A description/proof of that finite dimensional vectors spaces related by linear bijection are of same dimension
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of vectors space.
- The reader knows a definition of bijection.
- The reader admits the Clamer theorem.
Target Context
- The reader will have a description and a proof of the proposition that any 2 finite dimensional vectors spaces such that there is a linear bijection from one of them to the other are of the same dimension.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any finite dimensional vectors spaces, \(V_1\) and \(V_2\), such that there is any linear bijection, \(f: V_1 \rightarrow V_2\), the dimensions of \(V_1\) and \(V_2\) are the same.
2: Proof
There are a basis for \(V_1\), \((b_{11}, b_{12}, . . ., b_{1d_1})\), and a basis for \(V_2\), \((b_{21}, b_{22}, . . ., b_{2d_2})\). Any \(v_1 \in V_1\) is \(v_1 = {v_1}^i b_{1i}\). \(f (v_1) = {v_1}^i f (b_{1i})\). \(f (b_{1i}) = {f^j}_i b_{2j}\). \(f (v_1) = {v_1}^i {f^j}_i b_{2j} = {v_2}^i b_{2i}\). \(\begin{pmatrix} {f^1}_1 & . . . & {f^1}_{d_1} \\ . . . & . . . & . . . \\ {f^{d_2}}_1 & . . . & {f^{d_2}}_{d_1} \end{pmatrix} \begin{pmatrix} {v_1}^1 \\ . . . \\ {v_1}^{d_1} \end{pmatrix} = \begin{pmatrix} {v_2}^1 \\ . . . \\ {v_2}^{d_2} \end{pmatrix}\). By the Clamer theorem, if the rank of \(\begin{pmatrix} {f^j}_i \end{pmatrix}\), k, was \(k \lt d_1\), supposing that the theorem's condition for existence of solutions for \(v_1\) was satisfied (it had to be satisfied indeed), there would be solutions for \(v_1\) with \({v_1}^{k + 1} , . . ., {v_1}^{d_1}\) arbitrary, which would mean that \(f\) was not injective, so, \(k = d_1\), so, \(d_1 \le d_2\). By the Clamer theorem, if \(d_1 \lt d_2\), supposing that some rows and some columns had been already reordered according to the theorem, \(det \begin{pmatrix} {f^1}_1 & . . . & {f^1}_{d_1} & {v_2}^1 \\ . . . & . . . & . . . & . . . \\ {f^{d_1}}_1 & . . . & {f^{d_1}}_{d_1} & {v_2}^{d_1} \\ {f^{d_1 + 1}}_1 & . . . & {f^{d_1 + 1}}_{d_1} & {v_2}^{d_1 + 1} \end{pmatrix} = 0\) in order for any solution for \(v_1\) to exist, but it would not be possible for every possible \(v_2\), because the algebraic complement with respect to \({v_2}^{d_1 + 1}\) was not 0, so, the determinant could be deviated from 0 by choosing a \({v_2}^{d_1 + 1}\), which would mean that \(f\) was not surjective, so, \(d_1 = d_2\).
