Finding an invertible matrix P such that A=PJP^-1

[ysebastien]

Homework Statement

$$A=\left[\begin{array}{cccc}<br /> 5 & 4 & 2 & 1\\<br /> 0 & 1 & -1 & -1\\<br /> -1 & -1 & 3 & 0\\<br /> 1 & 1 & -1 & 2<br /> \end{array}\right]$$

1. Find the Jordan form of A
2. Find a matrix P such that A = PJP^-1

Homework Equations

The Attempt at a Solution

I've already found, and verified the Jordan form it is,

$$J=\left[\begin{array}{cccc}<br /> 4 & 1 & 0 & 0\\<br /> 0 & 4 & 0 & 0\\<br /> 0 & 0 & 2 & 0\\<br /> 0 & 0 & 0 & 1<br /> \end{array}\right]$$

from the characteristic polynomial $q(x) = (x-4)^{2}(x-2)(x-1)$.

To find P, I found a basis vector for each of null spaces of each eigenvalue.
$$\ker{(A-4I)}$$
Basis: $(1,0,-1,1)$

$$\ker{(A-4I)^{2}}$$
Basis: $(0,0,-1,1), (1,0,0,0)$

$$\ker{(A-2I)}$$
Basis: $(1,-1,0,1)$

$$\ker{A-I)}$$
Basis: $(-1,1,0,0)$

(these have all been verified with Mathematica)

Now I'm supposed to construct the columns of P by picking a set of 4 linearly independent vectors, which they all are. Here is my struggle. In choosing between the two basis vectors for $\ker{(A-4I)^{2}}$, choosing $(1,0,0,0)$ works fine i.e.

$$P=\left[\begin{array}{cccc}<br /> 1 & 1 & 1 & -1\\<br /> 0 & 0 & -1 & 1\\<br /> -1 & 0 & 0 & 0\\<br /> 1 & 0 & 1 & 0<br /> \end{array}\right]$$
but if I choose the other vector,

$$P=\left[\begin{array}{cccc}<br /> 1 & 0 & 1 & -1\\<br /> 0 & 0 & -1 & 1\\<br /> -1 & -1 & 0 & 0\\<br /> 1 & 1 & 1 & 0<br /> \end{array}\right]$$
and P doesn't satisfy $A=PJP^{-1}$. Oddly, if I take $(0,0,1,-1)$ it does work. Does anyone see if my logic is going awry somewhere? (0,0,1,-1) is just the negative of (0,0,-1,1) so shouldn't they in fact both work, satisfying A=PJP^-1?
Sorry for the long post.

 

[Ray Vickson]

Homework Statement

$$A=\left[\begin{array}{cccc}<br /> 5 & 4 & 2 & 1\\<br /> 0 & 1 & -1 & -1\\<br /> -1 & -1 & 3 & 0\\<br /> 1 & 1 & -1 & 2<br /> \end{array}\right]$$

1. Find the Jordan form of A
2. Find a matrix P such that A = PJP^-1

Homework Equations

The Attempt at a Solution

I've already found, and verified the Jordan form it is,

$$J=\left[\begin{array}{cccc}<br /> 4 & 1 & 0 & 0\\<br /> 0 & 4 & 0 & 0\\<br /> 0 & 0 & 2 & 0\\<br /> 0 & 0 & 0 & 1<br /> \end{array}\right]$$

from the characteristic polynomial $q(x) = (x-4)^{2}(x-2)(x-1)$.

To find P, I found a basis vector for each of null spaces of each eigenvalue.
$$\ker{(A-4I)}$$
Basis: $(1,0,-1,1)$

$$\ker{(A-4I)^{2}}$$
Basis: $(0,0,-1,1), (1,0,0,0)$

$$\ker{(A-2I)}$$
Basis: $(1,-1,0,1)$

$$\ker{A-I)}$$
Basis: $(-1,1,0,0)$

(these have all been verified with Mathematica)

Now I'm supposed to construct the columns of P by picking a set of 4 linearly independent vectors, which they all are. Here is my struggle. In choosing between the two basis vectors for $\ker{(A-4I)^{2}}$, choosing $(1,0,0,0)$ works fine i.e.

$$P=\left[\begin{array}{cccc}<br /> 1 & 1 & 1 & -1\\<br /> 0 & 0 & -1 & 1\\<br /> -1 & 0 & 0 & 0\\<br /> 1 & 0 & 1 & 0<br /> \end{array}\right]$$
but if I choose the other vector,

$$P=\left[\begin{array}{cccc}<br /> 1 & 0 & 1 & -1\\<br /> 0 & 0 & -1 & 1\\<br /> -1 & -1 & 0 & 0\\<br /> 1 & 1 & 1 & 0<br /> \end{array}\right]$$
and P doesn't satisfy $A=PJP^{-1}$. Oddly, if I take $(0,0,1,-1)$ it does work. Does anyone see if my logic is going awry somewhere? (0,0,1,-1) is just the negative of (0,0,-1,1) so shouldn't they in fact both work, satisfying A=PJP^-1?
Sorry for the long post.

If $(A-4I)^2 v = 0$ but $(A - 4I) v \neq 0$, then let $u = (A - 4I)v.$ These give $Av = u + 4v$ and $Au = 4u$. So, if you include u and v in a basis, they will give you the upper left Jordan block. The matrix P constructed from u, v and the other eigenvectors will do what you want.

 

[ysebastien]

Great, thanks.