Kernels and determinants of a matrix

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I read that an equation of the form Ax=0 has a solution iff the matrix A has non-trivial Kernel, which makes sense as if A had trivial kernel then x would be trivial as well, meaning that only the x={0} solution would exist, right?

Secondly, I read that in order for A to have a non-trivial kernel, we need detA=0. Why is this so?
 

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  • #2
jgens
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I read that an equation of the form Ax=0 has a solution iff the matrix A has non-trivial Kernel, which makes sense as if A had trivial kernel then x would be trivial as well, meaning that only the x={0} solution would exist, right?
This is more or less correct. I am not sure what the proper terminology is here, but it might be more proper to say something like:
An equation of the form Ax = 0 has a non-trivial solution if and only if the matrix A has non-trivial kernel.
I never really learned matrix algebra so maybe the x = 0 solution does not count or something, but it seems like you should add the non-trivial caveat for clarity.

Secondly, I read that in order for A to have a non-trivial kernel, we need detA=0. Why is this so?
There are several ways to look at this. Perhaps the simplest (although slightly unenlightening) way to see this is what follows: The determinant is multiplicative, so if A is invertible, then (det A)(det A-1) = 1 and this guarantees that neither of those guys can be zero. On the other hand if det A ≠ 0 then one can construct an inverse matrix. Just multiply the adjugate by (det A)-1 and you have your inverse.
 
  • #3
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The determinant is multiplicative, so if A is invertible, then (det A)(det A-1) = 1 and this guarantees that neither of those guys can be zero. On the other hand if det A ≠ 0 then one can construct an inverse matrix. Just multiply the adjugate by (det A)-1 and you have your inverse.
Although it makes sense, what you are saying sounds like det A≠0, whereas I was looking for det A=0.

How does your observation relate to my question about "requiring det A=0 in order to have a non-trivial kernel"?
 
  • #4
jgens
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Although it makes sense, what you are saying sounds like det A≠0, whereas I was looking for det A=0.
What happens in the det A = 0 case can be deduced from the det A ≠ 0 case. If you put some thought into, then I am sure you can figure it out.

How does your observation relate to my question about "requiring det A=0 in order to have a non-trivial kernel"?
It relates in a fairly obvious way.
 
  • #5
AlephZero
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You can interpret the product Ax as the sum of (the elements of x) times (the column vectors of A).

So, if Ax = 0 and x ≠ 0, the column vectors of A are linearly dependent, and therefore det A = 0.
 
  • #6
HallsofIvy
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I read that an equation of the form Ax=0 has a solution iff the matrix A has non-trivial Kernel, which makes sense as if A had trivial kernel then x would be trivial as well, meaning that only the x={0} solution would exist, right?
Actually this is NOT true. "Ax= 0" always has a solution: x= 0. It has non-trivial solution (a non-zero x such that Ax= 0) if and only if the kernel of A is non-trivial because the kernel of A is defined as the set such solutions. One is non-trivial if and only if the other is because they are, in fact, the same thing!

Secondly, I read that in order for A to have a non-trivial kernel, we need detA=0. Why is this so?
Matrix A has inverse if an only if it's determinant is non-0. If A has an inverse then we can multiply both sides of Ax= 0 by it to get [itex]A^{-1}Ax= A^{-1}0[/itex] of [itex]x= 0[/itex] so the kerne is trivial, consisting only of 0.
 

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