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Recipe

Verify V=U\oplus W and decompose a vector into its parts

Use when: you have two subspaces U and W and need to know whether V=U\oplus W, or need a specific vector's unique U-part and W-part once it holds

Goal: either a confirmed direct sum with a genuine unique decomposition, or a concrete reason it fails

Taught in Direct sums

Quick recipe

  1. Check that U+W reaches all of V
  2. Check that U\cap W contains only the zero vector
  3. If both checks pass, decompose a specific vector by matching components
  4. If either check fails, exhibit two different decompositions
  1. 1

    Confirm \dim U+\dim W equals \dim V, or directly verify every vector of V is some sum of a U-vector and a W-vector.

    Example

    For U=\operatorname{span}\{(1,0,0),(0,1,0)\} and W=\operatorname{span}\{(0,0,1)\} in \mathbb R^3: \dim U+\dim W=2+1=3=\dim\mathbb R^3.

    • \dim U+\dim W is less than \dim V — U+W can't be all of V, so V=U\oplus W already fails; stop here.
    • \dim U+\dim W equals or exceeds \dim V — continue to Step 2 and check the intersection.

    Why: a direct sum first requires an ordinary sum covering all of V; uniqueness cannot matter if U+W misses vectors.

  2. 2

    Solve for which vectors, if any, lie in both U and W at once; a direct sum requires that set to be exactly \{\mathbf 0\}.

    Example

    A vector (a,b,0) from U equals (0,0,c) from W only when a=b=c=0, so U\cap W=\{\mathbf 0\}.

    Why: uniqueness of the decomposition is equivalent to a trivial intersection, the exact fact the section's proof rests on.

  3. 3

    Write the target vector as a U-part plus a W-part, using the specific structure of U and W's own spanning sets to solve for the unknowns.

    Example

    For \mathbf v=(3,-2,5) with U the xy-plane and W the z-axis, matching components gives (3,-2,0)+(0,0,5).

    Why: since the sum is direct, this decomposition is guaranteed to exist and to be the only one, so solving it once suffices.

  4. 4

    Find a specific vector, then produce two genuinely different valid (\mathbf u,\mathbf w) pairs that both sum to it.

    Example

    (2,3,0) = (2,3,0)+(0,0,0), and also (2,3,0) = (1,2,0)+(1,1,0), both valid splits.

    Why: exhibiting two different splits is a concrete proof of non-uniqueness, stronger than only citing a nonzero intersection dimension.

Common mistakes

  • Checking only the intersection and skipping whether U+W actually reaches all of V.
  • Concluding a sum is direct just because the dimension arithmetic works out, without confirming the intersection is trivial.
  • Stopping at "the intersection is nonzero" instead of exhibiting an actual second decomposition when a counterexample is asked for.
  • Mixing up which vector's components label the U-part versus the W-part after solving.