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If m<n prove that y_1, ,y_m are linear functionals

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Homework Statement


Prove that if m<n, and if [tex]y_1,\cdots,y_m[/tex] are linear functionals on an n-dimensional vector space V, then there exists a non-zero vector x in V such that [tex][x,y_j]=0[/tex] for [tex]j=1,\cdots,m[/tex]. What does this result say about the solutions of linear equations?


Homework Equations



N/A

The Attempt at a Solution



Let [tex]X=\{x_1,\cdots,x_n\}[/tex] be a basis for V.
By a theorem in the book I know that there exists a uniquely determinet basis X' in V',
[tex] X'=\{y_1,\cdots,y_n\} [/tex] with the property that [tex][x_i,y_j]=\delta_{ij}[/tex].

Every [tex]x\in V[/tex] can be written as [tex] x=\xi_1 x_1+\cdots+\xi_n x_n[/tex].
If we let [tex]j=1,\cdots,m[/tex] we get,
[tex][x,y_j]=\xi_1[x_1,y_j]+\cdots+\xi_m[x_m,y_j][/tex] and since [tex][x_i,y_j]=\delta_{ij} [/tex], there exists some x such that [tex][x,y_j]=0[/tex].

The result says that the solution of linear equations consists of a homogeneous solution and a particular solution.

Any input is very welcome, Thanks!
 

Answers and Replies

  • #2
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Gentle bump.
 

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