AcademyVector Spaces
Academy
Subspaces
Level 1 - Math I (Physics) topic page in Vector Spaces.
Principle
A subspace is a subset of a vector space that is itself a vector space using the same addition and scalar multiplication. It is not enough to be a subset; the subset must stay closed under the linear operations.
Subspaces appear in physics when motion, states, or solutions are constrained to a linear set. A particle constrained to move along a line through the origin has a one-dimensional subspace of possible displacement vectors.
Notation
\(V\)
the parent vector space
\(W\)
a candidate subspace of V
\(W\subseteq V\)
every vector in W is also in V
\(\mathbf 0\)
the zero vector of V
\(\mathbf u\)
a vector in W
\(\mathbf v\)
another vector in W
\(\lambda\)
a real scalar
\(\mu\)
another real scalar
\(\lambda\mathbf u+\mu\mathbf v\)
a linear combination of two vectors in W
The operations on \(W\) are inherited from \(V\). A subspace does not invent new addition or scalar multiplication.
Method
- Confirm that \(W\subseteq V\), so every object in \(W\) is an object from the parent vector space.
- Check that \(W\) is non-empty. Usually this is done by checking \(\mathbf 0\in W\).
- Take arbitrary \(\mathbf u,\mathbf v\in W\) and show \(\mathbf u+\mathbf v\in W\).
- Take arbitrary \(\lambda\in\mathbb R\) and \(\mathbf u\in W\), then show \(\lambda\mathbf u\in W\).
- Equivalently, combine the last two tests by showing every linear combination \(\lambda\mathbf u+\mu\mathbf v\) lies in \(W\).
Rules
Linear-combination test
\[\mathbf u,\mathbf v\in W,\ \lambda,\mu\in\mathbb R\quad\Rightarrow\quad \lambda\mathbf u+\mu\mathbf v\in W\]
Plane through the origin
\[W=\{(x,y,z)\in\mathbb R^3:z=0\}\]
Translated line failure
\[\mathbf 0=(0,0)\notin\{(x,y)\in\mathbb R^2:y=1\}\]
- Every subspace contains the zero vector.
- Lines and planes through the origin are subspaces of coordinate spaces.
- A translated line or plane that does not pass through the origin is not a subspace.
- The intersection of subspaces is a subspace because a vector satisfying all the subspace conditions remains inside every intersected set.
Examples
Question
Show that
\[W=\{(x,y,z)\in\mathbb R^3:z=0\}\]
is a subspace of \[\mathbb R^3\]
Answer
The zero vector
\[(0,0,0)\]
has third component 0, so it lies in \(W\). Let \[\mathbf u=(u_1,u_2,0)\]
and \[\mathbf v=(v_1,v_2,0)\]
Then \[\lambda\mathbf u+\mu\mathbf v=(\lambda u_1+\mu v_1,\lambda u_2+\mu v_2,0)\]
The third component is still 0, so the linear combination is in \(W\).Checks
- Use the same operations as the parent vector space.
- Check that the zero vector is present before doing longer calculations.
- Avoid treating translated affine sets as subspaces.
- Test arbitrary elements; one successful numerical example does not prove closure.