Other positive definite properties and quotients of Hermite quadratic form
1. Other positive definiteness
1. Definition
Draw relevant conclusions about "negative definite", "semi-positive definite" and "semi-negative definite"
Note: If an H matrix (or its corresponding quadratic form) is neither semi-negative definite nor semi-positive definite, Then call this H matrix (or this quadratic f) indefinite .
2. Other positive determination methods
In "[Matrix Theory] Hermite Quadratic Form (2)" , we discussed the three theorems and non-equivalent conditions of positive definite quadratic forms. These properties and theorems can be naturally analogous to other positive definite judgment methods. on.
(1) Judgment method of negative qualitative
The following is an example of negative qualitative judgment . For the proof, please refer to the previous blog post, so I won’t repeat it here:
[Three Theorems]
- If a diagonal matrix is negative definite, then all diagonal elements of the diagonal matrix are negative numbers .
- If a matrix A is conjugated to a matrix B, and A is negative definite ↔ then B is also negative definite.
- If a matrix A is conjugated to a diagonal matrix Λ, then A is negative definite ↔ all diagonal elements on the diagonal matrix Λ are negative numbers .
[Five conditions]
- A is negative definite
- The eigenvalues of A are all less than zero
- A and- I conjugate contract
- There is a reversible matrix P such that A = -P H P
- The odd order main sub-expressions of A are all less than zero, and the even order main sub-expressions are all greater than zero .
[Note]
Most theorems and conditions are just an equivalent analogy. The conclusion about the "sequential principals" needs attention:
because A is negative definite ↔ -A is positive definite ↔ -all sequential principals of A are To be greater than zero,
examine the second-order main sub-expression of -A, and find that the negative signs in two rows (columns) can be proposed, and the negative negative makes the positive offset, so the second-order main sub-expression of -A and the second-order main sub-expression of A The values are the same.
[Example] Proof of quadratic form/matrix positive definiteness
In the proof of positive definiteness, because there are many theorems, properties and conditions that can be used, it is often inaccurate to choose the idea. You can basically refer to the following ideas for trial:
- First consider the definition of positive definiteness, which is the basis of mathematical concepts
- For the relevant proof of the H array, consider the "conjugate contract" relationship and the "diagonal" method
- Conclusions about using eigenvalues
[1]: According to the requirements of the title, first set up the positive definite matrix and the semi-positive definite matrix, and write the form of the matrix sum
[2]: According to the definition, substitute a certain matrix M into X 0 H MX 0 to perform symbolic derivation
[3 ]: The sum of a positive number and a non-negative number must be a positive number
(2) Judgment method of positive semi-definiteness
[Three Theorems]
[Five Conditions]
- Note that the matrix form given in the third point is the general formula of the canonical form of the positive semi-definite matrix
- In the fourth point, the matrix P neither restricts whether it is invertible , nor restricts its dimension and shape .
It can be verified now. For a matrix A = P H P, it is a positive semi-definite matrix (the corresponding quadratic form is positive semi-definite)
① To prove that A is an H matrix, there is obviously A H = A, no longer Discussion
② It is necessary to prove that for any vector X 0 (here, whether the vector X 0 is a zero vector does not affect the proof), X 0 H AX 0 ≥0
essentially uses the non-negativity of the inner product , so for the matrix P No restrictions are required.
- In the fifth point, the " order " restriction is removed for the main sub-style
Here is not a rigorous proof of the theorem of the fifth point. Let’s briefly discuss that “ sequential principal is greater than or equal to zero” can only be used as a necessary but insufficient condition for the matrix to be positive semi-definite
[insufficiency]: For the matrix shown below (diagonal matrix ), according to the diagonal elements, it is obvious that it is a semi-negative definite matrix, but the principal and sub-forms of all orders are ≥ 0
. The value is negative, thus demonstrating that the matrix is not positive semi-definite.
2. Rayleigh Quotient
In a series of discussions from similar standard forms to quadratic forms, we have introduced many new concepts and theorems, and they are all discussed around a very important concept-eigenvalues. But so far, we have not had a particularly good calculation method for eigenvalues, and the method of solving eigenvalues by solving characteristic equations is too troublesome.
Therefore, we introduce the concept of Rayleigh quotient to assist in solving the eigenvalues of the matrix .
1. Definition
2. Related Theorems
(1) Theorem description
(2) Theorem proof
The two maximum values of the eigenvalue correspond to the two maximum values of the Rayleigh quotient respectively. The proof ideas of the two are roughly the same, and only the minimum value is used to prove the following.
[1]: Because A is an n-th order H matrix , A has n pairwise orthogonal unit eigenvectors , and these n pairwise orthogonal unit eigenvectors belonging to C n space can form the C n space A set of orthonormal vector bases.
This is the nature of the H matrix. Readers who are not familiar with it can move to the blog post "[Matrix Theory] Hermite Quadratic Form (1)" , which explains and proves the nature of the H matrix.
[2]: Because X ∈ C n , X can also be expressed linearly with the basis of η 1 , η 2 ,..., η n , and the result of linear expression is substituted into X H AX for calculation.
According to the associative law of matrix multiplication , AX can be calculated first, because this set of bases η 1 , η 2 ,..., η n are exactly the eigenvalues of matrix A, so the X matrix is divided into columns and the matrix A is calculated The result of [λ 1 k 1 η 1 , λ 2 k 2 η 2 ,..., λ n k n η n ] can be obtained.
[3]: The X vector is in C n , and the result of AX (that is, a vector in the column space of matrix A) is also in C n . The operation of the form X H (AX) is to solve X in C n space The standard inner product of AX and AX ; according to our previous discussion of inner product , the inner product of two vectors can be converted into the inner product of coordinates under the same set of basis (η 1 , η 2 ,..., η n ) .
[4]: Because λ 1 is the smallest eigenvalue, we can scale all λ i
This step is also the only difference between the maximum and minimum proofs.
If you want to prove the conclusion of the maximum value, you only need to use λ n to amplify all λ i .
[5]: Σk i conjugate k i in the definition of the standard inner product is the inner product of the X vector and itself, that is, X H X
[6]: Synthesize 2-5 formula, that is, there is the inequality X H AX≥λ 1 X H X, and because X≠θ, X H X>0, divide both sides of the inequality by X H X at the same time , you get 6 boxes The conclusion.
Note that the establishment of the formula λ 1 ≤R(x) only shows that λ 1 is a lower bound of R(x) , but it does not mean that λ 1 is the lower bound (minimum value) of R(x).
We also need to prove that there is a certain X 0 that can satisfy R(X 0 ) = λ 1 .
[7]: found with [lambda] . 1 that corresponds to a feature vector [eta] . 1 , substitution operations prove to [lambda] . 1 is a R (X) of the infimum.
(3) Description
① This theorem only gives the idea of solving the two maximum values of the eigenvalues , and some of the eigenvalues in the middle can also be solved with the help of R(X), but the calculation is more complicated.
② The important condition for the establishment of the theorem is that the matrix A is an H matrix , because if it is not an H matrix, first of all there is no guarantee that R(X) is a real-valued function, let alone the solution of its most value.