Notes for AKT-140124/0:23:30: Difference between revisions
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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: |
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<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>. |
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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. |
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Revision as of 11:17, 14 June 2018
Let [math]\displaystyle{ \Lambda }[/math] be a symmetric, positive definite, non-singular square matrix. Then we have the following:
[math]\displaystyle{ \langle x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)\rangle = \lt x,\Lambda x\gt - \lt x, y\gt -\lt \Lambda^{-1}y, \Lambda x\gt + \lt \Lambda^{-1}y,y\gt }[/math].
We have [math]\displaystyle{ \lt \Lambda^{-1}y, \Lambda x\gt = \lt x,y\gt }[/math] and [math]\displaystyle{ \lt \Lambda^{-1}y,y\gt = \lt y,\Lambda^{-1}y\gt }[/math] since [math]\displaystyle{ \Lambda }[/math] is symmetric.
From the above, we see that [math]\displaystyle{ -\frac12 \lt x - \Lambda^{-1} y, \Lambda(x - \Lambda^{-1}y)\gt + \frac12\lt y,\Lambda^{-1}y\gt = -\frac12\lt x,\Lambda x\gt + \lt x, y\gt }[/math]
