Jordan Normal Form of Matrices

victoria13
Messages
7
Reaction score
0

Homework Statement



Find the Jordan Normal Form of

3,0,8
3,-1,6
-2,0,-5


Homework Equations





The Attempt at a Solution



I got the eigen values as lambda=-1

found 2 eigen vectors:

-2
0
1

and

0
1
0

now when i try and do (A+I)^2 I get a zero matrix so cannot seem to find the last column of the P matrix to then calculate P-1AP=J
 
Physics news on Phys.org
anyone?
 
If (A+I)^2=0 then ANY vector satisfies (A+I)^2. What you really want to get to the Jordan form is three vectors satisfying (A+I)v1=0, (A+I)v2=0, (A+I)v3=v2. Start by picking v3 outside of the eigenspace spanned by {v1,v2}. That determines the eigenvector v2. Now pick v1 to be any other linearly independent eigenvector.
 
Dick said:
If (A+I)^2=0 then ANY vector satisfies (A+I)^2. What you really want to get to the Jordan form is three vectors satisfying (A+I)v1=0, (A+I)v2=0, (A+I)v3=v2. Start by picking v3 outside of the eigenspace spanned by {v1,v2}. That determines the eigenvector v2. Now pick v1 to be any other linearly independent eigenvector.

ok I am not quite sure what you mean. when I set (A+I)v2=v3

using my v2= (2 0 -1)

i get a matrix that doesn't multiply to get a jordan form...
 
(A+I)v2=0 if v2=(2,0,-1). You have a two dimensional eigenspace spanned by {(-2,0,1),(0,1,0)}. You want to pick three vectors for the basis that satisfy the relations I gave in the last post. Since you want (A+I)v3=v2, you want to pick a v3 OUTSIDE of that eigenspace. Now define v2=(A+I)v3. Since (A+I)^2=0 that will automatically be in the eigenspace. Now pick the final v1 to be another linearly independent vector in the eigenspace. That's a Jordan basis. Just start doing it. You'll see.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top