Is There a Simpler Way to Construct a Linear Functional Given a Linear Operator?

[ihggin]

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

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

f(Tv)=cf(v)

f((T-cI)v)=0

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

Can you find a nonzero functional vanishing on this image?

 

[ihggin]

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

 

[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.