Determine if a matric is diagonalizable and diagonlize it

[DODGEVIPER13]

Homework Statement

Determine if the matrix is diagonalizable. If so, find matrices S and (symbol that looks similar to A) such that the given matrix equals S(weird symbol)S^(-1).

Homework Equations

C1X1+C2X2+...CnXn = 0

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 I am 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 can't use those but I have seen in some cases where that is not true so I am 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?

 

[DODGEVIPER13]

The forum gaarbled up what I put sorry but I don't know how to use brackets to tell the format to be correct

 

[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

 

[DryRun]

Determine if the matrix is diagonalizable. If so, find matrices S and (symbol that looks similar 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.

 

[Mark44]

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.

 

[Joffan]

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

 

[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

 

[DODGEVIPER13]

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

 

[HallsofIvy]

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.

 

[DODGEVIPER13]

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