Respuesta :

LucianoBordoli

Answer:

No, it isn't.

Step-by-step explanation:

We have [tex]V=IR^{2}[/tex] and let [tex]H[/tex] be the subset of [tex]V[/tex] of all points on the line

[tex]4x+3y=12[/tex]

We need to find if [tex]H[/tex] is a subspace of the vector space [tex]V[/tex].

In [tex]IR^{2}[/tex] all the possibilities for own subspace of the vector space [tex]IR^{2}[/tex] are :

  • [tex]IR^{2}[/tex] itself.
  • The vector [tex]0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right][/tex]
  • All lines in [tex]IR^{2}[/tex] that passes through the origin  (  [tex]0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right][/tex]  )

We know that [tex]H[/tex] is the subset of [tex]IR^{2}[/tex] of all points on the line [tex]4x+3y=12[/tex]

If we look at the equation, the point [tex]\left[\begin{array}{c}0&0\end{array}\right][/tex] doesn't verify it because :

[tex]4x+3y=12\\4(0)+3(0)=12\\0=12[/tex]

Which is an absurd. Therefore, [tex]H[/tex] doesn't contain the origin (and [tex]H[/tex] is a line in [tex]IR^{2}[/tex]). Finally, it can't be a vector space of [tex]V=IR^{2}[/tex]

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