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An observation...

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_3iMaJ

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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

New Member
_3iMaJ said:
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

New Member
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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