1. The problem statement, all variables and given/known data i) Find the Image and Kernel of A = (2,1)(-4,-2) (where each bracket is a row). ii) Calculate A^{2} and use i) to explain your result. 2. Relevant equations None 3. The attempt at a solution So I can do everything up to the very last bit (i think anyway). i) The Kernel = (1,-2) = Image. ii) A^{2} = 0 but this is where I don't know what to say. How do I use part i) to explain the 0 matrix found?
For i, you have the basic idea, but it needs some refinement. The kernel is not just one vector; there are an infinite number of vectors in the kernel of A. Likewise, the range is not just one vector either. Geometrically, A maps any vector along the line 2x + y = 0 to the zero vector. A maps any vector x not along the the line 2x + y = 0 to a vector along this line. IOW, if x is not in the kernel of A, A projects it onto this line. For ii, since A^{2}x = A(Ax)think about what A does to a vector x, and then think about what A does to a vector Ax.
OK i think i understand what you mean about the kernel. so applying to to another question, if I have the matrix: A=[{1,0,-2},{2,2,0},{0,3,6}] and I wanted to find the kernel, I'd reduce it down to: A=[{1,0,-2},{0,1,2},{0,0,0}] and thus the kernel is: Ker[A] = f[1,-1,2] where f is any number. Is that about right?
I think you got the 1's and 2's switched, i.e. [tex]\textrm{Ker}[A] = \{\vec{x}\in \Re^3\,|\, \vec{x} = f(2,-2,1), f \in \Re\}[/tex]
The point here is that the image is the kernel! For any vector v, Av is in the kernel of A so A(Av)= 0.