Linear algebra Eigenvalue related

In other words, x is a vector that satisfies Ax=λx. What do you get if you multiply this equation by I+2A^-1? I hope you'll see that this is a statement that (λ+2) is an eigenvalue of A+2I...In summary, A) The problem is asking for an orthogonal matrix P that diagonalizes the given matrix. The solution involves finding the eigenvalues and eigenvectors of the matrix and using them to construct the diagonal matrix D. P can then be constructed by using the eigenvectors as columns and ensuring they are orthogonal.B) The problem is asking for values of a that result in certain multiplicities and eigenvalues for the given matrix A. The
  • #1
Darkbalmunk
5
0

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.
 
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  • #2
Hi Darkbalmunk! :smile:

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

Darkbalmunk said:

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 [itex]X^2-X-a[/itex] 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?
 

FAQ: Linear algebra Eigenvalue related

1. What is an eigenvalue in linear algebra?

An eigenvalue is a scalar value that represents the scaling factor for a specific eigenvector in a linear transformation. It is an important concept in linear algebra as it helps to understand the behavior of a linear system and its stability.

2. How do you calculate eigenvalues?

The process of finding eigenvalues involves solving a characteristic equation, which is derived from the determinant of the matrix. In simpler terms, you need to find the values of lambda that satisfy the equation (A-lambda*I)x = 0, where A is the matrix, lambda is the eigenvalue, and I is the identity matrix.

3. What is the geometric interpretation of eigenvalues?

The geometric interpretation of eigenvalues is that they represent the amount of stretching or shrinking of a vector in a linear transformation. If the eigenvalue is positive, the vector is stretched, and if it is negative, the vector is shrunk. If the eigenvalue is 0, the vector is unchanged.

4. How are eigenvalues used in real-world applications?

Eigenvalues have various applications in fields such as physics, engineering, and computer science. They are used to analyze the stability of a system, compress data, and solve differential equations. They are also used in image and signal processing, such as in facial recognition algorithms.

5. Can a matrix have complex eigenvalues?

Yes, a matrix can have complex eigenvalues. This happens when the characteristic equation has complex roots. The eigenvalues of a matrix can be real or complex, depending on the matrix's properties. For example, a symmetric matrix will always have real eigenvalues, while a non-symmetric matrix can have complex eigenvalues.

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