Linear Algebra/ powers of a matrix

Wildcat
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Homework Statement



Let A = row 1 [3/2 ½] row 2 [-½ ½]
Find A^1000 without using a calculator or computer.

Homework Equations





The Attempt at a Solution


I found the Jordan canonical form of A to be [1 ½] [0 1] and P=[1 1] [-1 0] and P^-1=[0 -1]
[1 1]
A^1000= (PJP^-1)^1000 so all the PP^-1 cancel and leave you with PJ^1000P^-1 then raise the terms of J to the 1000th power. I tried it for the 2nd power and it did not work so I'm assuming it won't work for the 1000th. Does anyone see an error?
 
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It worked when I tried calculating A2 using your matrices. Maybe you simply made an arithmetic error when you multiplied it all out.
 
vela said:
It worked when I tried calculating A2 using your matrices. Maybe you simply made an arithmetic error when you multiplied it all out.

OK when I square A I get [2 1] [-1 0] then when I multiply the other form, I'm squaring all the entries of the Jordan matrix which gives me [1 ¼] [0 1] then when I multiply
PJ²P^-1 I get [5/4 ¼] [-¼ ¾] which does not equal A².
Can you tell what I am doing wrong??
 
Wildcat said:
OK when I square A I get [2 1] [-1 0]
That's correct.
then when I multiply the other form, I'm squaring all the entries of the Jordan matrix which gives me [1 ¼] [0 1]
That's not how you square a matrix. Square it the same way you squared A, using regular matrix multiplication.
then when I multiply
PJ²P^-1 I get [5/4 ¼] [-¼ ¾] which does not equal A².
Can you tell what I am doing wrong??
 
vela said:
That's correct.

That's not how you square a matrix. Square it the same way you squared A, using regular matrix multiplication.

OK then I get J² to be [1 ½] [0 1] then PJ²P^-1 is the original matrix A when I do the multiplication. What am I doing?
 
Can you get a decomposition A = XYX-1 where Y is a diagonal matrix? The power of a diagonal matrix is very easy!
 
You said you had

J=\begin{bmatrix} 1 & 1/2 \\ 0 & 1 \end{bmatrix}

so you should get

J^2=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}
 
Outlined said:
Can you get a decomposition A = XYX-1 where Y is a diagonal matrix? The power of a diagonal matrix is very easy!

That is what I thought that I had done, but now I'm not sure. Does Jordan Canonical form mean it is a diagonal matrix with zeros everywhere except the diagonal?
I had PJP^-1 as the matrices listed with J being the Jordan Canonical form. Is my J not Jordan Canonical Form?
 
Wildcat said:
That is what I thought that I had done, but now I'm not sure. Does Jordan Canonical form mean it is a diagonal matrix with zeros everywhere except the diagonal?
I had PJP^-1 as the matrices listed with J being the Jordan Canonical form. Is my J not Jordan Canonical Form?

OH MY GOSH! I was making a stupid mistake! Ugh! I'm so sorry I took up your time! But Thank you soooooo much!
 
  • #10
Wildcat said:
That is what I thought that I had done, but now I'm not sure. Does Jordan Canonical form mean it is a diagonal matrix with zeros everywhere except the diagonal?
I had PJP^-1 as the matrices listed with J being the Jordan Canonical form. Is my J not Jordan Canonical Form?
Jordan canonical form isn't a diagonal matrix. You can use it when you can't diagonalize the matrix, like in this case. If you try to diagonalize A, you'll find you can't because you can get only one eigenvector.
 
  • #11
Wildcat said:
OH MY GOSH! I was making a stupid mistake! Ugh! I'm so sorry I took up your time! But Thank you soooooo much!
Don't you hate it when that happens? :wink: Still, it's a relief when you finally realize it does actually make sense.
 
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