# Thread: Kernel and image of linear operator

1. ## Kernel and image of linear operator

Hi, I'm Kalish. I have a problem and a proposed solution. Please point out any errors!

Problem Statement: Determine the dimensions of the kernel and the image of the linear operator T on the space R^n defined by T(x_1,...,x_n)^t=(x_1+x_n,x_2+x_{n-1},...,x_n+x_1)^t.

My attempt: The dimension of the kernel is the number of vectors in the basis for [0,0,...,0]^t, or 0, because the dimension is 0. The dimension of the image = 1, because by the dimension formula, dim(image(T))=dim(V)-dim(ker T)=1-0=1.

This seems too easy to be true. Any thoughts? Thanks.

2. ## Re: Kernel and image of linear operator

You need to figure out where T maps the basis vectors of R^n. Suppose, for example, that n = 4. The first basis vector has coordinates (1, 0, 0, 0), and T(1, 0, 0, 0) = (1, 0, 0, 1). Next, T(0, 1, 0, 0) = (0, 1, 1, 0). Similarly, T(0, 0, 1, 0) = (0, 1, 1, 0) and T(0, 0, 0, 1) = (1, 0, 0, 1). Now, the dimension of the image is the dimension of span((1, 0, 0, 1), (0, 1, 1, 0), (0, 1, 1, 0), (1, 0, 0, 1)) = 2, so dim ker(T) = 4 - 2 = 2. Check also what happens when n = 5.

3. ## Re: Kernel and image of linear operator

Ok, I am getting that for the general case, if n is even, then dim(image(T)) = n/2, and if n is odd, then dim(image(T)) = (n+1)/2. Thus, the dimension of image(T) is the ceiling function of n/2, and the dimension of the kernel is dimV - dim(image(T)) = n-ceiling(n/2) = floor(n/2). Is this right?

4. ## Re: Kernel and image of linear operator

Well, as far as I can tell your reasoning appears to be "0 and 1 are the simplest numbers I know so I will just use them!" Yes, that is far too easy- and wrong. If the kernel has dimension 0, then the image has dimension n- 0= n, not 1. We are after all, dealing with $\displaystyle V=R^n$, of dimension n, NOT 1. Now, what reason do you have to say that the dimension of the kernel is 0?

If you don't understand the statement of the general theorem look at simple cases. (You can't prove a general theorem by looking at special cases but you can get an idea of what the theorem means and how to approach it.)

For example, if n= 2, we have $\displaystyle T(x_1, x_2)= (x_1+ x_2, x_2+ x_1)= (x_1+ x_2, x_2+ x_1)= (0, 0)$ so we have $\displaystyle x_1+ x_2= 0$ and $\displaystyle x_2+ x_1= 0$ which both reduce to $\displaystyle x_2= -x_1$. The kernel consists of all vectors of the form $\displaystyle (x_1, -x_1)= x_1(1, -1)$ which is spanned by {(1, -1)} and so has dimension 1, NOT 0. And, of course, the image has dimension 2-1= 1.

If n= 3, we have $\displaystyle T(x_1, x_2, x_3)= (x_1+ x_3, x_2+ x_2, x_3+ x_1)= (0, 0, 0)$ so that $\displaystyle x_1+ x_3= 0$ and $\displaystyle 2x_2= 0$. Obviously $\displaystyle x_2= 0$ and $\displaystyle x_3= -x_1$. The kernel consists of all vectors of the form $\displaystyle (x_1, 0, -x_1)= x_1(1, 0, -1)$ and so has dimension 1 again. But the image has dimension 3- 1= 2.

If n= 4, we have $\displaystyle T(x_1, x_2, x_3, x_4)= (x_1+ x_4, x_2+ x_3, x_3+ x_2, x_4+ x_1)= (0, 0, 0, 0)$ so that $\displaystyle x_1+ x_4= 0$ and $\displaystyle x_2+ x_3= 0$ so that [tex]x_4= -x_1/[tex] and $\displaystyle x_3=-x_2$. The kernel consists of all vectors of the form$\displaystyle (x_1, x_2, -x_2, -x_1)= (x_, 0, 0, -x_1)+ (0, x_2, -x_2, 0)= x_1(1, 0, 0, -1)+ x_2(0, 1, -1 0)$ and so has dimension 2. The image has dimension 4- 2= 2.

Now do you think you could guess what the dimension of the kernel and image would be if n= 5 or 6 or would you like to do the same as above? Can you guess the formulas for dimension of kernel and image in $\displaystyle R^n$? (You will have different formulas for n odd and even.)

5. ## Re: Kernel and image of linear operator

Thank you so much!! I understand perfectly well now. I was so hazy before this.