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Determine if a matric is diagonalizable and diagonlize it 
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#1
May712, 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  λ1 4  1 2   1 λ+2 Then I said (λ1)(λ+2)4 which equals λ^2+λ6 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 4x14x2 = 0 and x1x2 = 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? 


#2
May712, 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



#3
May712, 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



#4
May712, 06:35 PM

PF Gold
P: 836

Determine if a matric is diagonalizable and diagonlize it
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. 


#5
May712, 07:21 PM

Mentor
P: 21,258




#6
May712, 07:26 PM

P: 361

Is this your original matrix?
$$ \left( \begin{matrix} 1 & 4 \\ 1 & 2 \end{matrix} \right) $$ 


#7
May812, 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



#8
May912, 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.



#9
May912, 07:24 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,497

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. 


#10
May912, 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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