Solution space of AX = 0 will be equal to the the kernel of the mapping represented by A
If you solve the equation we get the solution space is of dimension 1 with basis = [1 -2 1 0]
Find vectors that span the kernel of matrix A.
A = [1 2 3 4 ; 0 1 2 3 ; 0 0 0 1] where ; is a new row.
I found that rref(A) = [1 0 -1 0 ; 0 1 2 0 ; 0 0 0 1] where ; is a new row.
Can anyone tell me where I go from here? Thanks
Actually you can work from either A itself of rref(A). The "kernel" of a matrix is the subspace of all vectors that it maps to 0. So if is a vector in the kernel, we must have .
That is the same as the system of equation a+ 2b+ 3c+ 4d= 0, b+ 2c+ 3d= 0, d= 0. You could, of course, reduce those to a- c= 0, b+ 2c= 0, d= 0, the row reduced form. From the first equation, a= c. From the second b= -2c. c can by any number, d must be 0. So any vector satisfying those equations, any vector in the kernel of A, must be of the form .
A basis for the kernel consists of that single vector , as aman_cc says.