- #1
buzzmath
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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
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