathbf{u}_1+\mathbf{u}_2\), that result must remain in \(U\), and when it forms \(\mathbf{w}_1+\mathbf{w}_2\), that result must remain in \(W\). If either input set is not a subspace, the proof no longer guarantees the required closure.
A related distinction is that \(U+W\) is not generally the union \(U\cup W\). The sum allows a vector to be assembled from two different subspaces; the union merely collects vectors already in at least one of them. For example, take the subspaces \(U=\{(x,0):x\in\mathbb{R}\}\) and \(W=\{(0,y):y\in\mathbb{R}\}\) of \(\mathbb{R}^2\). The vector \((1,1)\) is not in \(U\), because its second coordinate is not zero, and is not in \(W\), because its first coordinate is not zero. But it is in \(U+W\), since \((1,1)=(1,0)+(0,1)\).
Common Pitfall: Confusing Sum with Union
Wrong calculation: For the two coordinate axes just described, a student may write “\((1,1)\notin U\) and \((1,1)\notin W\), so \((1,1)\notin U+W\).” The conclusion uses the membership test for the union, not the sum. Correction: Membership in \(U+W\) asks whether there exist \(\mathbf{u}\in U\) and \(\mathbf{w}\in W\) whose sum is the target. Here \(\mathbf{u}=(1,0)\in U\), \(\mathbf{w}=(0,1)\in W\), and \(\mathbf{u}+\mathbf{w}=(1,1)\). Thus \((1,1)\in U+W\), even though it belongs to neither input subspace individually.
For a proof that a particular sum is a subspace, keep the logical tasks separate: establish that the inputs are subspaces of one ambient vector space, take arbitrary elements of the proposed sum for the closure checks, and exhibit the zero vector using one zero from each input. These steps prove the claim for every pair of subspaces, while examples help identify what the resulting set looks like in specific cases.
Check Your Understanding
Use the definition of \(U+W\) and the subspace conditions to justify each response.
- If \(U\) and \(W\) are subspaces of \(V\), write the expression that shows \(\mathbf{0}\in U+W\).
- Suppose \(\mathbf{x}=\mathbf{u}_1+\mathbf{w}_1\) and \(\mathbf{y}=\mathbf{u}_2+\mathbf{w}_2\), with the \(\mathbf{u}_i\) in \(U\) and \(\mathbf{w}_i\) in \(W\). Rewrite \(\mathbf{x}+\mathbf{y}\) as one vector from \(U\) plus one from \(W\).
- For the subspaces in the two-lines example, express \((4,2,0)\) as a vector in \(U\) plus a vector in \(W\).
- For the coordinate axes \(U=\{(x,0)\}\) and \(W=\{(0,y)\}\) in \(\mathbb{R}^2\), explain why \((2,-3)\) belongs to \(U+W\) but not to either subspace.
- Why does the proof that \(U+W\) is a subspace require \(U\) and \(W\) to lie in the same ambient vector space?
