Linear algebra Eigenvalue related

[Darkbalmunk]

Homework Statement

a) |-1 1 1|
| 1 -1 1| = A
| 1 1 -1|
Find an orthoginal matrix P that diagonalizes Ab) |0 1| What value of a is multiplicity 2, what value of a is eigen values -1 and 2
A = |a 1| what value of a does A have real eigenvaluesC) If A is a diagonizable matrix that has only one eigenvalue lambda
prove A = λ [Identity]

D) if lambda is a eigenvalue of matrix A then prove
i) λ ^2 is an eigenvalue of A^2
ii) 1/λ is an eigenvalue of A^-1
iii) λ + 2 is an eigenvalue of A+2I

Homework Equations

B) found determinate (X^2) - X - a
also b^2 = 4ac for ax^2 + bx + c
C) i found the form A-λ[identity] = 0
D) AX=λX

The Attempt at a Solution

A> when finding eigenvalue for the matrix i end up with a complicated polynomial
in the form of 4 - (3X^2) - (X^3) and i can't figure out how to factor it to find my eigenvalues and find the eigenvectors, also it asks for the orthoginal matrix P that diagonalizes A i know the eigenvectors make P but how do i make them orthoginal.
update: 4 = (3+λ)λ^2 i get λ = 1,-2,2 but my teacher says λ=1,-2 why is 2 not an eigenvector?

B> I found a to be -1/4 for multiplicity to be 2, and if a was 2 you factor to find eigenvalues to be -1 and 2, but does a = any real number for A to have real eigenvalues.

c) i really might have missed something cause i don't think it is as simple as taking A - λ Identity and prove that since lambda is scalar that i can just use algebra to say A = λ identity.
i found something what if i say A = PDP^-1 so if D = λ then can i say A = PλP^1 then A = λPP^-1 so A = λI.

D>
i was easy but
ii) AX = λX
if i A^-1 A X = A^-1 λ X
IX = λ A^-1 X
1/λ X = A^-1 X
did i do it correctly?

also for iii)
i was going to just add 2 to both sides but that was wrong, so i tried multiplying one side by (I + 2A^-1) that does the AX side but i have no idea how to go about this problem.

 

[micromass]

Hi Darkbalmunk!

That are a lot of exercises here, but I think most of them are alright.

The Attempt at a Solution

A> when finding eigenvalue for the matrix i end up with a complicated polynomial
in the form of 4 - (3X^2) - (X^3) and i can't figure out how to factor it to find my eigenvalues and find the eigenvectors, also it asks for the orthoginal matrix P that diagonalizes A i know the eigenvectors make P but how do i make them orthoginal.
update: 4 = (3+λ)λ^2 i get λ = 1,-2,2 but my teacher says λ=1,-2 why is 2 not an eigenvector?

Because 2 is not a solution to the equation 4 = (3+λ)λ². When I plug in 2 there, I get 4=28, which is false. Plugging in 1 and -2 does yield something true though. So the eigenvalues are 1 and -2.

Now you need to find some eigenvectors (don't care about orthogonality just yet). So find some eigenvectors. Since the matrix should be diagonalizable, you should find three eigenvectors. (Maybe the best thing to do is actually find the eigenspaces instead of the eigenvectors!)

_{By the way, I think you made the calculation of the characteristic polynomail far to difficult. If you calculate the characteristic polynomial the right way, then the characteristic polynomial is already factored. You did it a bad (but correct!) way so you have to worry about factoring a third degree equation...}

B> I found a to be -1/4 for multiplicity to be 2, and if a was 2 you factor to find eigenvalues to be -1 and 2, but does a = any real number for A to have real eigenvalues.

That seems alright. For the eigenvalues to be real, you'll need to make sure that $X^2-X-a$ has real roots, thus you'll need to find out for which a the discriminant is positive.

c) i really might have missed something cause i don't think it is as simple as taking A - λ Identity and prove that since lambda is scalar that i can just use algebra to say A = λ identity.
i found something what if i say A = PDP^-1 so if D = λ then can i say A = PλP^1 then A = λPP^-1 so A = λI.

Well, you can begin by showing that λ is an eigenvalue of A if and only if λ is an eigenvalue of the diagonalized matrix.

D>
i was easy but
ii) AX = λX
if i A^-1 A X = A^-1 λ X
IX = λ A^-1 X
1/λ X = A^-1 X
did i do it correctly?

Yes, that is correct. You may want to think why 1/λ makes sense though (i.e. why λ is not zero).

also for iii)
i was going to just add 2 to both sides but that was wrong, so i tried multiplying one side by (I + 2A^-1) that does the AX side but i have no idea how to go about this problem.

Well, if x is an eigenvector of A, then what is (A+2I)x?