# Determine if a matric is diagonalizable and diagonlize it

1. May 7, 2012

### DODGEVIPER13

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?

2. May 7, 2012

### DODGEVIPER13

The forum gaarbled up what I put sorry but I dont know how to use brackets to tell the format to be correct

3. May 7, 2012

### DODGEVIPER13

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. May 7, 2012

### sharks

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.

5. May 7, 2012

### Staff: Mentor

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.

6. May 7, 2012

### Joffan

$$\left( \begin{matrix} 1 & 4 \\ 1 & -2 \end{matrix} \right)$$

7. May 8, 2012

### DODGEVIPER13

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. May 9, 2012

### DODGEVIPER13

I have attached an image of my work it should be much clearer now sorry for the scratch out on one part.

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9. May 9, 2012

### HallsofIvy

Staff Emeritus
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.

10. May 9, 2012

### DODGEVIPER13

Ok thanks man I kinda figured it was right but the answer in the back scared me a bit