SOLUTION: if a=(1 1 k 0) the eigenvalues are -4 and 5 find the corresponding eigenvectors

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Question 1162515: if a=(1 1
k 0) the eigenvalues are -4 and 5 find the corresponding eigenvectors

Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20059) About Me  (Show Source):
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
The reason to find the eigenvectors is to diagonalize the matrix, so after
finding the eigenvectors, we'll go ahead and diagonalize the matrix A so we
can check to see that they are the correct eigenvectors.

matrix%281%2C3%2CA%2C%22%22=%22%22%2C%0D%0A%28matrix%282%2C2%2C1%2C1%2Ck%2C0%29%29%29

We are given the eigenvalues. For 

lambda=-4

abs%28matrix%282%2C2%2C1-%28-4%29%2C1%2Ck%2C0-%28-4%29%29%29%29%22%22=%22%220

abs%28matrix%282%2C2%2C1%2B4%2C1%2Ck%2C4%29%29%29%22%22=%22%220

abs%28matrix%282%2C2%2C5%2C1%2Ck%2C4%29%29%29%22%22=%22%220

20-k=0

So k=20.

We could do the same with the other eigenvector, but it will also give k=20.

So now we have

matrix%281%2C3%2CA%2C%22%22=%22%22%2C%0D%0A%28matrix%282%2C2%2C1%2C1%2C20%2C0%29%29%29

To diagonalize A we want to find matrices D and S so that we can write A as
 
matrix%281%2C3%2CA%2C%22%22=%22%22%2CSDS%5E%28-1%29%29

where D is the diagonal matrix with the two eigenvalues on the 
main diagonal:

D+=+%28matrix%282%2C2%2C-4%2C0%2C0%2C5%29%29

and the matrix S is 

S=%28matrix%281%2C2%2CV%5B1%5D%2CV%5B2%5D%29%29

where the V's are the two column eigenvectors for the two eigenvalues

We find V1 which is the eigenvector for the eigenvalue λ=-4.

We find solutions for

%28A-%28-4%29%5E%22%22I%29X=0

%28A%2B4I%29X=0





5x%5B1%5D%2Bx%5B2%5D=0
Divide thru by -18
x%5B2%5D=-5x%5B1%5D

We can take x1=1 and x2=-5

So 

v%5B1%5D=%28matrix%282%2C1%2C1%2C-5%29%29

Now we do the same for the other eigenvalue

---

We find solutions for

%28A-5I%29X=0





-4x%5B1%5D%2Bx%5B2%5D=0

x%5B2%5D=4x%5B1%5D

We can take x1=1 and x2=4

So 

v%5B2%5D=%28matrix%282%2C1%2C1%2C4%29%29

Now we have done that was asked for, for we have the two eigenvectors. So we
can stop here.  

v%5B1%5D=%28matrix%282%2C1%2C1%2C-5%29%29

v%5B2%5D=%28matrix%282%2C1%2C1%2C4%29%29

-------------------------------------

But let's check to make sure this matrix whose columns are the two
eigenvectors:

S=%28matrix%282%2C2%2C1%2C1%2C-5%2C4%29%29

provide the correct diagonalization of A.

Since the determinant of S is 9, to find S-1 we only need to swap the 
elements on the the main diagonal and change the signs of the other two
elements, then multiply by 1/9:



Then if we've done everything right, this will be true:

A=SDS%5E%28-1%29

So we check to see if this is the correct diagonalization of A:















So the eigenvalues are correct because 

A=SDS%5E%28-1%29

is the correct diagonalization of A.


Edwin