An observation... - Electronic Circuits Projects Diagrams Free

_3iMaJ · 30th November 2005, 11:09 PM

Something you take as intuitivly obvious really isn't that obvious at all...

A friend of mine and I were talking about solving for mesh currents (in a completely resistive circuit) VIA using a matrix. Now its routinely taken for granted that this matrix is invertable (IE non-singular). However, this is not intuitivly obvious. I was wondering if someone could show for the NxN case that the matrix MUST be non-singular, also would it then be positive definite? I believe that it is not only non-singular but it must also be positive definite...

I've shown for the 2x2 case that it MUST be non-singular, however I haven't found a way to show it generally. I'm still thinking about it though... :?

Optikon · 1st December 2005, 01:25 PM

Quote:

Originally Posted by _3iMaJ

Something you take as intuitivly obvious really isn't that obvious at all...

A friend of mine and I were talking about solving for mesh currents (in a completely resistive circuit) VIA using a matrix. Now its routinely taken for granted that this matrix is invertable (IE non-singular). However, this is not intuitivly obvious. I was wondering if someone could show for the NxN case that the matrix MUST be non-singular, also would it then be positive definite? I believe that it is not only non-singular but it must also be positive definite...

I've shown for the 2x2 case that it MUST be non-singular, however I haven't found a way to show it generally. I'm still thinking about it though... :?

If you've shown it for the 2x2 case, use method of induction to prove NxN case.

_3iMaJ · 1st December 2005, 01:28 PM

I considered that, however the determinant for a 2x2 matrix is very easy to solve for, anything larger than that and it gets difficult quickly.

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30th November 2005, 11:09 PM	(permalink)
_3iMaJ	An observation... Something you take as intuitivly obvious really isn't that obvious at all... A friend of mine and I were talking about solving for mesh currents (in a completely resistive circuit) VIA using a matrix. Now its routinely taken for granted that this matrix is invertable (IE non-singular). However, this is not intuitivly obvious. I was wondering if someone could show for the NxN case that the matrix MUST be non-singular, also would it then be positive definite? I believe that it is not only non-singular but it must also be positive definite... I've shown for the 2x2 case that it MUST be non-singular, however I haven't found a way to show it generally. I'm still thinking about it though... :?

1st December 2005, 01:28 PM	(permalink)
_3iMaJ	I considered that, however the determinant for a 2x2 matrix is very easy to solve for, anything larger than that and it gets difficult quickly.