# Determine if a matric is diagonalizable and diagonlize it

by DODGEVIPER13
Tags: determine, diagonalizable, diagonlize, matric
 P: 549 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?
 P: 549 The forum gaarbled up what I put sorry but I dont know how to use brackets to tell the format to be correct
 P: 549 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
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P: 836

## Determine if a matric is diagonalizable and diagonlize it

 Quote by DODGEVIPER13 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?: $\wedge$ - 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.
Mentor
P: 19,798
 Quote by sharks This "weird symbol" is it this?: $\wedge$ - 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.
 P: 328 Is this your original matrix? $$\left( \begin{matrix} 1 & 4 \\ 1 & -2 \end{matrix} \right)$$
 P: 549 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
 P: 549 I have attached an image of my work it should be much clearer now sorry for the scratch out on one part. Attached Thumbnails
 PF Patron Sci Advisor Thanks Emeritus P: 38,429 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 $\vec{V}$ which is incorrect- this is a matrix, not a vector.) $$\begin{bmatrix}1 & 4 \\ -1 & 1\end{bmatrix}$$ 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, $P^{-1}AP$ is diagonal. The matrix which you say is incorrect is perfectly correct. Using it as P will give you the diagonal matrix $$\begin{bmatrix}-3 & 0 \\ 0 & 1\end{bmatrix}$$ Using the matrix that you say is correct, $$\begin{bmatrix}4 & 1 \\ 1 & -1\end{bmatrix}$$ has the two columns (eigenvectors) reversed and so gives $$\begin{bmatrix}1 & 0 \\ 0 & -3\end{bmatrix}$$ which is also a diagonal matrix, just the eigenvalues in different places.
 P: 549 Ok thanks man I kinda figured it was right but the answer in the back scared me a bit

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