# Existence of linear functional

• ihggin
In summary: So the image of (T-cI) is a subspace of V with dimension dim(V)-1. Then you can extend the basis of this subspace to a basis of V. In summary, given a finite-dimensional vector space V and a linear operator T on V, if there exists a non-zero vector \alpha in V such that T\alpha = c\alpha for some scalar c, then there exists a non-zero linear functional f on V such that T^{t}f=cf, where T^{t}f=f\circ T is the transpose. This can be proven by choosing a basis for V that includes \alpha and defining f to be non-zero on a basis vector outside of the image of (T-c
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?

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?

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 $$x$$.

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.

Last edited:

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.

## What is a linear functional?

A linear functional is a mathematical function that takes as input a vector and outputs a scalar value. It is a type of linear transformation that maps a vector space to its underlying field.

## What is the significance of linear functionals in mathematics?

Linear functionals play an important role in many areas of mathematics, including functional analysis, linear algebra, and optimization. They are used to study the properties of linear transformations and help solve complex problems in various fields.

## How are linear functionals different from regular functions?

The main difference between linear functionals and regular functions is that linear functionals operate on vector spaces, while regular functions operate on real or complex numbers. Additionally, linear functionals must satisfy the properties of linearity, such as preserving addition and scalar multiplication.

## What are some examples of linear functionals?

Some common examples of linear functionals include dot product, integration, and differentiation. In dot product, the output is the scalar value of the projection of one vector onto another. In integration, the output is the area under a curve. In differentiation, the output is the slope of a tangent line at a specific point.

## How do linear functionals relate to the existence of dual spaces?

The existence of linear functionals is closely related to the concept of dual spaces. A dual space is a vector space consisting of all linear functionals on a given vector space. In other words, it is the space of all possible linear transformations from a vector space to its underlying field. The existence of dual spaces is crucial in many mathematical applications, such as optimization and functional analysis.

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