Solving Eigenvalue Questions for 2x2 Matrix & nxn Matrix

In summary, if an nxn matrix has a sum of all the entries that is 1, and all entries are positive, then the associated eigenvalue is less than or equal to 1 in absolute value terms.
  • #1
buzzmath
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0
I have two questions

1. If I have a 2x2 matrix A with entries a, b, c, d where a is the upper left corner, b upper right corner, c lower left, and d lower right. I have eigenvalues L1 and L2. I need to show that L1^2 + L2^2 <= a^2 + b^2 + c^2 + d^2. So far I've done this: I know det(A)=L1*L2 and that tr(A) = L1 + L2. So L1 + L2 = a + b square each side and get L1^2 + L2^2 +2*L1*L2 = a^2 + d^2 +2ad subtract 2*L1*L2 from each side and get L1^2 + L2^2 = a^2+d^2+2ad-2*L1*L2 where since L1*L2 is det(A) I have a^2+d^2 + 2ad-2ad+2bc thus L1^2 + L2^2 = a^2 + d^2 + 2bc. Now all I would really need to show is that 2bc <= b^2+c^2. I also need to know when this inequality is equal. Am I on the right track or does anyone have any advice?

2.If A is an nxn matrix where the sum of each row is 1 and all entires are positive. if v is an eigenvector of A with positive componentsShow the the associated eigenvalue is less than are equal to 1. Also, if we drop the requirement that the components of the eigenvector v be positive, is it still true that the associated eigenvalue is less than or equal to 1 in absolute value terms? Justify your answer. For this problem I was thinking I could consider the largest entry of v and take the corresponding entry in A. the largest that this A value could be is 1. If the value in v is less than 1 then the eigenvalue is then the most the eigenvalue could be is 1. If the value in v is 1 then the eigen value could be at most whatever the corresponding term in A is. If the value in v is greater than 1 then the eigenvalue is 1. I think I might be on the right track but I'm not really sure if this is a real clear/valid argument or if it's complete or how to put it together so it sounds lilke a clear proof. For the secone part I'm not really sure how to do it. Does anyone have any advice/help.

Thanks
 
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  • #2
1. Very good so far. For any b, c, [itex](b-c)^2= b^2- 2bc+ c^2\ge 0[/itex] so [itex]b^2+ c^2\ge 2bc[/itex]. Of course, equality holds when b= c.
 
  • #3
Thanks, I think I got the first one now. Any advice on the second one? Am I on the right track?
 

Related to Solving Eigenvalue Questions for 2x2 Matrix & nxn Matrix

1. How do I find the eigenvalues of a 2x2 matrix?

To find the eigenvalues of a 2x2 matrix, you can use the characteristic equation: det(A - λI) = 0. Set up the matrix A - λI and solve for λ, the eigenvalue. Repeat this process for the second eigenvalue.

2. Is there a specific method for solving eigenvalue questions for 2x2 matrices?

Yes, there is a specific method called the characteristic equation method. This involves setting up the characteristic equation and solving for the eigenvalues as mentioned in the previous answer.

3. Can the characteristic equation method be used for nxn matrices as well?

Yes, the characteristic equation method can be used for any square matrix. However, for larger matrices, it may become more complex and time-consuming.

4. What is the significance of finding eigenvalues for a matrix?

Finding eigenvalues can help in understanding the behavior and properties of the matrix. It is also useful in solving systems of linear equations and in diagonalizing matrices.

5. Are there any shortcuts or tricks for finding eigenvalues?

There are some shortcuts for finding eigenvalues of special types of matrices, such as symmetric or triangular matrices. However, for general matrices, there is no shortcut and the characteristic equation method must be used.

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