Question 1162515
<pre>
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(1,3,A,""="",
(matrix(2,2,1,1,k,0)))}}}

We are given the eigenvalues. For 

{{{lambda=-4}}}

{{{abs(matrix(2,2,1-(-4),1,k,0-(-4))))}}}{{{""=""}}}{{{0}}}

{{{abs(matrix(2,2,1+4,1,k,4)))}}}{{{""=""}}}{{{0}}}

{{{abs(matrix(2,2,5,1,k,4)))}}}{{{""=""}}}{{{0}}}

{{{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(1,3,A,""="",
(matrix(2,2,1,1,20,0)))}}}

To diagonalize A we want to find matrices D and S so that we can write A as
 
{{{matrix(1,3,A,""="",SDS^(-1))}}}

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

{{{D = (matrix(2,2,-4,0,0,5))}}}

and the matrix S is 

{{{S=(matrix(1,2,V[1],V[2]))}}}

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

We find V<sub>1</sub> which is the eigenvector for the eigenvalue λ=-4.

We find solutions for

{{{(A-(-4)^""I)X=0}}}

{{{(A+4I)X=0}}}

{{{((matrix(2,2,1^"",1^"",20^"",0^""))^""^""+(matrix(2,2,4^"",0^"",0^"",4^"")))(matrix(2,1,x[1],x[2]))=(matrix(2,1,0^"",0^""))}}}

{{{(matrix(2,2,5^"",1^"",20^"",4^""))(matrix(2,1,x[1],x[2]))=(matrix(2,1,0^"",0^""))}}}

{{{5x[1]+x[2]=0}}}
Divide thru by -18
{{{x[2]=-5x[1]}}}

We can take x<sub>1</sub>=1 and x<sub>2</sub>=-5

So 

{{{v[1]=(matrix(2,1,1,-5))}}}

Now we do the same for the other eigenvalue

---

We find solutions for

{{{(A-5I)X=0}}}

{{{((matrix(2,2,1^"",1^"",20^"",0^""))^""^""-(matrix(2,2,5^"",0^"",0^"",5^"")))(matrix(2,1,x[1],x[2]))=(matrix(2,1,0^"",0^""))}}}

{{{(matrix(2,2,-4^"",1^"",20^"",-5^""))(matrix(2,1,x[1],x[2]))=(matrix(2,1,0^"",0^""))}}}

{{{-4x[1]+x[2]=0}}}

{{{x[2]=4x[1]}}}

We can take x<sub>1</sub>=1 and x<sub>2</sub>=4

So 

{{{v[2]=(matrix(2,1,1,4))}}}

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

{{{v[1]=(matrix(2,1,1,-5))}}}

{{{v[2]=(matrix(2,1,1,4))}}}

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

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

{{{S=(matrix(2,2,1,1,-5,4))}}}

provide the correct diagonalization of A.

Since the determinant of S is 9, to find S<sup>-1</sup> 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:

{{{S^(-1)=expr(1/9)(matrix(2,2,4^"",-1^"",5^"",1^""))}}}

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

{{{A=SDS^(-1)}}}

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

{{{matrix(1,5,
A,
""="",
(matrix(2,2,1^"",1^"",20^"",0^"")),
""="",
(matrix(2,2,1^"",1^"",-5^"",4^""))*(matrix(2,2,-4,0^"",0^"",5))*(matrix(2,2,4/9,-1/9,5/9,1/9))

)}}}

{{{matrix(1,5,
A,
""="",
(matrix(2,2,1^"",1^"",20^"",0^"")),
""="",
(matrix(2,2,(1)(-4)+(1)(0)^"",(1)(0)+(1)(5)^"",(-5)(-4)+(4)(0)^"",(-5)(0)+(4)(5)^""))*(matrix(2,2,4/9,-1/9,5/9,1/9)))}}}

{{{matrix(1,5,
A,
""="",
(matrix(2,2,1^"",1^"",20^"",0^"")),
""="",
(matrix(2,2,-4^"",5^"",20^"",20^""))*(matrix(2,2,4/9,-1/9,5/9,1/9)))}}}

{{{matrix(1,5,
A,
""="",
(matrix(2,2,1^"",1^"",20^"",0^"")),
""="",
(matrix(2,2,-4^"",5^"",20^"",20^""))*(matrix(2,2,4/9,-1/9,5/9,1/9)))}}}

{{{matrix(1,5,
A,
""="",
(matrix(2,2,1^"",1^"",20^"",0^"")),
""="",
(matrix(2,2,-16/9+25/9,4/9+5/9,80/9+100/9,-20/9+20/9)))}}}

{{{matrix(1,5,
A,
""="",
(matrix(2,2,1^"",1^"",20^"",0^"")),
""="",
(matrix(2,2,9/9,9/9,180/9,0/9)))}}}

{{{matrix(1,5,
A,
""="",
(matrix(2,2,1^"",1^"",20^"",0^"")),
""="",
(matrix(2,2,1^"",1^"",20^"",0^"")))}}}

So the eigenvalues are correct because 

{{{A=SDS^(-1)}}}

is the correct diagonalization of A.


Edwin</pre>