User:IssaRice/Linear algebra/Riesz representation theorem: Difference between revisions

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Let's take the case where $V = R^{n}$ and the inner product is the usual dot product. What does the Riesz representation theorem say in this case? It says that if we have a linear functional $T : R^{n} \to R$ then we can write $T$ as $T v = v \cdot u$ for some vector $u \in R^{n}$ . But we already know (from the correspondence between matrices and linear transformations) that we can represent $T$ as a 1-by-n matrix. And v can be thought of as a n-by-1 matrix. And now the dot product is the same thing as the matrix multiplication!

If $σ = (e_{1}, \dots, e_{n})$ is the standard basis of $R^{n}$ and $(1)$ is the standard basis of $R$ , then $T v = [T v]^{(1)} = [T]_{(1)}^{σ} [v]^{σ} = [T]_{(1)}^{σ} \cdot [v]^{σ} = [v]^{σ} \cdot [T]_{σ}^{(1)} = v \cdot [T]_{σ}^{(1)}$ .

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	Let's take the case where <math>V = \mathbf R^n</math> and the inner product is the usual dot product. What does the Riesz representation theorem say in this case? It says that if we have a linear functional <math>T : \mathbf R^n \to \mathbf R</math> then we can write <math>T</math> as <math>Tv = v\cdot u</math> for some vector <math>u \in \mathbf R^n</math>. But we already know (from the correspondence between matrices and linear transformations) that we can represent <math>T</math> as a 1-by-n matrix. And v can be thought of as a n-by-1 matrix. And now the dot product is the same thing as the matrix multiplication!		Let's take the case where <math>V = \mathbf R^n</math> and the inner product is the usual dot product. What does the Riesz representation theorem say in this case? It says that if we have a linear functional <math>T : \mathbf R^n \to \mathbf R</math> then we can write <math>T</math> as <math>Tv = v\cdot u</math> for some vector <math>u \in \mathbf R^n</math>. But we already know (from the correspondence between matrices and linear transformations) that we can represent <math>T</math> as a 1-by-n matrix. And v can be thought of as a n-by-1 matrix. And now the dot product is the same thing as the matrix multiplication!

	If <math>\sigma = (e_1, \ldots, e_n)</math> is the standard basis, then <math>Tv = [Tv]^~~\sigma~~ = [T]_~~\sigma~~^\sigma [v]^\sigma = [T]_~~\sigma~~^\sigma \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^~~\sigma~~ = v \cdot [T]_\sigma^~~\sigma~~</math>.		If <math>\sigma = (e_1, \ldots, e_n)</math> is the standard basis of <math>\mathbf R^n</math> and <math>(1)</math> is the standard basis of <math>\mathbf R</math>, then <math>Tv = [Tv]^{(1)} = [T]_{(1)}^\sigma [v]^\sigma = [T]_{(1)}^\sigma \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</math>.