- #1

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- Homework Statement
- Prove that if ##y## is a linear functional on an ##n##-dimensional vector space ##V##, then the set of all those vectors ##x## such that ##[x, y] = 0## is a subspace of ##V##; what is the dimension of this subspace?

- Relevant Equations
- A function ##y : V \rightarrow \mathbb{R}## is a linear functional on ##V## if for all ##u, v \in V## and ##\alpha, \beta \in F##, we have ##[\alpha u + \beta v , y] = \alpha [u, y] + \beta [v, y]##.

We have the notation ##[v, y]## to mean ##y(v)##.

I am stuck on finding the dimension of the subspace. Here's what I have so far.

Proof: Let ##W = \lbrace x \in V : [x, y] = 0\rbrace##. We see ##[0, y] = 0##, so ##W## is non empty. Let ##u, v \in W## and ##\alpha, \beta## be scalars. Then ##[\alpha u + \beta v, y] = \alpha [u, y] + \beta [v, y] = \alpha \cdot 0 + \beta \cdot 0 = 0##. So, ##\alpha u + \beta v \in W##. This shows ##W## is a subspace of ##V##.

Next, we will find the dimension of ##W##. Let ##x_1, x_2, \dots, x_n## be a basis of ##V##. We can find a maximal subset of this basis ##z_1, \dots, z_k## such that ##[z_i, y] = 0## for all ##i##. We claim this is a basis for ##W##. Since ##u \in V##, there exists scalars ##\alpha_1, \dots, \alpha_n## such that ##u = \alpha_1 z_1 + \dots + \alpha_k z_k + \alpha_{k+1}x_{k+1} + \dots + \alpha_n x_n##. Then,

$$[u, y] = [\alpha_1 z_1 + \dots + \alpha_k z_k + \alpha_{k+1}x_{k+1} + \dots + \alpha_n x_n, y] = \alpha_{k+1} [x_{k+1}, y] + \dots + \alpha_n [x_n, y] = 0$$.

We want to somehow conclude ##\alpha_{k+1} = \dots = \alpha_n = 0##, but I'm not sure how to proceed. Can I have a hint, please?

Proof: Let ##W = \lbrace x \in V : [x, y] = 0\rbrace##. We see ##[0, y] = 0##, so ##W## is non empty. Let ##u, v \in W## and ##\alpha, \beta## be scalars. Then ##[\alpha u + \beta v, y] = \alpha [u, y] + \beta [v, y] = \alpha \cdot 0 + \beta \cdot 0 = 0##. So, ##\alpha u + \beta v \in W##. This shows ##W## is a subspace of ##V##.

Next, we will find the dimension of ##W##. Let ##x_1, x_2, \dots, x_n## be a basis of ##V##. We can find a maximal subset of this basis ##z_1, \dots, z_k## such that ##[z_i, y] = 0## for all ##i##. We claim this is a basis for ##W##. Since ##u \in V##, there exists scalars ##\alpha_1, \dots, \alpha_n## such that ##u = \alpha_1 z_1 + \dots + \alpha_k z_k + \alpha_{k+1}x_{k+1} + \dots + \alpha_n x_n##. Then,

$$[u, y] = [\alpha_1 z_1 + \dots + \alpha_k z_k + \alpha_{k+1}x_{k+1} + \dots + \alpha_n x_n, y] = \alpha_{k+1} [x_{k+1}, y] + \dots + \alpha_n [x_n, y] = 0$$.

We want to somehow conclude ##\alpha_{k+1} = \dots = \alpha_n = 0##, but I'm not sure how to proceed. Can I have a hint, please?