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Recipe

Construct a complement of U and project a vector along it

Use when: you have a subspace U and either need a valid complement W, or need to project a specific vector onto U along a given W

Goal: a verified complement W, or a vector's unique split into its U-part and its W-part along that complement

Taught in Complements and the projections they define

Quick recipe

  1. Extend a basis of U to a basis of the whole space
  2. Check the extended list is genuinely independent
  3. Take the span of the newly added vectors as the candidate W
  4. If a candidate W is given rather than constructed, verify it directly
  5. Project a specific vector onto U along the verified complement
  1. 1

    Start from a basis of U, then add independent vectors one at a time until the combined list spans V.

    Example

    \{(1,1,0)\} extends by adding (0,1,0) and (0,0,1), two vectors not already in \operatorname{span}\{(1,1,0)\}.

    Why: any basis extension works here, since the direct-sum definition only asks for SOME complement, not a specific one.

  2. 2

    Row-reduce the combined vectors and confirm the rank equals the number of vectors, meaning none is redundant.

    Example

    Row reduction of \{(1,1,0),(0,1,0),(0,0,1)\} reaches rank 3, so all three are independent.

    Why: a dependent "extension" would not actually enlarge the span, leaving W too small to be a real complement.

  3. 3

    Let W be exactly the span of the vectors added in Step 1, not including U's own original basis vectors.

    Example

    W=\operatorname{span}\{(0,1,0),(0,0,1)\}, using only the two added vectors, not (1,1,0) itself.

    Why: this is precisely how the existence proof builds a complement, guaranteeing V=U+W while independence forces U\cap W=\{\mathbf 0\}.

  4. 4

    Check that \dim U+\dim W equals \dim V, then confirm U\cap W=\{\mathbf 0\} by solving for any vector lying in both.

    Example

    For W=\operatorname{span}\{(1,-1,0),(0,0,1)\}: \dim U+\dim W=1+2=3=\dim\mathbb R^3, and solving a(1,1,0)=b(1,-1,0)+c(0,0,1) forces a=b=c=0.

    • \dim U+\dim W doesn't equal \dim V — W is not a complement; stop here.
    • the dimensions match but a nonzero vector solves the intersection system — still not a complement, despite matching dimensions.

    Why: both conditions together are exactly the definition of V=U\oplus W, and matching dimensions alone never guarantees a trivial intersection.

  5. 5

    Write the target vector as a U-vector plus a W-vector, solving for the unknown coefficients with the two subspaces' own spanning vectors.

    Example

    For \mathbf v=(3,2), U=\operatorname{span}\{(1,0)\}, W=\operatorname{span}\{(1,1)\}: solving (3,2)=(a,0)+c(1,1) gives c=2 and a=1.

    Why: this decomposition is exactly what a projection means, and it is guaranteed unique once V=U\oplus W is confirmed.

Common mistakes

  • Including one of U's own basis vectors inside the candidate W, so U\cap W isn't trivial.
  • Checking only the dimension count and skipping the intersection check, as if matching dimensions alone were enough.
  • Forgetting that changing W changes the projection, and reusing an old projection formula for a new complement.
  • Solving for the decomposition's coefficients using the wrong subspace's spanning vectors.