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

[ysebastien]

Homework Statement

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

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,

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

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.

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

<br /> P=\left[\begin{array}{cccc}<br /> 1 &amp; 0 &amp; 1 &amp; -1\\<br /> 0 &amp; 0 &amp; -1 &amp; 1\\<br /> -1 &amp; -1 &amp; 0 &amp; 0\\<br /> 1 &amp; 1 &amp; 1 &amp; 0<br /> \end{array}\right]<br />
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

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

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,

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

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.

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

<br /> P=\left[\begin{array}{cccc}<br /> 1 &amp; 0 &amp; 1 &amp; -1\\<br /> 0 &amp; 0 &amp; -1 &amp; 1\\<br /> -1 &amp; -1 &amp; 0 &amp; 0\\<br /> 1 &amp; 1 &amp; 1 &amp; 0<br /> \end{array}\right]<br />
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.