Linear Algebra Question..

[hank]

For matrices

A=
| -6 8 -2 |
|-5 6 -2 |
|6 -4 5 |

and B=
|-1 5 4 4|
|-2 6 4 4 |
|1 -4 -4 -8|
|0 2 4 8|

I found the minimum polynomials to be
mA(x) = (x-2)^2 (x-1) and mB(x) = (x-2)^2 (x-1) (x-4)

Using partial fraction expansions, I get

1/mA(x) = 1/(x-1) - 1/(x-2) + 1/(x-2)^2 and
1/mB(x) = (1/4)/(x-2) - (1/2)/(x-2)^2 - (1/3)/(x-1) + (1/12)/(x-4).

Now, I'm supposed to find polynomials p1, ..., p5 such that
1/mA(x) = p1(x)/(x-1) = p2(x)/(x-2)^2 and
1/mB(x) = p3(x)/(x-2)^2 + p4(x)/(x-1) + p5(x)/(x-4) by combining terms.

I'm not sure how to do that. Could someone maybe do 1/mA(x) for me so I can have an example to do 1/mB(x) to do on my own and for the rest of my homework problems?

 

[daon]

I'm not sure why one would do that (I guess Its been awhile for me and linear algebra), but it seems simple enough.

I'm sure you meant: "1/mA(x) = p1(x)/(x-1) + p2(x)/(x-2)^2" instead of "1/mA(x) = p1(x)/(x-1) = p2(x)/(x-2)^2" as the latter is only true for p1(x)=(x-1) and p2(x)=(x-2)^2 (ratios being 1)

We have 1/mA(x) = p1(x)/(x-1) + p2(x)/(x-2)^2 = 1/(x-1) - 1/(x-2) + 1/(x-2)^2

Combine the last two terms: 1/(x-1) + [-x+3]/(x-2)^2

Hence p1(x)=1, p2(x)=-x+3