Finding the eigenvector normalization for a 2×2 matrix

[g.lemaitre]

Homework Statement

Homework Equations

The Attempt at a Solution

I don't understand how $$u_1$$ = $$[1 -1]^T$$? By my reckoning $$u_1$$ =
$$\frac{v_1}{\parallel v_1 \parallel}$$
which is
$$\frac{-2}{\parallel -2+1 \parallel}, \frac{-2}{\parallel -2+1 \parallel}$$
which is
[-2, -2] not [1, -1]

 

[who_]

Could you clear up your tex? It is hard to understand what your actual question is. Use [ tex] and [ /tex] tags for tex. (Without the space!)

 

[SammyS]

Homework Statement

Homework Equations

The Attempt at a Solution

I don't understand how $u_1$ = $[1, -1]^T$?

By my reckoning $\displaystyle <br /> u_1=\frac{v_1}{\parallel v_1 \parallel}$

which is
$$\frac{-2}{\parallel -2+1 \parallel}, \frac{-2}{\parallel -2+1 \parallel}<br />$$
which is

[-2, -2] not [1, -1]

There.

I pretty much cleaned it up for you.

 

[who_]

I do not know what v_1 is, as it is not in the problem statement. But I can explain where u_1 comes from. The problem gives the matrix 2I-A. If you take u_1 and calculate (2I-A)u_1, then you get the 0 vector. Thus, u_1 lies in the null space of (2I - A). Since the null space of 2I - A has dimension one, that means that u_1 spans the null space of 2I - A, and so that 2Iu_1 = Au_1, meaning that u_1 is an eigenvector of A, and thus spans the eigenspace of A. Also, you do not have to normalize u_1, since the span of the normalized u_1 is the same as the span of u_1.

 

[g.lemaitre]

I do not know what v_1 is, as it is not in the problem statement. But I can explain where u_1 comes from. The problem gives the matrix 2I-A. If you take u_1 and calculate (2I-A)u_1, then you get the 0 vector. Thus, u_1 lies in the null space of (2I - A). Since the null space of 2I - A has dimension one, that means that u_1 spans the null space of 2I - A, and so that 2Iu_1 = Au_1, meaning that u_1 is an eigenvector of A, and thus spans the eigenspace of A. Also, you do not have to normalize u_1, since the span of the normalized u_1 is the same as the span of u_1.

Here is more of the question:

if u_1 spans the null space of 2I - A, then that does not explain how u_1 is $$[1 -1]^T$$

 

[g.lemaitre]

I think I understand what's going on now. Here's another example:

It looks like after you reduce things to row echelon form, you reduce it again so that one of the values = 1. here, x -3/4y = 0. then you put one of the variables on the other side of the equation, so that u_1 becomes $$[3/4 1 0]^T$$. But I would think it should be $$[1 3/4 0]$$

I'm not sure about u_2 however.

 

[vela]

You're looking for the solution to (2I-I)x=0, right? So solve
$$(2I-A)\vec{x} = \begin{pmatrix} -2 & -2 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = 0.$$ Forget about nullspaces and eigenspaces for the moment. You're just solving a system of equations.

 

[g.lemaitre]

I sort of understand what is going on.