Existence of linear functional

[ihggin]

Let [tex]V[/tex] be a finite-dimensional vector space over the field [tex]F[/tex] and let [tex]T[/tex] be a linear operator on [tex]V[/tex]. Let [tex]c[/tex] be a scalar and suppose there is a non-zero vector [tex]\alpha[/tex] in [tex]V[/tex] such that [tex]t \alpha = c \alpha[/tex]. Prove that there is a non-zero linear functional [tex]f[/tex] on [tex]V[/tex] such that [tex]T^{t}f=cf[/tex], where [tex]T^{t}f=f\circ T[/tex] is the transpose.

I tried the following: let [tex]B[/tex] be a basis for [tex]V[/tex] that contains [tex]\alpha[/tex] (we can do this since [tex]\alpha \neq 0[/tex]). Then define [tex]f[/tex] such that [tex]f(\alpha)=1[/tex] and [tex]f(b)=0[/tex] for all the other basis vectors. Extend the definition to arbitrary vectors using linearity of [tex]f[/tex]. So if [tex]v=\sum c_i b_i[/tex] for scalars [tex]c_i[/tex] and basis vectors [tex]b_i[/tex] with [tex]b_1= \alpha[/tex], we have [tex]f(v)=c_1[/tex]. However, I then ran into the problem that when I take [tex]f(Tv)[/tex], [tex]T[/tex] can map other basis vectors into vectors with components in [tex]\alpha[/tex], which messes my strategy up.

Is there some smart choice of basis vectors that can prevent this from happening? Or is this just not a good way of doing the problem?

 

[arkajad]

I would suggest, contemplate these equations and questions:

[tex]f(Tv)=cf(v)[/tex]

[tex]f((T-cI)v)=0[/tex]

Can [tex]T-cI[/tex] be invertible? How big is the image [tex](T-cI)V[/tex]? Can it be the whole of V?

Can you find a nonzero functional vanishing on this image?

 

[ihggin]

Thank you! So basically [tex](T-cI)\alpha=0[/tex] so the dimension of the kernel is greater than zero, which means the dimension of the image is less than that of [tex]V[/tex]. We can then take a vector [tex]x[/tex] that is not in the image and generate a basis from it. We then define [tex]f[/tex] so that it's zero on all the basis vectors except [tex]x[/tex].

 

[arkajad]

You almost got it. But you need to choose a basis in such a way that the dim(Im) vectors span the image and set your functional to vanish on these and not to vanish on at least one basis vector outside.

 

[blue_raver22]

Your approach is a good start, but there is a simpler way to construct the desired linear functional f. Since T\alpha = c\alpha, we can define f(v) = c for all v \in V such that Tv = c\alpha. This is a well-defined linear functional since V is finite-dimensional and T is a linear operator. Now, for any v \in V, we can write v = \sum c_i b_i for some basis vectors b_i and scalars c_i. Then T^{t}f(v) = T^{t}f(\sum c_i b_i) = T^{t}(c_1 b_1) = c_1 T^{t}(b_1) = c_1 f(b_1) = c_1 = f(v), as desired. Therefore, f is a non-zero linear functional on V such that T^{t}f = cf.