User:IssaRice/Linear algebra/Riesz representation theorem - Revision history

IssaRice at 04:21, 2 February 2021

2021-02-02T04:21:02Z

← Older revision		Revision as of 04:21, 2 February 2021
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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! In other words, the Riesz Representation Theorem (at least in this simple setting) is just saying that multiplying by a row vector is the same thing as dotting with that vector: <math>u^T v = v\cdot u</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]_\sigma^{(1)} [v]^\sigma = [T]_\sigma^{(1)} \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</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]_\sigma^{(1)} [v]^\sigma = [T]_\sigma^{(1)} \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</math>.

IssaRice at 23:46, 8 January 2019

2019-01-08T23:46:56Z

← Older revision		Revision as of 23:46, 8 January 2019
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	So in the case <math>V = \mathbf R^n</math> we can understand the Riesz representation theorem as saying something we already knew. What Riesz representation theorem does is extend this same sort of "representability" to all finite-dimensional inner product spaces V and all linear functionals <math>T : V \to \mathbf F</math>.		So in the case <math>V = \mathbf R^n</math> we can understand the Riesz representation theorem as saying something we already knew. What Riesz representation theorem does is extend this same sort of "representability" to all finite-dimensional inner product spaces V and all linear functionals <math>T : V \to \mathbf F</math>.

			[https://www.youtube.com/watch?v=LyGKycYT2v0&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=9 this video] talks about this for <math>\mathbf R^2</math>

IssaRice at 17:39, 6 January 2019

2019-01-06T17:39:11Z

← Older revision		Revision as of 17:39, 6 January 2019
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	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]_\sigma^{(1)} [v]^\sigma = [T]_\sigma^{(1)} \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</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]_\sigma^{(1)} [v]^\sigma = [T]_\sigma^{(1)} \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</math>.

	So in the case <math>V = \mathbf R^n</math> we can understand the Riesz representation theorem as saying something we already knew. What Riesz representation theorem does is extend this same sort of "representability" to all finite-dimensional V and all linear functionals <math>T : V \to \mathbf F</math>.		So in the case <math>V = \mathbf R^n</math> we can understand the Riesz representation theorem as saying something we already knew. What Riesz representation theorem does is extend this same sort of "representability" to all finite-dimensional inner product spaces V and all linear functionals <math>T : V \to \mathbf F</math>.

IssaRice at 17:36, 6 January 2019

2019-01-06T17:36:10Z

← Older revision		Revision as of 17:36, 6 January 2019
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	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]_\sigma^{(1)} [v]^\sigma = [T]_\sigma^{(1)} \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</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]_\sigma^{(1)} [v]^\sigma = [T]_\sigma^{(1)} \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</math>.

			So in the case <math>V = \mathbf R^n</math> we can understand the Riesz representation theorem as saying something we already knew. What Riesz representation theorem does is extend this same sort of "representability" to all finite-dimensional V and all linear functionals <math>T : V \to \mathbf F</math>.

IssaRice at 17:34, 6 January 2019

2019-01-06T17:34:08Z

← Older revision		Revision as of 17:34, 6 January 2019
Line 1:		Line 1:
	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 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]_\sigma^{(1)} [v]^\sigma = [T]_{(1)}~~^\sigma~~ \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</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]_\sigma^{(1)} [v]^\sigma = [T]_\sigma^{(1)} \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</math>.

IssaRice at 17:33, 6 January 2019

2019-01-06T17:33:50Z

← Older revision		Revision as of 17:33, 6 January 2019
Line 1:		Line 1:
	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 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>.		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]_\sigma^{(1)} [v]^\sigma = [T]_{(1)}^\sigma \cdot [v]^\sigma = [v]^\sigma \cdot [T]_\sigma^{(1)} = v \cdot [T]_\sigma^{(1)}</math>.

IssaRice at 17:33, 6 January 2019

2019-01-06T17:33:05Z

← Older revision		Revision as of 17:33, 6 January 2019
Line 1:		Line 1:
	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>.

IssaRice at 17:31, 6 January 2019

2019-01-06T17:31:15Z

← Older revision		Revision as of 17:31, 6 January 2019
Line 1:		Line 1:
	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</math>.		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>.

IssaRice at 17:30, 6 January 2019

2019-01-06T17:30:47Z

← Older revision		Revision as of 17:30, 6 January 2019
Line 1:		Line 1:
	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</math>.

IssaRice: Created page with "Let's take the case where $V = \mathbf R^n$ and the inner product is the usual dot product. What does the Riesz representation theorem say in this case? It says tha..."

2019-01-06T17:28:08Z

Created page with "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 tha..."

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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!

User:IssaRice/Linear algebra/Riesz representation theorem - Revision history

IssaRice at 04:21, 2 February 2021

IssaRice at 23:46, 8 January 2019

IssaRice at 17:39, 6 January 2019

IssaRice at 17:36, 6 January 2019

IssaRice at 17:34, 6 January 2019

IssaRice at 17:33, 6 January 2019

IssaRice at 17:33, 6 January 2019

IssaRice at 17:31, 6 January 2019

IssaRice at 17:30, 6 January 2019

IssaRice: Created page with "Let's take the case where V = \mathbf R^n and the inner product is the usual dot product. What does the Riesz representation theorem say in this case? It says tha..."

IssaRice: Created page with "Let's take the case where $V = \mathbf R^n$ and the inner product is the usual dot product. What does the Riesz representation theorem say in this case? It says tha..."