Using MATLAB to find matrix eigenvalue and eigenvector A=[1 5 1 3 3 1/5 1 6 4 2 1 1/6 1 3 4 1/3 1/4 1/3 1 2 1/3 1/2 1/4 1/2 1 ]

Using MATLAB to find matrix eigenvalue and eigenvector A=[1 5 1 3 3 1/5 1 6 4 2 1 1/6 1 3 4 1/3 1/4 1/3 1 2 1/3 1/2 1/4 1/2 1 ]


Input: x = [1 5 1 3 3; 1 / 5 1 6 4 2; 1 1 / 6 1 3 4; 1 / 3 1 / 4 1 / 3 1 2; 1 / 3 1 / 2 1 / 4 1 / 2 1] eig (x) output: ans = 6.3156 - 0.5309 + 2.7527i - 0.5309 - 2.7527i - 0.1269 + 0.4050i - 0.1269 - 0.4050i



Using MATLAB to find the eigenvector corresponding to the maximum eigenvalue of matrix
Matrix: a = [1 2 3 4; 1 / 2 1 2 3; 1 / 3 1 / 2 1 2; 1 / 4 1 / 3 1 / 2 1], find the eigenvector corresponding to the maximum eigenvalue of the matrix!


>> [d,v]=eig(A)
d =
-0.8135 -0.8493 -0.8493 -0.7038
-0.4826 0.0004 - 0.4268i 0.0004 + 0.4268i 0.5934
-0.2787 0.2498 - 0.0499i 0.2498 + 0.0499i -0.3592
-0.1661 0.0252 + 0.1758i 0.0252 - 0.1758i 0.1535
v =
4.0310 0 0 0
0 -0.0017 + 0.3533i 0 0
0 0 -0.0017 - 0.3533i 0
0 0 0 -0.0276
The maximum characteristic value is 4.0310
The first column of V is the corresponding eigenvector



Finding eigenvalues and eigenvectors of matrix (0 a 0 a)


┏0 a┓
Γ 0 a Γ is an upper triangular matrix with eigenvalues: λ 1 = 0, λ 2 = a
① , a = 0, λ 1 = λ 2 = 0, eigenvector: any (x, y) ′≠ 0
② A ≠ 0, λ 1 = 0, eigenvector: (k, 0) ′ (K ≠ 0)
λ 2 = a, eigenvector: (k, K) ′ (K ≠ 0)



Let a and B be 3-dimensional column vectors and satisfy the transpose * b = 5 of a, then what is the eigenvalue of transpose * a of matrix B
Eigenvalues of matrix b'a


B'a is not a matrix, so it should not be so simple. I guess ba'a is still 5 because ba'b = 5B = > its eigenvalue (because if the eigenvector is u and the eigenvalue is V, then ba'u = vu, but a'u



Let a = (α, 2, γ 2, 3, γ 3), B = (β, γ 2, γ 3), where α, β, γ 2, γ 3 are all three-dimensional column vectors, and / A / = 18. / B / = 2. Find / a-b/


The answer is 2, using the determinant property as shown in the figure. The economic mathematics team will help you solve it, please adopt it in time



On the problem of Linear Algebra: let α and β be 3-dimensional column vectors, and β ^ t be the transpose matrix of β. Then α * β ^ t is a matrix of order n with rank 1. Why?


I'm here to give you a proof: R (α * β ^ t) can be regarded as two matrices to multiply. A three-dimensional column vector is actually a matrix with three rows in one column. Then there is the formula R (AB) ≤ min (R (a), R (b)). However, R (a) r (b) is a non-zero one-dimensional column vector, and its rank is 1, and R (α * β ^ t) this



Let α 1 and α 2 be eigenvectors of matrix A belonging to different eigenvalues. It is proved that α 1 + α 2 are not eigenvectors of matrix A


It is proved that let α 1 and α 2 be the eigenvectors of a with different eigenvalues λ 1 and λ 2
Then a α 1 = λ 1 α 1, a α 2 = λ 2 α 2, and λ 1 ≠ λ 2
Suppose α 1 + α 2 are the eigenvectors of a belonging to the eigenvector λ
Then a (α 1 + α 2) = λ (α 1 + α 2)
So λ 1 α 1 + λ 2 α 2 = λ (α 1 + α 2)
So (λ - λ 1) α 1 + (λ - λ 2) α 2 = 0
Because the eigenvectors of a belong to different eigenvalues are linearly independent
So λ - λ 1 = 0, λ - λ 2 = 0
Therefore, λ = λ 1 = λ 2 is contradictory to λ 1 ≠ λ 2
So α 1 + α 2 is not the eigenvector of A



Teacher, how can I prove that for each eigenvalue, the linearly independent eigenvectors that a matrix can have will not exceed the multiplicity of this eigenvalue


This is more troublesome



Eigenvalue problems of symmetric positive definite matrices 4
Recently, I'm a little ahead of myself in mathematics. I can't think of some problems clearly
Now I know that a symmetric positive definite matrix of order n must have n positive eigenvalues
3. For a positive definite matrix A, can it have n nonnegative eigenvalues?
(positive definite matrices don't require symmetry, am I right?)
A lot of questions are asked, mainly to know when there will be a series of orthogonal eigenvectors to represent the whole space


For an asymmetric matrix A, its eigenvalues may appear imaginary numbers, but no matter how it is, there are always imaginary numbers
μ_ min



Is it necessary and sufficient for this matrix to be positive definite if all eigenvalues are positive?
Such as the title


Yes, it's necessary and sufficient. In addition, the order of the bamboo formula is greater than 0. The original formula y = XT a x > 0 is necessary and sufficient