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Determine if a matric is diagonalizable and diagonlize it

by DODGEVIPER13
Tags: determine, diagonalizable, diagonlize, matric
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DODGEVIPER13
#1
May7-12, 06:27 PM
P: 663
1. The problem statement, all variables and given/known data
Determine if the matrix is diagonalizable. If so, find matrices S and (symbol that looks simlar to A) such that the given matrix equals S(weird symbol)S^(-1).


2. Relevant equations
C1X1+C2X2+....CnXn = 0


3. The attempt at a solution
So what I did was take the matrix | 1 4 | and transform it to | λ-1 -4|
| 1 -2 | | -1 λ+2|

Then I said (λ-1)(λ+2)-4 which equals λ^2+λ-6 I found that the eigenvalues were -3 and 2 whic I then took and plugged -3 into the matrix equation that I transformed with the lamdas. Then I did this | -4 -4 | | x1 | |0|
| -1 -1 | | x2 | = |0|
which gave me two equations -4x1-4x2 = 0 and -x1-x2 = 0 but this is where im lost which one should I assign an abritray variavle to x1 or x2 I get that it is only to none pivot numbers and the second row are constants so you cant use those but I have seen in some cases where that is not true so im confused? Anyways solve that and I get v1 = |1 |
|-1|
and then I use the same procedure with the other eigen val and get v2 = |4|
|1|
I put those together and achieve | 1 4 |
|-1 1 |
this is incorrect however it is supposed to be | 4 1 |
| 1 -1 |
why is this and how do I know which eigenvalue gives me which eigenvector?
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DODGEVIPER13
#2
May7-12, 06:28 PM
P: 663
The forum gaarbled up what I put sorry but I dont know how to use brackets to tell the format to be correct
DODGEVIPER13
#3
May7-12, 06:30 PM
P: 663
Just slide the stuff to the left to the right so that it fits under where the rest of matrix is then you can read it again sorry

sharks
#4
May7-12, 06:35 PM
PF Gold
sharks's Avatar
P: 836
Determine if a matric is diagonalizable and diagonlize it

Quote Quote by DODGEVIPER13 View Post
Determine if the matrix is diagonalizable. If so, find matrices S and (symbol that looks simlar to A) such that the given matrix equals S(weird symbol)S^(-1).
This "weird symbol" is it this?: [itex]\wedge[/itex] - it's called vec

By the way, where is the full matrix? You should always post the entire question. If you don't know how to type it in LaTeX, maybe just do a screenshot or take a clear picture, and attach it to your post.
Mark44
#5
May7-12, 07:21 PM
Mentor
P: 21,216
Quote Quote by sharks View Post
This "weird symbol" is it this?: [itex]\wedge[/itex] - it's called vec
I don't think so. It's probably this symbol - ##\Lambda## - Uppercase Lambda. That makes more sense in this problem, since ##\Lambda## is the diagonal matrix, and it's entries on the main diagonal are the eigenvalues, λ1 and λ2.
Joffan
#6
May7-12, 07:26 PM
P: 361
Is this your original matrix?
$$
\left( \begin{matrix}
1 & 4 \\
1 & -2
\end{matrix} \right)
$$
DODGEVIPER13
#7
May8-12, 12:02 AM
P: 663
Thank you Mark44 for clarification on the symbol and thankyou sharks for now I will take a pic and submit it. And Joffan yes that is the matrix
DODGEVIPER13
#8
May9-12, 02:26 AM
P: 663
I have attached an image of my work it should be much clearer now sorry for the scratch out on one part.
Attached Thumbnails
EPSON001.jpg  
HallsofIvy
#9
May9-12, 07:24 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,317
Okay, you have eigenvector [1, -1] corresponding to eigenvalue -3 and eigenvector [4, 1] corresponding to eigevalue 1. Those are correct.

You then put them together to form matrix "P" (you have it labeled [itex]\vec{V}[/itex] which is incorrect- this is a matrix, not a vector.)
[tex]\begin{bmatrix}1 & 4 \\ -1 & 1\end{bmatrix}[/tex] and declare that it this is incorrect. Why? You need to understand that there are, in fact, an infinite number of different matrices, P, so that, for this matrix A, [itex]P^{-1}AP[/itex] is diagonal. The matrix which you say is incorrect is perfectly correct. Using it as P will give you the diagonal matrix
[tex]\begin{bmatrix}-3 & 0 \\ 0 & 1\end{bmatrix}[/tex]

Using the matrix that you say is correct,
[tex]\begin{bmatrix}4 & 1 \\ 1 & -1\end{bmatrix}[/tex]
has the two columns (eigenvectors) reversed and so gives
[tex]\begin{bmatrix}1 & 0 \\ 0 & -3\end{bmatrix}[/tex]
which is also a diagonal matrix, just the eigenvalues in different places.
DODGEVIPER13
#10
May9-12, 01:08 PM
P: 663
Ok thanks man I kinda figured it was right but the answer in the back scared me a bit


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