Raising the bar Correctness of the space dimension definition Archives

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 $c_{i}$

$\sum_{i=1}^{l-1}c_{i}g_{i}^{\prime}=\sum_{i=1}^{l-1}c_{i}g_{i}-\sum_{i=1}^{l-1}c_{i}\frac{a_{i,k}}{a_{lk}}g_{l}=0.$

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 $n=m.$