Vector space

[Guest]

Hey everybody i got a little problem with a math problem here it is if u can help me it will be wonderful tksss guys !
1/Let V be a vector space , f an endomorphism of V
2/Let [tex]\lambda[/tex] be the eigenvalue of f
3/ Let E as E = { v [tex]\in[/tex] V such as f(v) =[tex]\lambda \cdot[/tex] v}

Q: Show that E is a sub-vector space of V .

[HallsofIvy]

Start from the definitions! A subset, E, of vector space, V, is a subspace if and only if it is "closed under vector addition" and "closed under scalar multiplication". That is, if u and v are both in E then u+ v is also in E and if v is in E and a is a scalar then av is also in E. Also, in linear algebra, "endomorphisms" are linear functions.

Here, E is the set of all v such that [tex]f(v)= \lambda v[/tex].

If u and v are in E, so that [tex]f(u)= \lambda u[/tex] and [tex]f(v)= \lambda v[/tex], then [tex]f(u+ v)= f(u)+ f(v)= \lambda u+ \lambda v= \lambda(u+ v)[/tex]. Therefore u+ v is in E.

If v is in E and a is a scalar then [tex]f(av)= af(v)= a \lambda v= \lambda (av)[/tex]. Therefore av is in E.

[HallsofIvy]

One thing I should have included- the subspace must be non-empty. Typically that is easiest shown by showing that the 0 vector is in the set. Since f(0)= 0 for any endomorphism (linear transformation) and [tex]\lambda 0= 0[/tex], [tex]f(0)= \lambda 0[/tex] so the 0 vector is in the set.

(Some texts use "the 0 vector is in the subspace" rather than "the subspace is non-empty". They are equivalent. Obviously, if the 0 vector is in the subspace, it is non-empty. To go the other way, if the subspace is non empty, it contains some vector, v. Since the subspace is "closed under scalar multiplication, (-1)v is also in it. Since the subspace is "closed under vector addition", v+ (-1)v= v- v= 0 is in the subspace.)