Notes for AKT-140124/0:23:30 - Revision history

Leo algknt at 19:03, 18 June 2018

2018-06-18T19:03:11Z

← Older revision		Revision as of 15:03, 18 June 2018
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	Let <math>\Lambda</math> be a symmetric, positive definite, non-singular square matrix. Then we have the following:		Let <math>\Lambda</math> be a symmetric, positive definite, non-singular square matrix. Then we have the following:

	<math> \langle x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)\rangle = <x,\Lambda x> - <x, y> -<\Lambda^{-1}y, \Lambda x> + <\Lambda^{-1}y,y> </math>.		<math> \langle x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)\rangle = \langle x,\Lambda x \rangle - \langle x, y \rangle - \langle \Lambda^{-1}y, \Lambda x \rangle + \langle \Lambda^{-1}y,y \rangle </math>.

	We have <math><\Lambda^{-1}y, \Lambda x> = <x,y> </math> and <math><\Lambda^{-1}y,y> = <y,\Lambda^{-1}y></math> since <math>\Lambda</math> is symmetric.		We have <math> \langle \Lambda^{-1}y, \Lambda x \rangle = \langle x,y\rangle </math> and <math> \langle \Lambda^{-1}y,y \rangle = \langle y,\Lambda^{-1}y \rangle</math> since <math>\Lambda</math> is symmetric.

	From the above, we see that <math>-\frac12 <x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)> + \frac12<y,\Lambda^{-1}y> = -\frac12<x,\Lambda x> + <x, y></math>		From the above, we see that <math>-\frac12 \langle x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y) \rangle + \frac12 \langle y,\Lambda^{-1}y \rangle = -\frac12 \langle x,\Lambda x \rangle + \langle x, y \rangle</math>

Drorbn at 16:17, 14 June 2018

2018-06-14T16:17:11Z

← Older revision		Revision as of 12:17, 14 June 2018
Line 1:		Line 1:
	Let <math>\Lambda</math> be a symmetric, positive definite, non-singular square matrix. Then we have the following:		Let <math>\Lambda</math> be a symmetric, positive definite, non-singular square matrix. Then we have the following:

	<math> <x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)> = <x,\Lambda x> - <x, y> -<\Lambda^{-1}y, \Lambda x> + <\Lambda^{-1}y,y> </math>.		<math> \langle x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)\rangle = <x,\Lambda x> - <x, y> -<\Lambda^{-1}y, \Lambda x> + <\Lambda^{-1}y,y> </math>.

	We have <math><\Lambda^{-1}y, \Lambda x> = <x,y> </math> and <math><\Lambda^{-1}y,y> = <y,\Lambda^{-1}y></math> since <math>\Lambda</math> is symmetric.		We have <math><\Lambda^{-1}y, \Lambda x> = <x,y> </math> and <math><\Lambda^{-1}y,y> = <y,\Lambda^{-1}y></math> since <math>\Lambda</math> is symmetric.

Leo algknt: Created page with "Let $\Lambda$ be a symmetric, positive definite, non-singular square matrix. Then we have the following: $= ..."$

2018-06-06T05:23:01Z

Created page with "Let <math>\Lambda</math> be a symmetric, positive definite, non-singular square matrix. Then we have the following: <math> <x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)> = ..."

New page

Let <math>\Lambda</math> be a symmetric, positive definite, non-singular square matrix. Then we have the following:

<math> <x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)> = <x,\Lambda x> - <x, y> -<\Lambda^{-1}y, \Lambda x> + <\Lambda^{-1}y,y> </math>.

We have <math><\Lambda^{-1}y, \Lambda x> = <x,y> </math> and <math><\Lambda^{-1}y,y> = <y,\Lambda^{-1}y></math> since <math>\Lambda</math> is symmetric.

From the above, we see that <math>-\frac12 <x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)> + \frac12<y,\Lambda^{-1}y> = -\frac12<x,\Lambda x> + <x, y></math>

Notes for AKT-140124/0:23:30 - Revision history

Leo algknt at 19:03, 18 June 2018

Drorbn at 16:17, 14 June 2018

Leo algknt: Created page with "Let \Lambda be a symmetric, positive definite, non-singular square matrix. Then we have the following: = ..."

Leo algknt: Created page with "Let $\Lambda$ be a symmetric, positive definite, non-singular square matrix. Then we have the following: $= ..."$