Linear Algebra -- what is the relationship between positive definite quadratic form and semi positive definite quadratic form For example, is a positive definite quadratic form also a semi positive definite quadratic form? Why? X1 ^ 2 + x2 ^ 2 + X3 ^ 2, for any real vector x, x ^ tax > = 0, is it also a positive semidefinite quadratic form?

Linear Algebra -- what is the relationship between positive definite quadratic form and semi positive definite quadratic form For example, is a positive definite quadratic form also a semi positive definite quadratic form? Why? X1 ^ 2 + x2 ^ 2 + X3 ^ 2, for any real vector x, x ^ tax > = 0, is it also a positive semidefinite quadratic form?

The real quadratic form of positive definite requires x ^ ax > 0 for any [non-zero] vector x, and the semi positive definite requires ≥ 0, so the positive definite is semi positive definite, just as human beings are primates at the same time
If the space in question is three-dimensional, then x 1 ^ 2 + x 2 ^ 2 + x 3 ^ 2 is a positive definite quadratic form

Let the rank of matrix A of order n be 1, and prove that a ^ 2 = tr (a) a

Knowledge point: R (a) = 1 if and only if there are n-dimensional non-zero column vectors α, β, such that a = α, β ^ t
So there is a ^ 2 = (α β ^ t) (α β ^ t) = α (β ^ t α) β ^ t = (β ^ t α) α β ^ t = tr (a) a

In quantum mechanics, TR is trace, a, B and C are matrices. It is proved that: (1) tr (AB) = tr
In quantum mechanics, TR is trace, a, B and C are matrices. It is proved that: (1) tr (AB) = tr (BA); (2) tr (ABC) = tr (BCA) = tr (CAB)

You are so boring. Who gave you such a complicated formula derivation when you asked such a question
The first formula directly writes out the expression of (AB) II, and then sums it up. It is easy to prove that it is equal to the sum of (BA) II
The second is proved directly by the first formula. If a is regarded as a matrix and BC as a matrix, the first equation can be proved
Then consider B as a matrix and Ca as a matrix to prove the second formula

If f (E) = n, for any a and B, f (AB) = f (BA) and f (AA + BB) = AF (a) + BF (b), try to prove that f (a) = tr (a)
F is a function defined on P (n * n)

TR is a trace, which is the algebraic sum of the main diagonal elements of a matrix
In fact, this problem is about trace operation
The trace of two matrices satisfies tr (AB) = tr (BA), tr (AA + BB) = ATR (a) + BTR (b),
that is

Prove that tr (AB) = tr (BA)
Where a and B are not necessarily square matrices

Under what condition does matrix multiplication satisfy commutative law?
Such as the title

1: When one of the two matrices is a scalar matrix (the scalar matrix refers to the same number on the main diagonal which is not zero, and the other terms are all zero, it is a square matrix), then the matrix multiplication satisfies the commutative law
2: When two matrices are equal or one of them is a zero matrix, the matrix multiplication satisfies the commutative law, and the identity matrix is a quantity matrix
3: If a and B satisfy AB = a + B, then the product of a and B is commutative, that is, ab = ba

Multiplication of transposed matrix in linear algebra
What happens after (ABC) t unfolds?

First product then transpose, and finally, for each transpose, and vice versa
(C) T (b) t (a) t, the proving process is a bit complicated. Generally speaking, this can be remembered as a rule

Multiplication of linear algebraic matrix
Given that matrix A is a square matrix of order n, and the square of a is 2a, then the nth power of a =?

A^2=2A
A^3=AA^2=2A^2=2^2A
.
A^n=2^(n-1)A

Let a and B be matrices with M rows in linear algebra, and prove that Max {R (a), R (b)} ≤ R [(a, b)] ≤ R (a)+
linear algebra
Let a and B be matrices with M rows
max{R(A),R(B)}≤R[(A,B)]≤R(A)+R(B)

A. The column vector of B can be expressed linearly by the column vector of (a, b)
So r (a)

Proof of Linear Algebra: 1. Let a and B be matrices of order n, tangent to the square of a-2ab = E. prove that ab-ba + A is invertible

Certification:
A^2-2AB=E
A (A-2B)=E
It shows that a is reversible and the inverse of a is a - 2b
The deformation of the above formula gives B = (a ^ 2-e) / (2a)
It is simplified by ab-ba + a
Ab-ba + a = a (a ^ 2-e) / (2a) - (a ^ 2-e) a / (2a) + A
Ab-ba + a = a is obtained
Can be proved
Hope to adopt, thank you