4X4 real matrix eigenvalues

1. Mar 15, 2005

tanyas

Hi,

I need help on these questions for an assignment. I've been working on them for a couple of days and not getting anywhere. Any help would be appreciated...

1) A certain 4X4 real matrix is known to have these properties:
1. Two fo the eigenvalues of A are 3 and 2
2. the number 3 is an eigenvalue of the matrix A + 2I
3. det A =12

(i) what are the other two eigenvalues of the A?
(ii) what is the characteristic polynomial of A? of A'
(iii)what is the characteristic polynomial of A^-1

i guess (ii) and (iii) are easy once you get (i)

2) Let T(n) denote an nXn matrix such that for each tij,
tij= a if i=j
tij=b if i not equals j
(so basically a matrix with a's in the diagnol and b's as all the other elements
eg 3X3 a b b
b a b
b b a

(i) verify that T(n) = (a-b)I + bE where E is an nXn matirx of all 1's
(ii)find the eigenvalues of T
(iii) show that det T(n) = (a-b)^n-1 * (a + (n-1)b)

Thanks :)

2. Mar 15, 2005

matt grime

There is a general disdain for simply doing people's homework on these forums, especially questions whose answer is "look at the definition", ie Given a matrix M, what is the criterion for t to be an eigenvalue? That answers the first question.

How are you going about solving the second?

3. Mar 15, 2005

tanyas

hi,
im not asking for someone to solve the questions for me...just some hints as to which direction to go on...its a correspondence course...so its kind of hard to get help from the instructor(and there are no classmates to discuess with)...the lectures are all on tape...theres no visual so they're not very easy to follow and the textbook isnt much help either...ive tried looking up a few other books and internet resources but even then im having a problem with these questions
for the first ques...the only way i can think of id to construct a 4X4 general matrix and then use 1,2,3 to figure out its entries and then go from there...but that gets really messy and even then i dont think i'l get the solution that way....i'm wondering of im missing some property of determinants that may help
for the second ques i have absolutely no idea where to begin...
would be really grateful if u could give me a couple of pointers

4. Mar 15, 2005

cronxeh

det is a multiple of all eigenvalues
trace is a sum of all eigenvalues

5. Mar 15, 2005

tanyas

got it! thanks a lot :)
just wondering...where do you use property 2 then...because you can solve the whole question using just 1 and 3 then..

6. Mar 15, 2005

cronxeh

2. the number 3 is an eigenvalue of the matrix A + 2I
A*x = lambda*x

(A+2I)*x=lambda*x
(A+2I-lambda)*x=0
(A+2-3)*x=0
(A-1)*x=0

So, det(A-1) is the characteristic polynomial of A
det(A-1) is also characteristic polynomial of A' ( since det(A')=det(A) )
det(A^-1)=1/det(A-1)
Characteristic polynomial of A^-1 is 1/det(A-1) (I think..)

det A= 2*2*3 = 12
trace A = 2+2+3 = 7

-----snip-------
never mind most of this stuff below, i was trying to find matrix A itself....

det(A+2I)=(a11+2)(a22+2)(a33+2)(a44+2) + (a12)(a23)(a34)(a41) + (a13)(a24)(a31)(a42) + (a14)(a21)(a32)(a43) -(a14)(a23)(a32)(a41) -(a11+2)(a24)(a33+2)(a42) -(a12)(a21)(a34)(a43) -(a13)(a22)(a31)(a44+2) = 0 --> (equation 1)

trace(A+2) = trace(A) + trace(2) = 7+2 = 9

So.. 9 = a11+a22+a33+a44+8
a11+a22+a33+a44=1 --> (equation 2)

If you use other eigenvalue (lambda=2) you have:
A*x=lambda*x
A*x=2*x
A*x-2*x=0
(A-2I)*x=0
det(A-2I) = 0

det(A-2I) = (a11-2)(a22-2)(a33-2)(a44-2) + (a12)(a23)(a34)(a41) + (a13)(a24)(a31)(a42) + (a14)(a21)(a32)(a43) - (a14)(a23)(a32)(a41) - (a11-2)(a24)(a33+2)(a42) - (a12)(a21)(a34)(a43) - (a13)(a22)(a31)(a44-2) = 0 --> (equation 3)
trace(A-2I) = trace(A) - trace(2) = 7 - 2 = 5
trace(A-2I)= a11+a22+a33+a44-8 = 5

a11+a22+a33+a44-8 = 5 -->> (equation 4)
-----snip-------

Last edited: Mar 15, 2005
7. Mar 15, 2005

cronxeh

This is that time when I regret not buying that cheap whiteboard from staples.. :grumpy:

I cant find A !@\$

We need a topology/ring expert here

8. Mar 16, 2005

matt grime

No, you seem to have made a few errors there.

The characteristic polynomial of A is Det(A - xI) (or sometimes Det(xI-A), it doesn't really matter).

set P(x)= Det(A-xI)

What you wrote down isn't even a polynomial, so let's asssume we've corrected that.

t is an eigenvalue of A if and only if Det(A-tI) = 0

Det(A^{-1} - xI) most certainly isn't 1/Det(A-xI) - this isn't even a polynomial.

However, by cayley's theorem we know that P(A)=0. Can you think of a polynomial that A^{-1} satisfies? If that's not the way your brain works:

Det(A^{-1} - xI) = Det(A^-1)Det(1-Ax) = x^nDet(A^{-1})(Det(1/x - A) let y=1/x, and can you see how to get P into it?

In the specific case where A is the 4x4 above, I have an issue with the question, since we do not know that A has 4 eigenvalues, and need not. In fact the question is phrased so that we cannot make any deductions about "both" the other eigenvalues. Certainly, since 3 is an eigenvalue of A+2I, we know that Det(A+2I-3I) = 0= Det(A-I), so 1 is an eigen value. Now, if A had 4 distinct eigenvalues their product would equal 12, ie 1*2*3*d-12, where d is the last value to find, however, solving d=2, so we get a "repeated" eigenvalue, so actually there are only 3 eigenvalues, 1,2,3 and 2 occurs with multiplicty two in the characteristic polynomial (though it may not have two eigen vectors).

Hence the P(x) = (x-1)(x-2)^2(x-3)

I think A' means transpose, right? Use Cayleys theorem again to get a degree 4 polynomial that A' satisfies from P(x) (Hint, it's P(x)).

As for 2,

T = (a-b)I + bE

What is the entry at position i,j on each side of the equals sign? Is it the same?

Suppose v is an e'vector of T, that T(v) = kv.

kv=Tv=(a-b)v+ bEv

So Ev = ((k+b-a)/b)v

ie v is an eigenvactor of E

conversely, show any eigenvector of E is an eigen vector of T.

Can you find any eigen vectors of E (there are a lot, in fact there is a basis of eigen vectors) let me start you off with (1,1,1,..,1) and (1,-1,0,0,,...,0)