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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!