# Expressing Matrix Power as linear combination

1. Aug 19, 2015

### SteliosVas

1. The problem statement, all variables and given/known data

Okay I am given a matrix A = [2 1 ; 3 4]

The first step is to find numbers of a and b such that A2 + aA + bI = [0 0; 0 0]

I is an identity matrix (2x2).

Part B - After that is says to use the result of the above to express A5 as a linear combination of A and I

2. Relevant equations

Okay I am pretty sure for the first part it is just quite simply squaring A, putting the letter a and b in front of the respective matrix and multiplying.
Than equalling to 0 you have 2 unknowns and 4 equations solve for A and B.
As for the second part I am not sure if I should be using the characteristic polynomial or/and eigen values?

3. The attempt at a solution

Okay so for the first part i got 4 equations once eventually done the computations as:

2a + b + 17 = 0
a + 6 = 0
3a+18=0
4a+b+19=0

Solving I get a = -6 and b = 5

Now for part B I am really stuck. . If i calculate the eigen values then, they are also the eigen values for A5.. Because I + A + A2.... is an infinite series.. with a common ratio... Really stuck sure there is an easier way to look at it,

2. Aug 19, 2015

### andrewkirk

Write $A^5$ as a product of powers of A, each no greater than 2.
You have an equation that enables you to write $A^2$ as a linear function of A.
So by repeated substitution for $A^2$ in that product you should be able to get $A^5$ down to a linear function of A.

3. Aug 20, 2015

### SteliosVas

Do you mean A2 * A2 * A?

Sorry could you explain a little more?

4. Aug 20, 2015

### andrewkirk

Yes that's what I mean. Now use the equation you derived in Part A to replace each of those two instances of $A^2$ by a linear combination of A and I. Then expand and collect terms. Can you see a way forward from there?

5. Aug 20, 2015

### SteliosVas

Okay so I sort of understand where you are going but let me clarify....

So we know the value of a and the value of b... We also know that for the Matrix A2+a*A + b*I we get it equal to a 2x2 zero's matrix.
Since we know the values of a and b we can plug those into the equation that I got which was A2 * -6A + 5I = [0 0 ; 0 0]
So basically do I need to replace A2 with A*A than use that result to multiply again by A and repeat?

EDIT:

Okay maybe now I understood, do you mean multiply A1 by a than A2 by a and so on?

Last edited: Aug 20, 2015
6. Aug 20, 2015

### andrewkirk

No. That matrix on the right-hand side of your equation is just the zero matrix. So your equation is $A^2-6A+5I=0$. Now make $A^2$ the subject of the equation and you'll have a formula for $A^2$ that is a linear function of $A$ and $I$.

7. Aug 20, 2015

### SteliosVas

Okay fantastic now I get you!!

So the real equation for A2 = 6A - 5I

So since we have the equation of A2 we can multiply that by another linerar combination of A2 to obtain A4 than again multiply by A to get A5?

Last edited: Aug 20, 2015
8. Aug 20, 2015

### RUber

That is right, but the linear combination means you can only have constants times A and constants times I.
Continue substituting until all higher powers of A are gone and you will be left with cA -DI = A^5.
Remember to check your answer with a computer or calculator since repeated multiplications and additions provide many opportunities for errors.