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Glacius
Linear algebraConcept reference

Bases

A basis of a subspace is a linearly independent list of vectors that spans that subspace.

1 min read1 figureSources & further reading
On this page 7 sections
  1. Overview
  2. A basis spans a specified subspace and is linearly independent
  3. The list (1,0), (0,1), (1,1) spans the plane
  4. An output subspace is spanned by p=(1,0,1) and q=(0,1,1)
  5. Key takeaway
  6. Sources & further reading
  7. Concept connections

01A basis spans a specified subspace and is linearly independent#

A basis spans a specified subspace and is linearly independent. Every vector in that space can be built from it, with no redundant direction.

Two arrows from the origin: u=(1,0) points right and v=(1,1) points right and up. Equal horizontal and vertical unit scales show different directions.Two arrows from the origin: u=(1,0) points right and v=(1,1) points right and up. Equal horizontal and vertical unit scales show different directions.
Figure 1Two arrows from the origin: u=(1,0) points right and v=(1,1) points right and up. Equal horizontal and vertical unit scales show different directions.
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For u=(1,0) and v=(1,1), any (s,t) can be built with these weights:

(s,t)=(s−t)u+tv(s,t)=(s-t)u+tv

If au+bv=(0,0), the second coordinate forces b=0; then a=0. These vectors span the plane and are independent.

Check your reasoning

For S=R2S=\mathbb R^2, is B={(1,0), (0,2)}B=\{(1,0),\,(0,2)\} a basis?

  1. ANo: independent only.
  2. BYes: both conditions.
  3. CNo: dependent.
Show answer and explanation
Yes: both conditions.

The two independent axes span the plane.

02The list (1,0), (0,1), (1,1) spans the plane#

The list (1,0), (0,1), (1,1) spans the plane. Its third vector is the sum of the first two, so the list is dependent. It fails the basis test.

Check your reasoning

Repair this basis claim for R2\mathbb R^2: {(2,0), (0,1), (4,0)}\{(2,0),\,(0,1),\,(4,0)\}.

  1. A{(2,0), (0,1)}\{(2,0),\,(0,1)\}
  2. B{(0,1)}\{(0,1)\}
  3. C{(2,0), (4,0)}\{(2,0),\,(4,0)\}
Show answer and explanation
{(2,0), (0,1)}\{(2,0),\,(0,1)\}

Remove (4,0), twice (2,0).

03An output subspace is spanned by p=(1,0,1) and q=(0,1,1)#

An output subspace is spanned by p=(1,0,1) and q=(0,1,1). Replacing p by p+q keeps a basis: recover p by subtracting q. No direction is lost or added.

p=(p+q)−qp=(p+q)-q
Check your reasoning

U={u,v}U=\{u,v\} is a basis for an output space. Is B={u+v,2u+2v}B=\{u+v,2u+2v\} also a basis?

  1. AYes: both conditions.
  2. BNo: dependent.
  3. CNo: independent only.
Show answer and explanation
No: dependent.

The second vector is twice the first.

Key takeaway

A basis must span the specified space and contain no redundant vector. Spanning alone is not enough.

  • Verify a basis for a specified finite-dimensional subspace.

Sources & further reading

  1. [1]

Reference this concept

Link to this page, a section, or an individual figure.

Glacius. “Bases.” Math behind ML. /learn/la-basis