Jul 19

Correctness of the space dimension definition

Correctness of the space dimension definition

This proof has been taken from I. M. Gelfand, Lectures in Linear Algebra, 4th edition, 1970 (in Russian).

Lemma. Let f_1,...,f_k be some vectors. Suppose that vectors g_1,...,g_l are linearly independent and belong to the span \text{span}(f_{1},...,f_{k}). Then l\leq k.

Proof. We prove the lemma by induction. Let k=1. Then l=1 (linear dependence for an empty set of vectors is not defined) and the inequality l\leq k is trivial.

Suppose the statement holds for k-1 and let us prove it for k. Since g_{1},...,g_{l}\in \text{span}(f_{1},...,f_{k}), we have

(1) g_{1}=a_{11}f_{1}+...+a_{1k}f_{k} \\... \\  g_{l}=a_{l1}f_{1}+...+a_{lk}f_{k}.

If all of a_{1k},...,a_{lk} are zero, we have g_{1},...,g_{l}\in \text{span}(f_{1},...,f_{k-1}) and then by the induction assumption l\leq k-1 and, trivially, l\leq k.

Thus, we can assume that not all of a_{1k},...,a_{lk} are zero. Suppose a_{lk}\neq 0. Then we can solve the last equation in (1) for f_{k}:

(2) f_{k}=\frac{1}{a_{lk}}g_{l}-\frac{1}{a_{lk}}\sum_{j=1}^{k-1}a_{l,j}f_{j}.

Plugging this equation in the first equation in (1) we get

g_{1}=a_{11}f_{1}+...+a_{1,k-1}f_{k-1}+a_{1k}\left( \frac{1}{a_{lk}}g_{l}-\frac{1}{a_{lk}}\sum_{j=1}^{k-1}a_{l,j}f_{j}\right) .

Send g_{l} to the left side; the remaining expression on the right side is a linear combination of f_{1},...,f_{k-1}; exact expressions of the coefficients b_{1,j} of this linear combination don't matter. The result will be g_{1}-\frac{a_{1k}}{a_{lk}}g_{l}=\sum_{j=1}^{k-1}b_{1,j}f_{j}. After doing the same with the first l-1 equations of (1) we get the system

g_{1}-\frac{a_{1k}}{a_{lk}}g_{l}=\sum_{j=1}^{k-1}b_{1,j}f_{j},...,\ \  g_{l-1}-\frac{a_{l-1,k}}{a_{lk}}g_{l}=\sum_{j=1}^{k-1}b_{l-1,j}f_{j}.

This shows that the vectors

g_{1}^{\prime }=g_{1}-\frac{a_{1k}}{a_{lk}}g_{l},...,g_{l-1}^{\prime }=\  g_{l-1}-\frac{a_{l-1,k}}{a_{lk}}g_{l}

belong to \text{span}(f_{1},...,f_{k-1}). If they are linearly independent, we can use the induction assumption to conclude that l-1\leq k-1, which will prove l\leq k.

Suppose with some c_{i}


By the assumed linear independence of g_{1},...,g_{l} this implies c_{1}=...=c_{l-1}=0, so the system g_{1}^{\prime },...,g_{l-1}^{\prime } is linearly independent.

Theorem. The definition of the space dimension is correct.

Proof. We write \dim (L)=n if a) L contains n linearly independent vectors x_{1},...,x_{n} and b) L=\text{span}(x_{1},...,x_{n}). We need to prove that any system with properties a) and b) has the same number of vectors. Suppose y_{1},...,y_{m} is another such system. Since y_{1},...,y_{m} belong to \text{span}(x_{1},...,x_{n}), by the lemma m\leq n. Similarly, n\leq m. So n=m.

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