Linear algebra: Let λ1=-1,λ2=λ3=1 be the eigenvalue of the real symmetric matrix A of third order, and let p1={0,1,1} be the eigenvector of λ1=-1 for A. The eigenvector of A belonging to eigenvalue λ2=λ3=1 is obtained, and the symmetry matrix A is obtained. Let the eigenvectors x={ x1, x2, x3} be transposed. That's why. One of the solutions is p2={1,0,0} transpose p2={0,1,-1} transpose. Why not let p2={1,1,1} transpose? Is it linear? How to judge the correlation if it is two vectors? I only know three vectors...

Linear algebra: Let λ1=-1,λ2=λ3=1 be the eigenvalue of the real symmetric matrix A of third order, and let p1={0,1,1} be the eigenvector of λ1=-1 for A. The eigenvector of A belonging to eigenvalue λ2=λ3=1 is obtained, and the symmetry matrix A is obtained. Let the eigenvectors x={ x1, x2, x3} be transposed. That's why. One of the solutions is p2={1,0,0} transpose p2={0,1,-1} transpose. Why not let p2={1,1,1} transpose? Is it linear? How to judge the correlation if it is two vectors? I only know three vectors...

The first problem is that the eigenvectors belonging to different eigenvalues are orthogonal to each other. Therefore, the eigenvectors belonging to 1 and -1 are orthogonal to each other. Assuming that the eigenvectors belonging to 1 are (x, y, z), then: y+z=0, x is arbitrary.

Given that A is a third-order real symmetric matrix with three eigenvalues, can we know how many eigenvectors there are for each eigenvalue? Does a real symmetric matrix of order 3 have three eigenvalues?

The third-order matrix must have three eigenvalues, because the eigenvalue equation |E-A|=0 is a cubic equation with one element, there must be three roots, but there may be multiple roots.Therefore, the three eigenvalues may have the same value.Each eigenvalue has infinite eigenvectors, and the eigenvectors corresponding to each eigenvalue constitute a linear space whose dimension (...

Given that P is a eigenvector of a real symmetric matrix A, the corresponding eigenvalue =?

Let a be the eigenvalue of A and P be the eigenvector of A belonging to the eigenvalue a
Then AP=aP
So P^TAP=aP^TP
Since P is the eigenvector, P=0, so P^TP >0
So a =(P^TAP)/(P^TP)

Let A be a real symmetric matrix of order n and P be an invertible matrix of order n. Given that the n-dimensional column vector α is the eigenvector of A belonging to the eigenvalue λ, then the eigenvector of [P^(-1) AP ]^T belonging to the eigenvalue λ is (). A.[ P^(-1)]αB.[ P^T ]αC. PαD.{[ P^(-1)]^T }α Let A be a real symmetric matrix of order n and P be an invertible matrix of order n. Given that the n-dimensional column vector α is the eigenvector of A belonging to the eigenvalue λ, then the eigenvector of [P^(-1) AP ]^T belonging to the eigenvalue λ is (). A.[ P^(-1)]αB.[ P^T ]αC.PαD.{[ P^(-1)]^T }α

By known A α=λα
So P^TA (P^T)^-1 P^Tα=λP^Tα
So P^TA (P^-1)^T P^Tα=λP^Tα
So (P^-1AP)^T P^Tα=λP^Tα
(B) Correct

Let A be a real symmetric matrix of order n and P be an invertible matrix of order n. Given that the n-dimensional column vector α is the eigenvector of A belonging to the eigenvalue λ, then the eigenvector of matrix (P-1AP) T belonging to the eigenvalue λ is (). A.P-1α B. PTα C.Pα D.(P-1) Tα

It is known that that n-dimensional column vector a is the eigenvector of A belonging to the eigenvalue λ,
Then: Aα=,(P-1AP) T=PTA (PT)-1,
Multiply both sides of the equation by PTα, i.e.
(P-1AP) T (PTα)= PTA [(PT)-1PT ]α= PTAα=λ(PTα),
Therefore: B.

The n-dimensional column vector a is known to be a eigenvector of A belonging to the eigenvalue λ,
Then: Aα=,(P-1AP) T=PTA (PT)-1,
Multiply both sides of the equation by PTα, i.e.
(P-1AP) T (PTα)= PTA [(PT)-1PT ]α= PTAα=λ(PTα),
Selected from: B.

The n-dimensional column vector a is known to be a eigenvector of A belonging to the eigenvalue λ,
Then: Aα=,(P-1AP) T=PTA (PT)-1,
Multiply both sides of the equation by PTα, i.e.
(P-1AP) T (PTα)= PTA [(PT)-1PT ]α= PTAα=λ(PTα),
Therefore: B.

Let α be the eigenvector of n-order symmetric matrix A corresponding to eigenvalue λ, and find the eigenvector of matrix ((P^-1) AP)^T corresponding to eigenvalue λ

Write P as Q for ease of writing
Let β=Qα
((P^-1) AP)^T=QA (Q^-1)
((P^-1) AP)^Tβ=QA (Q^-1) Qα=QAα=λQ α=λβ
So its eigenvector corresponding to the eigenvalue λ is β, i.e.(P^T)α

For ease of writing, P is transposed as Q
Let β=Qα
((P^-1) AP)^T=QA (Q^-1)
((P^-1) AP)^Tβ=QA (Q^-1) Qα=QAα=λQ α=λβ
So its eigenvector corresponding to the eigenvalue λ is β, i.e.(P^T)α