What is the associated eigenvalue?

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In summary, V is an eigenvector of the nxn matrix A with an eigenvalue of 4. This means that Av=4v. When multiplied by the matrix A^2+2A+3I, the result is A(Av)+2Av+3v which simplifies to (4Av)+8v+3v. Therefore, the associated eigenvalue for A^2+2A+3I is 27.
  • #1
riordo
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Homework Statement



V is an eigenvector of the nxn matrix A, with a eigenvalue of 4. explain why V is a eigenvector of A^2+2A+3I. what is the associated eigenvalue?

Homework Equations





The Attempt at a Solution



is the eigenvalue of A^2+2A+3I=21?
 
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  • #2
What does it mean that V is an eigenvector of A with eigenvalue 4?

Answer that, and then ask yourself what happens when you multiply the matrix (A^2+2A+3I) by V from the right.

The answer for the eigenvalue is not 21.
 
  • #3
I don't understand how A^2+2A+3I is populated in what I assume is a 2x2 matrix? If you tell me that it will help me a lot. I don't understand how a polynomial populates a matrix
 
  • #4
to answer you question. A is a matrix (nxn) v is a nonzero vector in R^n. Av is a scalar multiple of lamda, Av=landav. lamda is the eigenvalue. the unknown here is A either its a matrix that i do know the vectors of or it's a variable that populates the matrix that is used to calculate the eigenvalue...right? so as you say A^2+2A+3I is the matrix how does it populated matrix...what is its dimensions?
 
  • #5
I don't understand your use of the work "populated".

Here, A is an nxn matrix. And as you probably know, multiplying two nxn matrices or adding two nxn matrices, or multiplying an nxn matrix by a scalar gives out another nxn matrix. So (A^2+2A+3I) is an nxn matrix.

Now, what is (A^2+2A+3I)v, knowing that Av=4v?
 
  • #6
eigenvalue

looks like it equals 4...this has been helpful. I've had a misconception that coefficients in A^2+2A+3I would be entered into the nxn matrix in some manner..e.g. [[A^2, 2], [3, 0]] or if i knew the vectors of A i could try to calculated the eigenvalue. so is 4 the correct solution?
 
  • #7
No, it isn't.

Use the fact that multiplication between matrix and vector is distributive. Meaning that for A, B two nxn matrices and u in R^n, (A+B)u=Au+Bu.
 
  • #8
Saying that v is an eigenvector of A with eigenvalue 4 means that Av= 4v.

What is (A2+ 2A+ 3I)v= A(Av)+ 2Av+ v?
 
  • #9
eigenvalue

ok i get what you've been saying about using the distributive rule (thanks) A^2v+2Av+3v (is 3v correct 3I*v=3v). so are we at this point A^2v+2Av+3v=4v? if yes then what? thanks for your help.
 
  • #10
Why "A^2v+2Av+3v=4v"? What you know is that Av=4v.

So use that to write A^2v+2Av+3v = A(4v)+2(4v)+3v=... (you finish)
 
  • #11
riordo said:
ok i get what you've been saying about using the distributive rule (thanks) A^2v+2Av+3v (is 3v correct 3I*v=3v). so are we at this point A^2v+2Av+3v=4v? if yes then what? thanks for your help.

No, I didn't say that at all. Av= 4v. What is A(Av)= A(4v)= 4Av? What is 2Av?
 
  • #12
A(4v)+2(4v)+3v=4(Av)+8v+3v=4(Av)+8v+3v=4(4v)+8v+3v=16v+8v+3v=27v does that make sense?
 
  • #13
Plenty.
 
  • #14
thank you. i'll do better with the next question!
 

1. What is an eigenvalue?

An eigenvalue is a scalar value that represents a factor by which a linear transformation changes a vector. In other words, it is a number that characterizes a certain transformation.

2. Why are eigenvalues important?

Eigenvalues are important because they provide information about the behavior and properties of a linear transformation. They also help in solving systems of linear equations and understanding the stability of a system.

3. How do you calculate eigenvalues?

To calculate eigenvalues, you need to first find the characteristic polynomial of the matrix or linear transformation. Then, solve for the roots of the polynomial to find the eigenvalues.

4. What is the significance of the associated eigenvalue?

The associated eigenvalue is significant because it tells us how much a vector is scaled or stretched by a linear transformation. It also helps in understanding the behavior and stability of the system.

5. Can eigenvalues be negative or complex?

Yes, eigenvalues can be negative or complex, depending on the matrix or linear transformation. In fact, complex eigenvalues are often associated with oscillatory behavior in systems, while negative eigenvalues can indicate instability.

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