# how do i..

• Oct 8th 2006, 06:51 PM
kgpretty
how do i..
How do I find a matrix P, where there is a matrix A, such that P^(-1)AP is diagonal??
• Oct 8th 2006, 09:12 PM
CaptainBlack
Quote:

Originally Posted by kgpretty
How do I find a matrix P, where there is a matrix A, such that P^(-1)AP is diagonal??

Look up the eigen decompositon either here or here

RonL
• Oct 9th 2006, 03:29 AM
kgpretty
Quote:

Originally Posted by CaptainBlack
Look up the eigen decompositon either here or here

RonL

Thankyou very much. I like the explanation wikipedia gave better.
• Oct 9th 2006, 04:08 AM
kgpretty
Another question asks, for a matrix A, show that it satisfies it's characteristic equation.

Looking at the example on Wikipedia, would the characteristic equation be (Y - 3)(Y - 2)(Y - 1)?

If it is, how do I complete the above question?
• Oct 9th 2006, 04:42 AM
CaptainBlack
Quote:

Originally Posted by kgpretty
Another question asks, for a matrix A, show that it satisfies it's characteristic equation.

Looking at the example on Wikipedia, would the characteristic equation be (Y - 3)(Y - 2)(Y - 1)?

If it is, how do I complete the above question?

The charateristic equation is:

det(A-lambda I) = 0

(as a polynomial in lambda). Now as 3, 2, 1 are the roots of this equation
we know that if we expand the above expression we will have:

(lambda-3)(lambda-2)(lambda-1)=0

for the charateristic equation, and as A should satisfy this we should find
that:

(A-3)(A-2)(A-1)=0

where there is an impicit multiplication of the 3, 2, 1 and 0 by I.

This expands to:

(A-3)(A-2)(A-1) = (A-3)(A^2 - 3A +2) = A^3 - 6A^2 + 11A -6

The powers of A are simply A^2=AA, A^3=A(AA), and the 6 is short hand
for 6 I.

RonKL
• Oct 9th 2006, 05:08 AM
kgpretty
Does the question stop at A^3 - 6A^2 + 11A -6
Or do I have to break down the above equation even further. Like find A^3, 6A^2 and so on, then do the addition and substraction??
• Oct 9th 2006, 07:10 AM
CaptainBlack
Quote:

Originally Posted by kgpretty
Does the question stop at A^3 - 6A^2 + 11A -6
Or do I have to break down the above equation even further. Like find A^3, 6A^2 and so on, then do the addition and substraction??

I'm a bit confused by what you need to show. With a particular matrix you
can compute the powers and substitute them into the equation and show
that the matrix satisfies it.

For a general matrix you could show this is in general true.

If D is the diagonal matrix then as each element on the diagonal is an eigen
value of A, we have tivialy that D satisfies the charateristic equation.

Also as P D P^-1 = A, and P D^2 P^-1 = P D P^-1 P D P^-1 = A^2,
and so on for higher powers, so P D^n P^-1 = A^n and we have:

if

sum_{r=0,n} a_r x^r = 0

is a polynomial satisfied by D, then it is also satisfied by A, since:

P sum_{r=0,n} a_r D^r P^-1 = sum_{r=0,n} a_r P D^r P^-1

..................=sum_{r=0,n} a_r A^r = 0.

RonL
• Oct 10th 2006, 03:31 PM
kgpretty
Ok, I think I get it. Thankyou.

However I have a question for which I am completely stumped.
First, it asked you to find the eigenvectors and eigenvalues for matrix A.
Code:

```Matrix A: |1 3| |2 6| Here's what I did: |1 3| |2 6|x  = Yx |1 3| |2 6|x - Yx = 0 |1-Y      3  | |  2    6 - Y|x    = 0 .: (1 - Y)(6 - Y) - 6 = 0 6 + Y^2 - 6Y - 1Y - 6 = 0 Y^2 - 7Y + 0 = 0 (Y + 0)(Y - 7) = 0 When Y = 0 |1 3| |x1| |2 6| |x2| = 0 Since |1 3| | 3| |2 6| |-1| = 0 | 3| |-1| is an eigenvector corresponding to the eigenvalue 0. When Y = 7 |1-7    3| |2    6-7|x = 0 |-6  3| |x1| | 2  -1| |x2| = 0 Since |-6  3| |1| | 2  -1| |2| = 0 |1| |2| is an eigenvector corresponding to the eigenvalue 7. Good, so I got the eigenvalues and vectors. The question then asked to find P for which P^-1AP is a diagonal matrix. P = |3  1|     |-1  2| P-1 = |3  1|       |-1  2| = | 3  1|1  0|   |-1  2|0  1| = |1  0|2  -1|   |0  1|1  3| = |2  -1|   |1  3| .: P^-1AP = |2  -1|  |1  3|  | 3  1|         |1  3|  |2  6|  |-1  2|           = |0  0| |3  1|             |7  21| |-1  2|```
So you see my problem, P^-1AP is not equal to a diagonal.... All the other problems I did the same way gave me a diagonal.
• Oct 10th 2006, 07:10 PM
CaptainBlack
Quote:

Originally Posted by kgpretty
Ok, I think I get it. Thankyou.

However I have a question for which I am completely stumped.
First, it asked you to find the eigenvectors and eigenvalues for matrix A.
Code:

```Matrix A: |1 3| |2 6|   Here's what I did: |1 3| |2 6|x  = Yx   |1 3| |2 6|x - Yx = 0   |1-Y      3  | |  2    6 - Y|x    = 0   .: (1 - Y)(6 - Y) - 6 = 0 6 + Y^2 - 6Y - 1Y - 6 = 0 Y^2 - 7Y + 0 = 0 (Y + 0)(Y - 7) = 0   When Y = 0 |1 3| |x1| |2 6| |x2| = 0   Since |1 3| | 3| |2 6| |-1| = 0   | 3| |-1| is an eigenvector corresponding to the eigenvalue 0.     When Y = 7 |1-7    3| |2    6-7|x = 0   |-6  3| |x1| | 2  -1| |x2| = 0   Since |-6  3| |1| | 2  -1| |2| = 0   |1| |2| is an eigenvector corresponding to the eigenvalue 7.     Good, so I got the eigenvalues and vectors. The question then asked to find P for which P^-1AP is a diagonal matrix. P = |3  1|     |-1  2|   P-1 = |3  1|       |-1  2|   = | 3  1|1  0|   |-1  2|0  1|   = |1  0|2  -1|   |0  1|1  3|   = |2  -1|   |1  3|   .: P^-1AP = |2  -1|  |1  3|  | 3  1|         |1  3|  |2  6|  |-1  2|             = |0  0| |3  1|             |7  21| |-1  2|```
So you see my problem, P^-1AP is not equal to a diagonal.... All the other problems I did the same way gave me a diagonal.

If:

P = [3 1]
.....[-1 2]

P^-1 =.[4/10 2/10]
...........[2/10 6/10]

Ooops, I appear to have inverted the wrong matrix some how, the inverse
should be:

P^-1= [2/7 -1/7]
.........[1/7...3/7]

RonL
• Oct 11th 2006, 05:42 AM
kgpretty
Quote:

Originally Posted by CaptainBlack
If:

P = [3 1]
.....[-1 2]

P^-1 =.[4/10 2/10]
...........[2/10 6/10]

RonL

How did you arrive at
P^-1 = [4/10 2/10]
[2/10 6/10]
• Oct 11th 2006, 06:52 AM
CaptainBlack
Quote:

Originally Posted by kgpretty
How did you arrive at
P^-1 = [4/10 2/10]
[2/10 6/10]

some sort of brain fade has occured, I must have inverted the wrong matrix
(this must be the case as I multiplied what I inverted by this "inverse" and
got the identity matrix)

Checking this, the real inverse is

P^-1= [2/7 -1/7]
.........[1/7...3/7]

obtained by Gaussian elimination with multiple right hand sides.

There are many ways of inverting a 2x2 matrix, there is Gaussian elimination,
Cramers rule, inspection, ..

Code:

``` Demo of Gaussian elimination ----------------------------   Initial set-up:    3 1  |  1  0 -1 2  |  0  1   multiply the second row by 3 and add the first:    3 1  |  1  0  0 7  |  1  3   multiply the first row by 7 and subtract the second:   21 0  |  6 -3  0 7  |  1  3   Divide the first row through by 21, and the second by 7:    1 0  |  2/7  -1/7  0 1  |  1/7  3/7   So the inverse the the matrix:   [2/7  -1/7] [1/7  3/7]   Check:   >  p=[3,1;-1,2]             3            1           -1            2 >ip=[2/7,-1/7;1/7,3/7]     0.285714    -0.142857     0.142857      0.428571 > > >ip.p             1            0             0            1 >```
RonL
• Oct 11th 2006, 06:06 PM
kgpretty
Quote:

Originally Posted by CaptainBlack
some sort of brain fade has occured, I must have inverted the wrong matrix
(this must be the case as I multiplied what I inverted by this "inverse" and
got the identity matrix)

Checking this, the real inverse is

P^-1= [2/7 -1/7]
.........[1/7...3/7]

obtained by Gaussian elimination with multiple right hand sides.

There are many ways of inverting a 2x2 matrix, there is Gaussian elimination,
Cramers rule, inspection, ..

Code:

``` Demo of Gaussian elimination ----------------------------   Initial set-up:    3 1  |  1  0 -1 2  |  0  1   multiply the second row by 3 and add the first:    3 1  |  1  0  0 7  |  1  3   multiply the first row by 7 and subtract the second:   21 0  |  6 -3  0 7  |  1  3   Divide the first row through by 21, and the second by 7:    1 0  |  2/7  -1/7  0 1  |  1/7  3/7   So the inverse the the matrix:   [2/7  -1/7] [1/7  3/7]   Check:   >  p=[3,1;-1,2]             3            1           -1            2 >ip=[2/7,-1/7;1/7,3/7]     0.285714    -0.142857     0.142857      0.428571 > > >ip.p             1            0             0            1 >```
RonL

I got the same matrix elements for p^-1 as you did.
You have shown that ip.p is a diagonal matrix. However, the question asked to show P^-1AP is a diagonal and that's where I got stumped because it isn't.