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:


<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 = \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>

Latest revision as of 14:03, 18 June 2018

Let be a symmetric, positive definite, non-singular square matrix. Then we have the following:

.

We have and since is symmetric.

From the above, we see that