Let the eigenvalues of the third-order matrix be 1,0, - 1, and the corresponding eigenvectors be (122), (2-21), (- 2-12) What does diag mean downstairs?

Let the eigenvalues of the third-order matrix be 1,0, - 1, and the corresponding eigenvectors be (122), (2-21), (- 2-12) What does diag mean downstairs?

Diag (1,0, - 1) is a third-order diagonal matrix with three diagonal elements of 1,0, - 1
Let this matrix be a and P = (1, 2 - 2), then AP = P * diag (1,0, - 1)
(2 -2 -1)
(2 1 2),
So a = P * diag (1,0, - 1) * P ^ {- 1}
Calculated
A=(-1/3 0 2/3)
(0 1/3 2/3)
(2/3 2/3 0)

Find the eigenvalues and corresponding eigenvectors of matrix A = [1,1,1; 1,2,3; 3,2,1]

The eigenvalues are - 1,0,5, and the corresponding eigenvectors are (0, - 1,1), (1, - 2,1), (6,13,11)

The sum of the elements in each row of the third order matrix A is 3, and 1 - 1 {0} {- 1} is the solution of AX = 0. Find the eigenvalue and eigenvector of A

I see! Because the sum of the elements in each row of the third-order matrix A is 3, so a (1,1,1) ^ t = 3 (1,1,1) ^ T. that is, 3 is the eigenvalue of a, and (1,1,1) ^ t is the eigenvector of a belonging to eigenvalue 3. And because (1,0,1) ^ t, (- 1, - 1,0) ^ t are the solutions of AX = 0, and they are linearly independent, so 0 is of A

It is known that 0,1, - 1 are the eigenvalues of the third-order matrix A, and ξ 1, ξ 2 and ξ 3 are the corresponding eigenvectors. If P = (3 ξ 3,2 ξ 2, ξ 1), then p ^ - 1AP=

It is known that the three eigenvalues of 3-order real symmetric matrix A are 1, - 1,0, and how to find the eigenvectors corresponding to 1, - 1

From the eigenvectors of - 1 and 1, according to the orthogonality of the eigenvectors of the real symmetric matrix, the eigenvectors corresponding to 0 are obtained. The three eigenvectors are arranged in turn to form the similarity transformation matrix P. then from PAP-1 = a, a can be obtained, where P-1 is the inverse matrix of P and a is the diagonal matrix with three eigenvalues arranged in turn

The three eigenvalues of 3-order real symmetric matrix A are 2,5,5, and the eigenvectors of a belonging to eigenvalue 2 are (1,1,1)
Then the eigenvector of a belonging to eigenvalue 5 is?

The eigenvectors of real symmetric matrices belonging to different eigenvalues are orthogonal to each other
So the eigenvector of a belonging to eigenvalue 5 is orthogonal to (1,1,1)
That is, X1 + x2 + X3 = 0
The basic solution system: A1 = (1, - 1,0) ', A2 = (1,0, - 1)'
So the eigenvector of a belonging to eigenvalue 5 is
K 1A 1 + K 2A 2, K 1, K 2 are arbitrary constants which are not all zero

Let the eigenvalues of the third-order real symmetric matrix a be 1,1, - 1, and the corresponding eigenvectors of eigenvalue 1 have (1,1,1), (2,2,1), then find the matrix A

Because the eigenvectors of symmetric matrix belong to different eigenvalues are orthogonal
So let the eigenvector of eigenvalue-1 be (x1, X2, x3) ^ t
Then X1 + x2 + X3 = 0
2x1+2x2+x3=0
The fundamental solution system of the equations is ζ 3 = (1, - 1,0) ^ t
So the eigenvector of eigenvalue - 1 is C (1, - 1,0) ^ t, and C is a nonzero constant
Order p=
1 2 1
1 2 -1
1 1 0
Then p is reversible and P ^ - 1AP = diag (1,1, - 1)
So there is a = pdiag (1,1, - 1) P ^ - 1=
0 1 0
1 0 0
0 0 1
Note: in order to avoid finding the inverse of P, the eigenvectors of eigenvalue 1 can be orthogonalized, and then the three vectors can be unitized

Is the multiplicity of eigenvalues of a matrix equal to the number of corresponding eigenvectors
I see a problem in exercise book is not equal, a double root has only one eigenvector
But students say that the multiplicity of eigenvalues of a matrix is equal to the number of corresponding eigenvectors
What the hell is going on?

This is the problem of diagonalization of matrix
Generally, there are: the number of eigenvectors ≤ the multiplicity of eigenvalues
The sufficient and necessary condition of diagonalization is that the multiplicity of eigenvalues is equal to the number of eigenvectors of corresponding eigenvalues

What are the special properties of eigenvalues and eigenvectors of real symmetric matrices?

The eigenvalues of real symmetric matrices are all real numbers
Eigenvectors belonging to different eigenvalues are orthogonal
K multiple eigenvalues have K linearly independent eigenvectors

Eigenvalues of n-order real symmetric matrix
The book shows that the eigenvalues of n-order real symmetric matrix must be real numbers by simple proof. If we expand the characteristic polynomial, are they all linear factors (in the range of real numbers), and there is no quadratic factor that cannot be disassembled because the discriminant is less than 0?

Yes