Question 1161347
<pre>
We diagnalize the matrix:

{{{matrix(1,3,M,""="",
(matrix(2,2,4,4,-2,10)))}}}

We find the eigenvalues

{{{abs(matrix(2,2,4-lambda,4,-2,10-lambda)))}}}{{{""=""}}}{{{0}}}

{{{(4-lambda)(10-lambda)-(-2)(4)=0}}}

{{{40-14lambda+lambda^2+8=0}}}

{{{lambda^2-14lambda+48=0}}}

{{{(lambda-6)(lambda-8)=0}}}

λ-6=0;  λ-8=0
  λ=6;    λ=8

by writing it as 

{{{matrix(1,3,M,""="",PDP^(-1))}}}

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

{{{D = (matrix(2,2,6,0,0,8))}}}

and the matrix P is 

{{{P=(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 eigengvector for the eigenvalue λ=6.

We find solutions for

{{{(M-6I)X=0}}}

{{{((matrix(2,2,4^"",4^"",-2^"",10^""))^""^""-(matrix(2,2,6^"",0^"",0^"",6^"")))(matrix(2,1,x[1],x[2]))=(matrix(2,1,0^"",0^""))}}}

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

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

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

So 

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

Now we do the same for the other eigenvalue

---

We find solutions for

{{{(M-8I)X=0}}}

{{{((matrix(2,2,4^"",4^"",-2^"",10^""))^""^""-(matrix(2,2,8^"",0^"",0^"",8^"")))(matrix(2,1,x[1],x[2]))=(matrix(2,1,0^"",0^""))}}}

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

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

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

So 

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

So

{{{P=(matrix(2,2,2,1,1,1))}}}

And since the determinant of P is 1, to find P<sup>-1</sup> we only
need to swap the elements on the the main diagonal and change the
signs of the other two elements"

{{{P^(-1)=(matrix(2,2,1,-1,1,2))}}}

Then 

{{{M=PDP^(-1)}}}

So

{{{M^n=(PDP^(-1))^n=(PDP^(-1))(PDP^(-1))(PDP^(-1))*""*""*(PDP^(-1))}}}

to n factors

{{{M^n=(PDP^(-1))^n=PD(P^(-1)P)D(P^(-1)P)D(P^(-1)*P)*""*""*DP^(-1)=PDIDIDI*""*""DP^(-1)}}}

{{{M^n = P*D^n*P^(-1)}}}

Any power of a diagonal matrix is the matrix whose elements are that
power of the elements, so we have the final answer as:

{{{matrix(1,5,
M^n,
""="",
(matrix(2,2,4^"",4^"",-2^"",10^""))^n,
""="",
(matrix(2,2,2^"",1^"",1^"",1^""))*(matrix(2,2,6^n,0^"",0^"",8^n))*(matrix(2,2,1^"",-1^"",-1^"",2^""))

)}}}

Edwin</pre>