# Thread: Dimension of the kernel

1. ## Dimension of the kernel

In the past, I've had hardly any answers to questions I've asked here, so I really hope someone is able to take a stab at this one!

Here's the question as presented on my homework (sorry, I'm not sure how to get all the symbols as they're supposed to appear):

Show that if L: V_F -> W_F is a linear transformation and if Ker L has dimension 2, then if M is a matrix representing L (for some basis B of V_F and some basis B' of W_F) then also det(M-[lambda]I) has the form [lambda]^2(p(x)) where p(x) is a polynomial.

I think I understand what the kernel and dimension are, but I have no clue how to find either one. I'm also not sure what bearing this has on the characteristic polynomial (which is what I'm pretty sure the second half of the problem is). Long story short, my teacher doesn't teach and I'm having to learn all this stuff on my own. Anybody care to help?

2. Originally Posted by scosgurl
In the past, I've had hardly any answers to questions I've asked here, so I really hope someone is able to take a stab at this one!

Here's the question as presented on my homework (sorry, I'm not sure how to get all the symbols as they're supposed to appear):

Show that if L: V_F -> W_F is a linear transformation and if Ker L has dimension 2, then if M is a matrix representing L (for some basis B of V_F and some basis B' of W_F) then also det(M-[lambda]I) has the form [lambda]^2(p(x)) where p(x) is a polynomial.

I think I understand what the kernel and dimension are, but I have no clue how to find either one. I'm also not sure what bearing this has on the characteristic polynomial (which is what I'm pretty sure the second half of the problem is). Long story short, my teacher doesn't teach and I'm having to learn all this stuff on my own. Anybody care to help?
Since $\ker (L)$ has dimension two it means there exists two linearly independent vectors $v_1,v_2\in V_F$ so that $L(v_1) = 0, L(v_2) = 0$. Thus, we see that $0$ is an eigenvalue for $v_1$ and $0$ is an eigenvalue for $v_2$. This means if we write $k_1,k_2,...,k_n$ as a complete list of repeated eigenvalues for $L$ then they are the roots of the charachteristic polynomial for $L$. Let $f(x)$ be the charachteristic polynomial for $L$ then $f(x) = (x-k_1)(x-k_2)...(x-k_n)$. However, two of these eigenvalues, say $k_1,k_2$ are $0$, and so $f(x) = x^2p(x)$ where $p(x) = (x-k_3)...(x-k_n)$.