homomorphisms and basis

hjortur · #1 $Old$ November 3rd, 2009, 02:40 PM

Let V be 3-dimensional, and f be a homomorphism on V such that $f^3=0$
but $f^2\ne0$ . Let $\vec b \in V$ such that $f^2(\vec b)\ne0$ .

Show that $<\vec b, f(\vec b), f^2(\vec b)>$ is a basis for V. Also find the matrix representing f with respect to this basis.

Now the first question is what does $f^3=0$ mean?
Is it that $f(f(f(x)))=0$ for all x (constant 0)? If so is it not correct to write $f^3=\vec 0$ ?
I have no idea how to start this one.

Bruno J. · #2 $Old$ November 3rd, 2009, 03:49 PM

Hello!

You have the right interpretation for the meaning of $f^3$ . Perhaps an arrow above 0 would have been appropriate but it's not absolutely necessary.

To show that $\{b, f(b), f^2(b)\}$ is a basis, we need to show that they are linearly independent. The statement will follow since the dimension of the vector space is 3. Suppose there are constants, not all $0$ , such that $c_1b+c_2f(b)+c_3f^2(b)=0$ . First apply $f^2$ to both sides of this equation to show that we must have $c_1=0$ ; next apply $f$ to both sides to show that $c_2=0$ also. What do you conclude?

By the way, to write $\{\}$ in LaTeX, use \{ and \}.

hjortur · #3 $Old$ November 3rd, 2009, 04:54 PM

Would I be correct then when I find the matrix representation of f with respect to that basis like this?

Determine the action of the map on the basis
$\vec b \mapsto f(b)$ , $f(\vec b) \mapsto f^2(\vec b)$ and $f^2(\vec b)\mapsto \vec 0$ . Hence the matrix is

$f: \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$

Is this correct?

Bruno J. · #4 $Old$ November 3rd, 2009, 05:05 PM

Yup that is good!