It is proved that the eigenvalues of real antisymmetric matrices are zero or pure imaginary numbers It doesn't matter if it's a bit wordy. I must understand it

As long as we can prove that the eigenvalues of Hermite matrix are real numbers
If h is a Hermite matrix, (C, x) is the characteristic pair of H, that is, HX = CX, then C = x * HX / (x * x) is a real number
Next, a is an anti Hermite matrix if and only if IA is a Hermite matrix, so the eigenvalues of the anti Hermite matrix are on the imaginary axis, and the real anti symmetric matrix is of course an anti Hermite matrix
Of course, you can also directly process AX = CX to get conj (c) = - C, which is the same as Hermite matrix, but it's necessary to understand all the above things

How to prove that the eigenvalue of real anti symmetric matrix can only be zero or pure imaginary number?
The eigenvalues of real antisymmetric matrices can only be zero or pure imaginary numbers
How to prove it?

Proof:Suppose A is a reel skew-symmetric matrix,and λ is a eigenvalue of A.
That is,Aα=λα （α=(a1,a2,...,an)'）
We multply by (α conjugate) 'on both sides
(α conjugate) 'a α = (α conjugate)' λ α = λ (α conjugate) 'α
on the other hand
(α conjugation) 'a α = (α conjugation)' (- a ') α = - (conjugation of a α)' α = - (λ α conjugation) 'α
So λ (α conjugate) 'α = - (λ α conjugate)' α = - λ (α conjugate) 'α
so λ=-λ
we suppose λ=a+bi
that is a=0
λ=0 or λ=bi

Let a be an orthogonal matrix of order n, then what is the determinant of a

Because a is an orthogonal matrix
So a (a ^ t) = E
The determinant on both sides is: | a | a ^ t | = 1
And | a ^ t | = | a|
So | a | & # 178; = 1
A = ± 1
Answer: | a | = 1 or - 1

The eigenvalues of orthogonal matrix are——

The eigenvalues of orthogonal matrices are complex numbers with modulus 1, and the conjugate complex roots appear in pairs
Conversely, any complex number satisfying the above conditions can be used as the eigenvalue of orthogonal matrix
It's just a hoax. Just give me an example
A=
0 0 1
1 0 0
0 1 0

Is it necessary to prove the invertible matrix AB = e or Ba = e?
What about orthogonal matrices?

Just prove one of them
If AB = e
Then | a | B | = E
So | a ≠ 0, | B ≠ 0
So a and B are reversible
And B = a ^ - 1 is obtained from ab = E and a ^ - 1 on both sides
Right multiply B ^ - 1 on both sides to get a = B ^ - 1

If matrix A has an eigenvalue of 0, then det (a ^ 3) =?

Knowledge point: deta is equal to the product of all eigenvalues of A
So deta = 0
So det (a ^ 3) = (DET (a)) ^ 3 = 0

A matrix of order n is invertible. Why is zero not its eigenvalue
Just explain why,

Let the eigenvalue be in and the eigenvector be a
(I-A) a = 0;
If in = 0;
Then | a | = 0;
A irreversible

Given that matrix A is an invertible second order matrix and a ^ 2 = a, then the eigenvalue of a is?

Let λ be the eigenvalue of a, then λ ^ 2 - λ is the eigenvalue of a ^ 2-a
And a ^ 2-A = 0
So λ ^ 2 - λ = 0
So λ (λ - 1) = 0
So λ = 1 or λ = 0
Because a is reversible, the eigenvalue of a is not equal to 0
So the eigenvalue of a is 1

How to find eigenvalue and eigenvector of matrix in MATLAB

a=[1 1/4;4 1]
a =
1.0000 0.2500
4.0000 1.0000
>> [v,d]=eig(a)
v =
0.2425 -0.2425
0.9701 0.9701
d =
2 0
0 0
According to the calculation process of this problem, eig is the command to find the eigenvalue and eigenvector, V is the eigenvector, is the column vector, D is the eigenvalue matrix, the main diagonal element is the eigenvalue, corresponding to the column of the eigenvector

How to get eigenvector from eigenvalue of matrix in MATLAB

Example: a = [1 2 3]
4 5 6
7 8 9]
In the command window, enter: [V, D] = eig (a), and the result is as follows:
v =
-0.2404 -0.6747 0.5185
-0.5469 -0.2339 -0.7890
-0.8019 0.7001 0.3296
d =
15.5553 0 0
0 -1.4194 0
0 0 -0.1359
It means:
The corresponding eigenvectors of eigenvalue 15.5553 are (- 0.2404, - 0.5469, - 0.8019)
The corresponding eigenvectors of eigenvalue-1.4194 are (- 0.6747, - 0.2339,0.7001)
The corresponding eigenvectors of eigenvalue -0.1359 are (0.5185, - 0.7890,0.3296)

It is proved that the eigenvalues of real antisymmetric matrices are zero or pure imaginary numbers It doesn't matter if it's a bit wordy. I must understand it