Subspace Verification

From Subsets to Subspaces

In any vector space $V$, we can consider various subsets of its vectors. However, not all subsets are equally interesting from a linear algebra perspective. We care most about subsets that preserve the algebraic structure of the parent space—that is, subsets where vector addition and scalar multiplication "play nice" and don't lead you outside the subset. These special subsets are called subspaces.

Let's begin with a video that provides excellent geometric intuition for this idea.

The Subspace Test: A Powerful Shortcut

As you saw in the video and as we previewed in the last lesson, we don't need to re-verify all ten vector space axioms to prove a subset $U$ is a subspace. Since $U$ is part of a larger vector space $V$, most of the axioms (like commutativity and associativity) are inherited automatically. We only need to verify the properties that ensure $U$ is a self-contained vector space.

This leads to the crucial Subspace Test.

To summarize, a subset $U$ of a vector space $V$ is a subspace if and only if it satisfies these three conditions:

1. Contains the Zero Vector: The zero vector $0$ of $V$ must be in $U$.
2. Closure under Addition: For any two vectors ${\mathbf{u}}_1$ and ${\mathbf{u}}_2$ in $U$, their sum ${\mathbf{u}}_1+{\mathbf{u}}_2$ must also be in $U$.
3. Closure under Scalar Multiplication: For any vector $\mathbf{u}$ in $U$ and any scalar $c$, the product $c\mathbf{u}$ must also be in $U$.

This three-point checklist is the tool we will use for all our proofs moving forward.

Proving a Subset Is a Subspace

To prove that a set $U$ is a subspace, we must show that all three conditions of the Subspace Test hold for arbitrary elements of $U$.

Let's see this in action with a concrete example.

A More Abstract Example: The Null Space

The Subspace Test applies not just to subsets of ${\mathbb{R}}^n$, but to more abstractly defined sets as well. One of the most important subspaces you'll encounter is the null space of a matrix $A$, which is the set of all vectors $\mathbf{x}$ that solve the homogeneous equation $A\mathbf{x}=0$. This concept is central to understanding the solution structure of linear systems.

Let's walk through the proof that the null space is always a subspace.

The proof structure is identical to the video example, just applied to a more general setting:

1. Check for $0$: Is $A0=0$? Yes. So, $0$ is in the null space.
2. Check Closure under Addition: If ${\mathbf{x}}_1$ and ${\mathbf{x}}_2$ are in the null space (meaning $A{\mathbf{x}}_1=0$ and $A{\mathbf{x}}_2=0$), is ${\mathbf{x}}_1+{\mathbf{x}}_2$ in the null space? We check: $A({\mathbf{x}}_1+{\mathbf{x}}_2)=A{\mathbf{x}}_1+A{\mathbf{x}}_2=0+0=0$. Yes.
3. Check Closure under Scalar Multiplication: If $\mathbf{x}$ is in the null space ($A\mathbf{x}=0$) and $c$ is a scalar, is $c\mathbf{x}$ in the null space? We check: $A(c\mathbf{x})=c(A\mathbf{x})=c0=0$. Yes.

Since all three conditions hold, the null space is a subspace.

Proving a Subset Is Not a Subspace

To prove that a subset is not a subspace, you only need to provide one specific counterexample that shows the failure of any one of the three conditions.

1. Easiest Check: Does the subset contain the zero vector? If not, you're done. For example, the set of vectors $(x,y)$ in ${\mathbb{R}}^2$ where $x+y=1$ is not a subspace because $(0,0)$ is not in the set ($0+0\ne 1$).

2. Finding a Counterexample for Closure: If the zero vector is included, you must find a case where addition or scalar multiplication fails.

Let's watch a clear demonstration of this.

Another Common Counterexample: Determinants

Consider the vector space $M_{2\times 2}(\mathbb{R})$ of all 2x2 matrices. Let $U$ be the subset of matrices where the determinant is zero. Is $U$ a subspace?

1. Zero Vector: The zero vector in this space is the zero matrix $\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$. Its determinant is 0, so it's in $U$.

2. Closure under Addition: Let's test this. Consider these two matrices in $U$:

   $$A=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),\det (A)=0$$ $$B=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right),\det (B)=0$$

   Their sum is:

   $$A+B=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$$

   The determinant of their sum is $\det (I)=1$, which is not 0. Therefore, $A+B$ is not in $U$.

Since we found a counterexample for closure under addition, $U$ is not a subspace.

Conclusion

In this lesson, we have developed the primary tool for identifying the fundamental building blocks of vector spaces.

Here are the key takeaways:

• A subspace is a non-empty subset of a vector space that is itself a vector space under the same operations.
• The Subspace Test is a three-condition check that provides an efficient way to prove or disprove that a subset is a subspace. The conditions are: containing the zero vector, closure under addition, and closure under scalar multiplication.
• To prove a set is a subspace, you must show that all three conditions hold for arbitrary elements.
• To disprove that a set is a subspace, you only need to provide a single, concrete counterexample demonstrating that one of the three conditions fails.

Preview of the next lesson:
We have seen that subspaces can be defined by equations (like planes or null spaces). An equally powerful way to define a subspace is by constructing it from a set of vectors. In the next lesson, we will explore the concept of span—the set of all possible linear combinations of a given set of vectors—and prove that the span of any set of vectors is always a subspace.