1. ## Linear Algebra Question

Hi everyone,

I cannot figure out this simply worded question. It asks if there exists a Linear mapping L: R4 -> R4 in which the kernel of L is equal to the Range of L. (Ker (L) = Range (L))

I have tried many possibilities, but am unable to find anything that works.

Some insight would be much Appreciated. Thanks!

2. ## Re: Linear Algebra Question

If this problem talked about "$R^3\to R^3$" (or, more generally, $R^n\to R^n$ with n odd) the answer would be, very quickly "no" because of the "rank-nullity" theorem- the rank and nullity of a linear operator must add up to the dimension of the range space. If Ker(L)= Range(L), they would have to have the same dimension and so the sum of dimensions would be an even number, not odd.

That also tells us that, in this case, with dimension 4, we must look for a linear transformation whose kernel has dimension 2, and whose range has dimension 2- so that 2+ 2= 4. What would be the simplest mapping of <a, b, c, d> into a four-vector with kernel of dimension 2?

3. ## Re: Linear Algebra Question

THe simplest mapping would be just taking <a,b,c,d> and outputting only <a,b, 0,0>? BUt that would make the basis for the kernel <(0,0,1,0), (0, 0, 0, 1)>, and the basis for the range < (1,0,0,0), (0,1,0,0)>

4. ## Re: Linear Algebra Question

Okay, now "tweak" it a bit. What about moving a and b?