From bases to projections

This foundation bridge supplies the finite-dimensional linear algebra used to begin the averaging proof of complete reducibility. It belongs to the programme’s existing linear-algebra foundation; it is not a replacement for that course.

The basis arguments below are adapted from Jim Hefferon’s Linear Algebra, at source revision df2262e089a02651c127f1dd12649c4622ee1383. The complete selected source statements and proofs, their source locators, and the author’s grant are supplied with this lesson. This adaptation retains Creative Commons Attribution–ShareAlike 2.5 terms; it is not CC0 and is not relicensed as GFDL. See the author’s grant, acknowledgements, exact selected source proofs, and reuse terms.

Adaptation, scope comparison, examples and solutions by OpenAI GPT-6 Astra in Codex, Ultra reasoning effort, October 2026. Self-checked by the AI writing this bridge; no human or independent AI review is claimed. The original source remains unchanged. This lesson corrects an already-recorded coefficient sign, makes the field condition explicit, and expands the finite termination and projection arguments. These are standard results, not claimed discoveries.

What to know first

We use field and vector-space axioms, finite sums, finite induction and elementary finite-set counting. A field has distinct elements $0$ and $1$; every nonzero scalar has an inverse. A vector space $V$ over a field $k$ is an abelian additive group with scalar multiplication satisfying the vector-space axioms.

For $S\subseteq V$, its span consists of finite linear combinations of elements of $S$; the empty combination is zero. The set is linearly independent when every relation involving finitely many distinct elements has all coefficients zero. A basis is a linearly independent spanning set. Throughout, $V$ has a finite basis. No topology, inner product, algebraic closure or characteristic-zero assumption is used.

Linear combinations of linear combinations remain linear combinations, by distributivity. Thus a span is a subspace. A subset of an independent set is independent, since a relation in the subset is also one in the larger set. Finally, coordinates in a basis are unique: subtracting two expressions gives a relation in the basis and therefore zero differences between all coefficients. We will use these elementary consequences explicitly.

Adding one independent vector

Lemma. If $S$ is independent and $v\notin S$, then $S\cup\{v\}$ is independent exactly when $v\notin\operatorname{span}(S)$.

Proof. If $v=\sum_{i=1}^r a_i s_i$, combine repeated terms so the $s_i$ are distinct. The equation

$$\sum_{i=1}^r a_i s_i-v=0$$

is a nontrivial relation on distinct elements of $S\cup\{v\}$: the coefficient of $v$ is $-1\ne0$. Conversely, any nontrivial relation on that union must involve $v$, because $S$ is independent. Write it as

$$\sum_{i=1}^r a_i s_i+bv=0.$$

If $b=0$, independence of $S$ forces all other coefficients to vanish, a contradiction. Consequently $b\ne0$, and

$$v=-\sum_{i=1}^r b^{-1}a_i s_i\in\operatorname{span}(S).$$

This proves both implications. It also covers $S=\varnothing$: the only dependent singleton is $\{0\}$. $\square$

The minus sign in the last display repairs the existing source finding B40-VS2-SOURCE-003. Omitting it did not change the source’s span-membership conclusion, but it made the displayed equality wrong outside characteristic two.

Exchanging a basis vector

Lemma. Let $B=(b_1,\ldots,b_n)$ be a basis, and write $v=\sum_{j=1}^n c_jb_j$. If $c_i\ne0$, replacing $b_i$ by $v$ gives another basis.

Proof. Suppose

$$d_i v+\sum_{j\ne i}d_jb_j=0.$$

Substitution expresses zero in the original basis. Its $b_i$-coefficient is $d_ic_i$, so $d_i=0$. The other coefficients are then $d_j=0$. Thus the replacement family is independent. Moreover,

$$b_i=c_i^{-1}v-\sum_{j\ne i}c_i^{-1}c_jb_j.$$

Every old basis vector therefore belongs to the span of the replacement family. That family spans $V$, and hence is a basis. $\square$

Why finite dimension bounds independence

Theorem. If $V$ has a basis with $n$ elements, every independent subset of $V$ has at most $n$ elements. Consequently all bases of $V$ have the same size.

Proof. If $n=0$, then $V=\{0\}$, whose only independent subset is empty. Suppose $n>0$ and, towards a contradiction, that an independent set contains distinct vectors $s_1,\ldots,s_{n+1}$.

Start with the given basis $B_0$. Inductively suppose that $B_r$ is a basis containing $s_1,\ldots,s_r$ and $n-r$ of the original basis vectors, where $0\le r<n$. Express $s_{r+1}$ in $B_r$. Some coefficient of one of the remaining original vectors must be nonzero: otherwise $s_{r+1}$ would be in the span of $s_1,\ldots,s_r$, contradicting independence. The exchange lemma replaces such an original vector by $s_{r+1}$. This constructs $B_{r+1}$.

After $n$ exchanges, $(s_1,\ldots,s_n)$ is a basis. It spans $s_{n+1}$, again contradicting independence. Thus there cannot be $n+1$ distinct elements in an independent set. This proves the bound without presupposing that the set spans or is already finite.

If another basis has $m$ elements, the bound gives $m\le n$. Applying the same argument with that basis gives $n\le m$. Hence $m=n$; their common value is $\dim_k V$. $\square$

This is the source’s exchange argument with its induction written directly for an arbitrary independent set. It avoids using the existence of a minimal-sized basis when proving the independent-set bound.

Extending a basis and choosing a complement

Theorem. Every independent subset of $V$ extends to a basis of $V$. Every subspace $U\subseteq V$ has a finite basis, which extends to one of $V$.

Proof. An independent set $S$ has at most $n=\dim_k V$ elements. If it does not span $V$, choose a vector outside its span. The first lemma says that adjoining it preserves independence. Repeat whenever the span is still not $V$. Each step increases the finite cardinality by one. The preceding theorem makes more than $n-|S|$ additions impossible, so the process must reach a basis.

For a subspace $U$, start with the empty set and choose each new vector in $U$ outside the current span. These sets are independent in $V$, so the same bound forces termination in a spanning independent set of $U$. Extend this basis to one of $V$ by the first part. Only finitely many choices are needed. If $U=0$, its basis is empty; if $U=V$, there are no extra basis vectors. $\square$

Corollary. Every subspace of $V$ is the image of a linear projection. More precisely, there are linear maps $i:U\to V$ and $p:V\to U$ with $i$ the inclusion and $pi=1_U$. The endomorphism $p_0=ip$ satisfies

$$p_0^2=p_0,\qquad\operatorname{im}p_0=U, \qquad V=U\oplus\ker p_0.$$

Proof. Choose a basis $(u_1,\ldots,u_r)$ of $U$ and extend it to $(u_1,\ldots,u_r,w_1,\ldots,w_{n-r})$ of $V$. Unique coordinates define

$$\begin{aligned} v&=\sum_{i=1}^r a_i u_i+\sum_{j=1}^{n-r}b_jw_j,\\ p(v)&=\sum_{i=1}^r a_i u_i. \end{aligned}$$

Coordinates of a sum are the sums of the coordinates, and coordinates of a scalar multiple are the scalar multiples of the coordinates, again by uniqueness. Therefore $p$ is linear. It fixes $U$, which gives $pi=1_U$, $p_0^2=p_0$, and $\operatorname{im}p_0=U$. Its kernel is the span of the $w_j$. For every $v$,

$$v=p_0v+(v-p_0v),\qquad p_0(v-p_0v)=0.$$

If $u\in U\cap\ker p_0$, then $u=p_0u=0$. This proves the direct sum. $\square$

There is generally no distinguished choice of $p_0$. In the displayed adapted basis its matrix is $\operatorname{diag}(I_r,0)$. A different complement can give a different projection onto the same subspace.

The exact step used in representation theory

In Representations and complete reducibility, Theorem 2.3 begins by extending a basis of an invariant subspace $U$, then defining a projection $p_0$. The preceding corollary supplies exactly this construction, over every field in that theorem, including finite fields. Exercise 5 applies the same construction to its isotypic subspace $S$. No inner product or orthogonal projection is needed for either initial construction.

The representation-theoretic averaging has an additional condition: $|G|$ must be invertible in $k$. If $U$ is invariant, put

$$P=\frac{1}{|G|}\sum_{g\in G}\rho(g)p_0\rho(g^{-1}).$$

Here each summand maps into $U$ and acts as the identity on $U$. Hence $P(V)\subseteq U$ and $P|_U=1_U$. Conjugation by $\rho(h)$ permutes the summands by $g\mapsto hg$, so $P$ commutes with the group action. Also $P^2=P$, because $P$ fixes its image. The decomposition $v=Pv+(v-Pv)$ gives the invariant complement $\ker P$. The requirement that $|G|$ be invertible belongs here, not to basis extension.

This comparison applies to the exact lesson source SHA-256 659942BDF47420E6665D59C4E80CF973867730D3B231D7E9C80AB8177BD83712, at Theorem 2.3 and Exercise 5. It checks only the finite-basis/projection prerequisite and its use. It does not certify the rest of the lesson, its other prerequisites, the isotypic identification, or the complete course. The original consumer lesson and its review state are unchanged.

Worked example and exercises

Example. In $k^2$, take $U=k(1,0)$. For every $a\in k$, the vector $(a,1)$ is a complement basis vector, and the resulting projection is

$$p_a(x,y)=(x-ay,0),\qquad [p_a]=\begin{pmatrix}1&-a\\0&0\end{pmatrix}.$$

Its kernel is $k(a,1)$. Direct substitution gives $p_a^2=p_a$. These distinct projections show why choosing a subspace does not canonically choose a complement.

Exercise 1. Over a field, extend $u=(1,1,0)$, $v=(0,1,1)$ to a basis of $k^3$, and write a projection onto their span.

Solution. A relation $au+bv=0$ gives $a=0$ in the first coordinate and $b=0$ in the third. The vector $e_2=(0,1,0)$ is outside their span: the same two coordinates would force both coefficients to vanish. Thus $(u,v,e_2)$ is a basis. In fact

$$\begin{aligned} (x,y,z)&=xu+zv+(y-x-z)e_2,\\ p_0(x,y,z)&=(x,x+z,z). \end{aligned}$$

This formula is valid in every characteristic. It fixes $u,v$, kills $e_2$, and satisfies $p_0^2=p_0$.

Exercise 2. Why does basis extension not by itself prove that every invariant subspace has an invariant complement in characteristic $p$?

Solution. Over a field of characteristic $p>0$, let a generator of $C_p$ act on $k^2$ by

$$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}=I+N.$$

Since $N^2=0$, induction gives $(I+N)^m=I+mN$, and thus $A^p=I$. The line $U=ke_1$ is invariant. Any complementary line has a generator $w=(a,1)$. If it were invariant, $Aw=(a+1,1)$ would equal $cw$. The second coordinate forces $c=1$, while the first then forces $1=0$, impossible. Ordinary linear complements exist; invariant ones do not in this example. Averaging cannot repair them because the scalar $|C_p|=p$ is zero in $k$.

Exercise 3. Which part of the proof prevents it from asserting basis extension for arbitrary infinite-dimensional spaces?

Solution. The finite bound on the size of an independent set makes the process of adding vectors terminate. With no finite spanning basis, this argument supplies no such bound and therefore no termination claim. An infinite-dimensional theorem needs a different argument and any additional set-theoretic assumptions must be stated. Nothing in this bridge claims that theorem.

Sources and changes

Hefferon, Linear Algebra, vector-spaces chapter: the add-vector lemma in src/vs/vs2.tex, lines 492–525; exchange, equal basis size, independent-set bound and basis extension in src/vs/vs3.tex, lines 1657–1779, 1818–1831 and 1857–1874. The programme’s existing foundation reader preserves the source edition and its editorial notes. The author’s site supplies the broader course.

The adaptation above keeps the same algebraic argument, spells out the zero-dimensional case and finite termination, proves the subspace and projection consequences, and gives examples with solutions. The requirement $0\ne1$ also respects the already-recorded source-context finding B40-FIELDS-SOURCE-001. It is not a newly discovered defect. The field-general formulation is justified by the displayed algebraic proof, not merely by changing the word “real” in a source statement.

All adapted teaching and added exposition on this page use CC BY-SA 2.5. The separately linked representation-theory lesson retains its own source, credit and licence. No change to it or to the canonical Stacks text is made here.