Question

Let A be the general real 2 x 2 symmetric matrix. Prove directly using the discriminant of the characteristic polynomial that A has real eigenvalues.

Answer 1

Let A = { (a,b),(b,c) ) then -> DET {(a-L,b),(b,c-L)} = 0
---> (a-L)(c-L) - b^2 =0 --> L^2 -L(a+c)-b^2+ac =0
--->(a+c)/2 +or-1/2 SQRT( (a+c)^2 -4(ac-b^2)) = eigenvalues
clearly (a+c)/2 is bigger than 1/2 SQRT( (a+c)^2 -4(ac-b^2)) so it doesnt matter if you subtract or add it to (a+c)/2 the result is always positive

Answer 2

I probably should know this, but "discriminant of the characteristic polynomial" just meant for me to find the det of the the matrix?

Answer 3

I've taken 2 linear algebra classes and ive never heard of "discriminant" this is an eigenvalue problem so yea... just means the determinant of the (A-I*L) matrix = characteristic polynomial.

Answer 4

and proving this is saying that A will always have real eigenvalues? Or are there special cases when this is not true?

Answer 5

This proves that is A is a real and symmetric matrix then it will always have real eigenvalues

Answer 6

if*

Answer 7

key being symmetric right, because if its not symmetric then it might not have eigenvalues

Answer 8

right, b/c if it is not symmetric then you cant take the det of it right?

Answer 9

you can take the determinant regardless if A is symmetric or not. You can always always always take the determinant.

Answer 10

What will change is the eigenvalues

Answer 11

of course in my proof i set L=eigenvalue

Answer 12

right... the second part of this problem says...
Show that A has equal eigenvalues only if A is a scalar multiple of I(2)

Answer 13

do i do the same process multiply the (lambda)I matrix by a scalar r?

Answer 14

Of course. You might not have covered this but if A is a scalar multiple of I where I = {(1,0),(0,1)} then A is a diagonal matrix meaning that it only has values on its diagonal... in this case the eigenvalues are the diagonal elements... you can check this for yourself just consider some constans a then set it: A= a*I = {(a,0),(0,a)} take the determanant of (A-L*I) = DET {(a-L,0),(0,a-L)} = (a-L)^2 =0 obviously L=a is the only solution..... then you have 2 equal eigenvalues

Answer 15

if i want to verify that eigenvectors are orthogonal, do I just make sure that the equation
Av=Lv
and make sure that holds, does this prove orthogonality?
**This is for a different question

Answer 16

If you want to check if the eigenvectors are orthogonal then you take their inner product (in your case the dot product). Lets say you have eigenvectos E1 and E2.... assuming the eigenvectors are REAL you take the transpose of one and matrix multiply it with the other. If either of the eigenvectors are complex then take the complex conjugate of the complex eigenvector, then transpose it... then matrix multiply it with the other eigenvector

Answer 17

In my problem the two eigenvectors that I found was (-1/2, 1) and (2,1)... so I just take their product and if it equals zero... in this case which it does, then they are orthogonal to each other... or orthogonal to the matrix?

Answer 18

orthogonal to each other

Answer 19

awesome. thank you for all your help. this course gives me alot of trouble, so I appreciate all the help.

Answer 20

no problem

Answer 21

please select my answer if you think it was the best ( :

Answer 22

I did :-)