AcademySubspaces And Dimension
Academy
Subspaces Of Rn
Level 1 - Linear Algebra topic page in Subspaces And Dimension.
Principle
A subspace of \(\mathbb R^n\) is a subset that behaves like a vector space using the usual vector addition and scalar multiplication. It must contain the zero vector and be closed under the two vector operations.
Subspaces are the linear sets that pass through the origin: lines through the origin, planes through the origin, and higher-dimensional analogues.
Notation
\(W\)
a subset of \mathbb R^n being tested as a subspace
\(\mathbf 0\)
the zero vector in \mathbb R^n
\(\mathbf u,\mathbf v\)
vectors in W
\(\lambda\)
a real scalar
\(\lambda\mathbf u+\mathbf v\)
a linear combination used in the subspace test
The Core Test
To prove that \(W\subseteq\mathbb R^n\) is a subspace, check these three conditions:
- \(\mathbf 0\in W\).
- If \(\mathbf u,\mathbf v\in W\), then \(\mathbf u+\mathbf v\in W\).
- If \(\mathbf u\in W\) and \(\lambda\in\mathbb R\), then \(\lambda\mathbf u\in W\).
Equivalently, combine closure under addition and scalar multiplication into one linear-combination test:
Subspace linear-combination test
\[\mathbf u,\mathbf v\in W,\;\lambda,\mu\in\mathbb R\Longrightarrow \lambda\mathbf u+\mu\mathbf v\in W\]
If any one condition fails, \(W\) is not a subspace.
Worked Cases
Question
Show that \(W=\{(x,y,z)\in\mathbb R^3:x+2y-z=0\}\) is a subspace of \(\mathbb R^3\).
Answer
First check the zero vector: \(0+2(0)-0=0\), so \(\mathbf0\in W\). Let \(\mathbf u=(x_1,y_1,z_1)\in W\) and \(\mathbf v=(x_2,y_2,z_2)\in W\). Then \(x_1+2y_1-z_1=0\) and \(x_2+2y_2-z_2=0\). For scalars \(\lambda,\mu\), compute the defining expression for \(\lambda\mathbf u+\mu\mathbf v\): \(\lambda x_1+\mu x_2+2(\lambda y_1+\mu y_2)-(\lambda z_1+\mu z_2)\). Group terms: \(\lambda(x_1+2y_1-z_1)+\mu(x_2+2y_2-z_2)=\lambda0+\mu0=0\). Therefore every linear combination stays in \(W\), so \(W\) is a subspace.
Examples
Question
Is \(L=\{t(2,-1,3):t\in\mathbb R\}\) a subspace of \(\mathbb R^3\)?
Answer
Yes. The zero vector occurs when \(t=0\). If \(\mathbf u=s(2,-1,3)\) and \(\mathbf v=t(2,-1,3)\), then \(\lambda\mathbf u+\mu\mathbf v=(\lambda s+\mu t)(2,-1,3)\), which is still in \(L\). Therefore \(L\) is a subspace.