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.