A Reliable Workflow for Subspace Problems
A subspace question often begins with a set described in a different language: equations on coordinates, conditions on polynomial coefficients, restrictions on matrix entries, or rules about function values and sequence terms. The key is to identify the ambient vector space and its operations before testing the set. Then check whether the zero element is included and whether the defining conditions survive linear combinations. A condition that looks restrictive can still define a subspace; the crucial distinction is whether it is homogeneous.
Unless stated otherwise, the scalar field in this workshop is \(\mathbb{R}\). In \(\mathbb{R}^n\), addition and scalar multiplication are the usual coordinatewise operations. For polynomials, matrices, functions, and sequences, the operations are also defined entrywise or pointwise. For example, if \(f\) and \(g\) are real-valued functions, then \((af+bg)(x)=af(x)+bg(x)\) for every \(x\).
In practice, the combined linear-combination check is often the most efficient: choose arbitrary members of the proposed set and arbitrary real scalars, combine them, and verify every defining condition again. To show that a set is not a subspace, one valid failure is enough. It may fail to contain zero, or a specific pair of members and scalars may produce a combination outside the set.
Why Homogeneous Constraints Behave Well
A constraint is homogeneous when its required value is zero. Examples include \(x-2y+z=0\), \(p(1)=0\), and \(a_{11}+a_{22}=0\). Such conditions are preserved by linear combinations because multiplying zero by a scalar and adding zero still gives zero. The following result makes this reasoning precise.
Proof. By linearity, \(L_i(\mathbf{0})=L_i(0\mathbf{v})=0L_i(\mathbf{v})=0\) for each \(i\), so \(\mathbf{0}\in S\). Now take \(\mathbf{u},\mathbf{v}\in S\) and \(a,b\in\mathbb{R}\). For every index \(i\), linearity gives \(L_i(a\mathbf{u}+b\mathbf{v})=aL_i(\mathbf{u})+bL_i(\mathbf{v})=a\cdot0+b\cdot0=0\). Thus \(a\mathbf{u}+b\mathbf{v}\) satisfies every constraint and belongs to \(S\). The Subspace Test proves that \(S\) is a subspace. \(\square\)
This proposition is a way to organize a calculation, not a shortcut that makes the checks optional. For a particular set, identify what each constraint evaluates to and verify that the evaluation respects linear combinations. In coefficient-based and pointwise settings, writing down those evaluations explicitly prevents hidden assumptions about the operations.
Worked Example: Two Homogeneous Equations in \(\mathbb{R}^3\)
Classify \(S=\{(x,y,z)\in\mathbb{R}^3:x-2y+z=0,\ y+z=0\}\), and describe its members.
Both conditions are homogeneous linear equations in the coordinates, so the proposition applies. To describe the solutions, set \(z=t\). The second equation gives \(y=-t\). Substituting this into the first gives \(x-2(-t)+t=x+3t=0\), so \(x=-3t\). Therefore every member has the form \((-3t,-t,t)\), where \(t\in\mathbb{R}\).
Check: for \((-3t,-t,t)\), the left side of the first condition is \(-3t-2(-t)+t=0\), and the left side of the second is \(-t+t=0\). In particular, \(t=0\) gives the zero vector. The homogeneous-constraint proposition verifies closure for all real linear combinations, so \(S\) is a subspace.
Translate the Conditions into the Right Language
When the ambient space changes, the principle stays the same, but the meaning of a constraint changes. For polynomials, evaluate the polynomial or its coefficients. For matrices, express the condition using entries. For functions and sequences, use their pointwise operations and check the relevant values or coordinates. The next examples make those translations explicit.
Worked Example: Polynomial Conditions in \(P_3\)
Let \(P_3\) be the real vector space of polynomials of degree at most \(3\), including the zero polynomial. Classify \(S=\{p\in P_3:p(1)=0,\ p'(0)=0\}\).
Write an arbitrary polynomial as \(p(x)=a_0+a_1x+a_2x^2+a_3x^3\). Its derivative is \(p'(x)=a_1+2a_2x+3a_3x^2\), so \(p'(0)=a_1\). The second condition therefore requires \(a_1=0\). The first condition becomes \(a_0+a_1+a_2+a_3=0\), hence \(a_0=-a_2-a_3\). Every polynomial in the set is consequently \(p(x)=a_2(x^2-1)+a_3(x^3-1)\), with \(a_2,a_3\in\mathbb{R}\).
Check: evaluating this expression at \(1\) gives \(a_2(1-1)+a_3(1-1)=0\). Its derivative is \(2a_2x+3a_3x^2\), whose value at \(0\) is \(0\). The zero polynomial occurs when \(a_2=a_3=0\). Evaluation at \(1\) and differentiation followed by evaluation at \(0\) are linear operations, so both constraints are homogeneous linear constraints. The proposition shows that \(S\) is a subspace of \(P_3\).
Worked Example: A Matrix Condition on Entries
Classify the set \(S\) of real \(2\times2\) matrices satisfying \(a_{11}+a_{22}=0\) and \(a_{12}=a_{21}\).
Write a general matrix as \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\). The first condition gives \(d=-a\); the second gives \(c=b\). Hence the matrices in \(S\) are exactly \(\begin{bmatrix}a&b\\b&-a\end{bmatrix}\), where \(a,b\in\mathbb{R}\).
For a closure check, let \(\mathbf{A}=\begin{bmatrix}a&b\\b&-a\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}c&d\\d&-c\end{bmatrix}\) be in \(S\), and let \(r,s\in\mathbb{R}\). Then \(r\mathbf{A}+s\mathbf{B}=\begin{bmatrix}ra+sc&rb+sd\\rb+sd&-ra-sc\end{bmatrix}\). The diagonal entries sum to \(ra+sc-ra-sc=0\), and the off-diagonal entries are equal. Also, choosing \(r=s=0\) gives the zero matrix. Thus the defining constraints and the Subspace Test show that \(S\) is a subspace.
Check: take \(\mathbf{A}=\begin{bmatrix}1&2\\2&-1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}3&-1\\-1&-3\end{bmatrix}\). Both satisfy the conditions. Their sum is \(\begin{bmatrix}4&1\\1&-4\end{bmatrix}\), whose diagonal entries sum to \(4-4=0\) and whose off-diagonal entries agree.
Worked Example: Function Values at Fixed Inputs
Let \(V\) be the real vector space of all functions \(f:\mathbb{R}\to\mathbb{R}\), with pointwise operations. Classify \(S=\{f\in V:f(0)=0,\ f(2)=3f(1)\}\).
Take \(f,g\in S\) and \(a,b\in\mathbb{R}\). At zero, \((af+bg)(0)=af(0)+bg(0)=a\cdot0+b\cdot0=0\). At the other two inputs, \((af+bg)(2)=af(2)+bg(2)=a(3f(1))+b(3g(1))=3(af(1)+bg(1))=3(af+bg)(1)\). Thus every linear combination satisfies both defining conditions. The zero function also satisfies them, so \(S\) is a subspace.
Check: the function \(f(x)=x+x^2\) is a nonzero member: \(f(0)=0\), \(f(1)=2\), and \(f(2)=6=3f(1)\). The computation above checks arbitrary linear combinations, not just this example function.
Worked Example: Coordinate Conditions on Sequences
Let \(V\) be the real vector space of all real sequences \(\mathbf{a}=(a_1,a_2,a_3,\dots)\), with coordinatewise operations. Classify \(S=\{\mathbf{a}\in V:a_1=2a_2,\ a_4=0\}\).
If \(\mathbf{a},\mathbf{b}\in S\) and \(r,s\in\mathbb{R}\), then \((r\mathbf{a}+s\mathbf{b})_1=ra_1+sb_1=2ra_2+2sb_2=2(r\mathbf{a}+s\mathbf{b})_2\). Also, \((r\mathbf{a}+s\mathbf{b})_4=ra_4+sb_4=0\). Both conditions hold for the zero sequence as well. Therefore \(S\) is a subspace.
Check: the sequence \(\mathbf{a}=(1,\tfrac12,0,0,0,\dots)\) belongs to \(S\), since \(1=2(\tfrac12)\) and its fourth term is \(0\). If \(\mathbf{b}=(2,1,5,0,0,\dots)\), then \(\mathbf{a}+\mathbf{b}=(3,\tfrac32,5,0,0,\dots)\): its first term is \(3=2(\tfrac32)\), and its fourth term remains \(0\).
When a Constraint Rules Out a Subspace
A condition with a nonzero required value behaves differently from a homogeneous condition. For example, \(x+y=1\) cannot hold at the zero vector. A nonlinear condition may include zero and still fail closure: the zero check is necessary, but not sufficient.
Proof. If \(c=0\), the set is \(\{\mathbf{v}\in V:L(\mathbf{v})=0\}\), which is a subspace by the homogeneous-constraint proposition. If \(c\ne0\), linearity gives \(L(\mathbf{0})=0\), so \(\mathbf{0}\) does not belong to the level set \(L(\mathbf{v})=c\). Since every subspace must contain \(\mathbf{0}\), the level set is not a subspace. These cases prove both directions. \(\square\)
Worked Example: An Affine Plane That Misses Zero
Classify \(S=\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\}\).
The zero vector gives \(0+0+0=0\ne1\), so \(S\) does not contain the zero vector and is not a subspace. The level-set proposition applies with \(L(x,y,z)=x+y+z\) and \(c=1\).
Check: \((1,0,0)\) and \((0,1,0)\) both belong to \(S\), but their sum \((1,1,0)\) has coordinate sum \(2\), not \(1\). Thus closure under addition fails as well.
Common Mistake: Zero Is Not the Whole Test
A frequent error is to check that a set contains zero and then assume it is a subspace. Consider \(S=\{(x,y)\in\mathbb{R}^2:xy=0\}\), the union of the two coordinate axes. The zero vector is in \(S\), and so are \((1,0)\) and \((0,1)\). But \((1,0)+(0,1)=(1,1)\), and its coordinate product is \(1\cdot1=1\ne0\). The set is not closed under addition, so it is not a subspace.
The incorrect calculation would stop after observing that \((0,0)\in S\) and conclude “subspace.” The correction is to finish the Subspace Test: either verify closure for arbitrary members or find one counterexample. In this case the displayed sum is a concrete failure. It also illustrates why a product constraint such as \(xy=0\) should not be treated like a homogeneous linear equation: the product of coordinates does not preserve addition in the way a linear condition does.
For a systematic solution, write the conditions in their original language first, then test them under the operations of the ambient space. The following checklist helps keep the argument complete.
Specify whether the elements are coordinate vectors, polynomials, matrices, functions, or sequences, and use that space’s operations.
Substitute the zero element into every defining condition. If it fails even one, the set is not a subspace.
For a proposed subspace, take arbitrary members and arbitrary real scalars, form their linear combination, and check every condition.
Give a failed zero check or explicit members and a linear combination that leaves the set.
Homogeneous equations are especially manageable because the zero vector satisfies them and linearity carries them through every linear combination. In contrast, a nonzero required value immediately rules out zero, while nonlinear conditions demand a direct closure check. Keeping these distinctions visible makes problems across different vector spaces instances of the same careful test.
Check Your Understanding
For each set, state the ambient space and justify the classification by checking zero and closure, or by giving a specific failure.
- In \(\mathbb{R}^2\), classify \(\{(x,y):2x-y=0\}\).
- In \(P_2\), classify the polynomials satisfying \(p(0)=0\) and \(p(1)=2\).
- In the real \(2\times2\) matrices, classify the matrices satisfying \(a_{11}=a_{22}\) and \(a_{12}=0\).
- In the space of all real-valued functions on \(\mathbb{R}\), decide whether the condition \(f(0)=1\) defines a subspace.
- In the space of real sequences, test whether \(a_1+a_2=0\) and \(a_3=0\) define a subspace.
