# Eigenvalues proof

Gold Member

## Homework Statement

Let A=LU and B=UL, where U is an upper triangular matrix and L is a lower triangular matrix. Demonstrate that A and B have the same eigenvalues.

Not sure.

## The Attempt at a Solution

I know that if I can show that A and B are similar (so if I can find a matrix P such that P^(-1)AP=B) they have the same eigenvalues. But I didn't find P yet, nor do I know really how to search efficiently for P.

Another route I've thought of is to write det (I*lambda-A)=0 gives the same values for lambda as det (I*lambda-B)=0. I've thought of using det (A)=det (LU)=det L * det U = det U * det L = det B... but still can't reach anything I find useful.
Any tip is greatly appreciated.

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lanedance
Homework Helper
how about multiplying thorugh by the inverse of L.. you may need to show it exists

Try to write A in terms of B (or the other way around), and see what you get!

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lanedance
Homework Helper
hint was probably enough

Gold Member
LU=U^(-1)ULU=LU so the equality is true. Notice that LU=A and UL=B. I have that P=U, thus LU is similar to UL (A similar to B) so they share the same eigenvalues.

Now I must justify the use of the inverse of U. Well I believe its element on the diagonal are all non zero. So if the matrix is nxn, the span of its column vector is R^n so it is invertible, hence U^-1 exists.

Is that well justified?

Dick
Homework Helper
LU=U^(-1)ULU=LU so the equality is true. Notice that LU=A and UL=B. I have that P=U, thus LU is similar to UL (A similar to B) so they share the same eigenvalues.

Now I must justify the use of the inverse of U. Well I believe its element on the diagonal are all non zero. So if the matrix is nxn, the span of its column vector is R^n so it is invertible, hence U^-1 exists.

Is that well justified?
It would be if the problem stated that the elements of U along the diagonal are nonzero. Does it? Other versions of this problem I've found say that L or U is a UNIT triangular matrix. Did you miss a word in the problem statement?

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Gold Member
It would be if the problem stated that the elements of U along the diagonal are nonzero. Does it? Other versions of this problem I've found say that L or U is a UNIT triangular matrix. Did you miss a word in the problem statement?
You pointed out a very interesting thing to me. In my assignment it isn't specified. But most of the exercises assigned are taken from Kincaid's book on numerical analysis. I just checked there and it clearly states L to be UNITARY.
If only they had stated this in my assignment I could have guessed that I had to take the inverse of L. :) Well I'm not 100% sure but probably.
So basically the problem is solved now... thanks guys.

Dick