![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Given that an eigenvalue of matrix A = 1a23 is - 1, find another eigenvalue of matrix A and an eigenvector belonging to λ
The characteristic polynomial of matrix A = 1a23 is f (λ) = (λ - 1) (λ - 3) - 2A, and a = 4 is obtained from F (- 1) = 0. Let f (λ) = 0, then λ = - 1 or λ = 5. By solving the equations {, (5 − 1) x − 4Y = 0 − 2x + (5 − 3) y = 0, we can get a set of nonzero solutions {, x = 1y = 1, so another eigenvalue of matrix A is 5, and an eigenvector belonging to 5 is e = 11
Which God can help me to calculate the maximum eigenvalues of the following two matrices and the corresponding eigenvectors? Thank you!
1\x050.667\x050.429
1.5\x051\x050.667
2.333\x051.5\x051
1\x050.25\x051/1.5\x052.333
4\x051\x051.5\x059
1.5\x051/1.5\x051\x054
1/2.333\x051/9\x051/4\x051
>> [d,v]=eig([1\x050.667\x050.4291.5\x051\x050.6672.333\x051.5\x051])d =-0.3408 -0.1698 + 0.2940i -0.1698 - 0.2940i-0.5172 -0.2602 - 0.4463i -0.2602 + 0.4463i-0.7851 0.7860 0.7860 v =3.0008 0 0 0 -0.0...
Given that the three eigenvalues of the third-order matrix are 1, - 1,2, then the eigenvalues of a ^ 2 + 2A + 3E are
The eigenvalue of a ^ 2 + 2A + 3E is
1.1²+2+3=6
2.（-1）²-2+3=1-2+3=2
3.2²+2×2+3=4+4+3=11
Namely
The eigenvalues are: 6,2,11
RELATED INFORMATIONS
- 1. Given that a is a matrix of order n and satisfies the equation A2 + 2A = 0, it is proved that the eigenvalue of a can only be 0 or - 2
- 2. Let a be a matrix of order 4 × 3, C = AAT, then | C|=
- 3. It is known that the fourth-order square matrix a satisfies | A-E | = 0, square matrix B = a ^ 3-3a ^ 2, BB ^ t = 2E, and | B|
- 4. Let the eigenvalues of the fourth-order square matrix a be 1 / 2,1 / 3,1 / 4,1 / 5, then | a ^ - 1-e | =?
- 5. Let the eigenvalues of the third-order square matrix a be - 1, - 2, - 3, and find a *, a & # 178; + 3A + E
- 6. Proof: let n-order square matrix a satisfy a ^ 2 = a, and prove that the eigenvalue of a is 1 or 0
- 7. Let a be a 3-order square matrix with eigenvalues of 1,2, - 3, find the eigenvalues of a ^ 2-3a + A ^ - 1 + 2E, and | a ^ 2-3a + A ^ - 1 + 2e |, hope to give the process
- 8. Let n square matrix a satisfy a ^ 2 = A and E be the identity matrix of order n, and prove that R (a) + R (A-E) = n
- 9. Let a be a square matrix of order n, and satisfy AA ^ t = E and | a | = - 1, prove that the determinant | e + a | = 0 My question is why |A| |E+A'| = |A| |(E+A)'| = |A| |E+A|
- 10. Let a be a square matrix of order n, AA = a, and prove that R (a) + R (A-E) = n
- 11. It is known that a = (α, β, γ), B = (α + β + γ, α + 2 β + 4 γ, α + 3 β + 9 γ), where α, β and γ are 3-dimensional sequence vectors, | a | = m, find | B|
- 12. If a is a 3-order square matrix, α is a 3-dimensional column vector If a is a 3-order square matrix, α is a 3-dimensional column vector, and the group of constant vectors α, a α, a & # 178; α is linearly independent, and a & # 179; α = 5A α - 3A & # 178; α, we prove that the matrix B = (α, a α, a ^ 4 α) is invertible My idea is: we can transform the topic into α, a α, a & # 178; α linearly independent, a α, a & # 178; α, a & # 179; α linearly related, and prove α, a α, a ^ 4 α linearly independent. But we still haven't done it
- 13. The existence matrix has a double root eigenvalue, which only corresponds to the eigenvalue of a linearly independent eigenvector
- 14. It is proved that if n-order matrix a satisfies AAT = E and | a | = - 1, then matrix a must have an eigenvalue of - 1
- 15. A is a nonzero matrix of order n. given a ^ 2 + a = 0, can we deduce that - 1 is an eigenvalue of a?
- 16. Let the sum of the elements in each column of the invertible matrix A of order 3 be 4, and find an eigenvalue of A
- 17. Given that the eigenvalues of matrix A of order 3 are 1,2,3, find | a * + A ^ 2 + 3E| Such as the title
- 18. Let a and B be matrices of order n, ab = a + B. It is proved that: (1) A-E and B-E are invertible; (2) AB = ba
- 19. It is proved that AB is similar to ba
- 20. (1) Given cos0 = - 1 / 2, find the value of sin0, tan0 (0 is the letter with a cross in the middle) (2) given tan0 = 2. Find sin0-2cos0 / 3cos0 + (1) Given cos0 = - 1 / 2, find the value of sin0 and tan0 (0 is the letter with a cross in the middle) (2) known tan0 = 2. Find the value of sin0-2cos0 / 3cos0 + 5sin0 (0 is explained above)