When Do Complex Matrices Yield Infinite Solutions?

[transgalactic]

i remmember my teacher mentioned that when we have a matrix with a parameter
and i am asked to find for which values there are infinite number of answers
no answer or a single answer

she said that if the space is defined as C then we must carry on
and split into complex numbers

i can't find a question that makes that

can you give me an example to work on??

 

[quantumdude]

I can't figure out what you're asking.

 

[HallsofIvy]

i remmember my teacher mentioned that when we have a matrix with a parameter
and i am asked to find for which values there are infinite number of answers
no answer or a single answer

she said that if the space is defined as C then we must carry on
and split into complex numbers

i can't find a question that makes that

can you give me an example to work on??

I doubt your teacher said that, exactly! I suspect that your teacher was talking about a matrix equation, like Ax= b where A depends on a parameter. If the determinant of A is non-zero, remember, A has an inverse so multiplying by that inverse on both sides of the equation gives A^-1Ax= x= A^-1b, a single solution.

If the determinant of A is 0, however, either of two things can happen. If the determinant of A is 0, then the kernel of A has dimension greater than 0 and so the dimension of the image A is less than the dimension of the space. If b happens to lie in the image of A, then there will be an infinite number of solutions. If b is not in the image of A, then there is no solution.

If the vector space is over the complex numbers (not "defined as C") then the entries in the vectors b and x, as well as A can be complex numbers. She may have said that you can split them into real and imaginary parts but surely not "split into complex numbers".