The module length of vector A+B = the module length of vector A-B =2 times the module length of vector A The angle between the first two terms is? The module length of vector A+B = the module length of vector A-B =2 times the module length of vector A The angle between the first two terms of the formula is?

The module length of vector A+B = the module length of vector A-B =2 times the module length of vector A The angle between the first two terms is? The module length of vector A+B = the module length of vector A-B =2 times the module length of vector A The angle between the first two terms of the formula is?

120 Degrees
The translation process by vector operation can be solved in a parallelogram, from the first two equal, the diagonal is equal, is rectangular, and the diagonal is twice the length of one side

How to Calculate the Module Problem of Vector A Subtracting Vector B

If both vector A and vector B are known
Direct coordinate minus one minus, square sum root
If not, only the module/length of A, B is known, and the inner product A·B is known
Then |A-B |^2=|A |^2 B |^2-2 A·B
Then substitute |A|,|B|, A·B on the right
Then open the root.

If both vector A and vector B are known
Direct coordinate minus one minus, square sum root
If we don't know, we only know the module/length of A, B and the inner product A·B.
Then |A-B |^2=|A |^2 B |^2-2 A·B
Then substitute |A|,|B|, A·B on the right
Then open the root.

How to calculate the vector |a+b|

0

Given that A is a real symmetric matrix of order n, for any n-dimensional vector X, there is X'(the transposition of X) AX=0. It's all XTAX=0.

That's not right. A is a 0 matrix. How can we multiply A by the inverse? Isn't that bullshit?
First of all, if A is a real symmetric matrix of order n, then A must be similar to a diagonal matrix. Let B=P^(-1) AP, P^(-1) be the inverse of P, then A=PBP^(-1), for any n-dimensional vector X, there is X'AX=0, then we can deduce that all diagonal elements of B are 0, that is, B=0. According to A=PBP^(-1), we can know A=0.

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

From the eigenvectors of -1 and 1, according to the orthogonality of the eigenvectors of the real symmetric matrix, the corresponding eigenvectors of 0 are obtained, and the three eigenvectors are arranged in sequence to form a similar transformation matrix p. Then, from PaP-1=A, we can obtain A, where P-1 is the inverse matrix of P, and a is the diagonal matrix composed of three eigenvalues arranged in sequence. Do you understand?

From the eigenvectors of -1 and 1, according to the orthogonality of the eigenvectors of the real symmetric matrix, the corresponding eigenvectors of 0 are obtained, and the three eigenvectors are arranged in sequence to form a similar transformation matrix p. Then, from PaP-1=A, we can obtain A, where P-1 is the inverse matrix of P, and a is the diagonal matrix composed of three eigenvalues arranged in sequence.

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.