Will the rank of transpose matrix change compared with the original matrix

Will the rank of transpose matrix change compared with the original matrix

Let the transpose of matrix a be at
be
Rank (a) = rrank (a) = crank (a) rank = row rank = column rank
also
rrank(A)=crank(AT)
crank(A)=rrank(AT)
So rank (at) = rrank (at) = crank (a) = rank (a)
Therefore,

How to judge whether two matrices are similar?

For example, if the similar matrix must have the same eigenvalue, the same trace, the same determinant, then a can only be excluded, B, C, D and matrix A have the same trace, the same determinant and the same eigenvalue

Miss Liu, how to judge whether two matrices are similar?
Matrix:
2 1 -1
A= 1 2 1
-1 1 2
2 0 1
B=-1 3 1
2 0 1
It is easy to know that a can be diagonalized, and its eigenvalues are: 3,3,0
Why is diagonal matrix expressed as:
3 0 0
0 3 0
0 0 0
instead of
0 0 0 3 0 0
0 3 0 or 0 0 0
0 0 3 0 0 3
Please give me some advice!

These expressions are OK
It's just a habit to put positive numbers before, negative numbers after, and 0 at the end
It should be noted that the column vector (eigenvector) of the invertible matrix P must correspond to the column of the eigenvalue!

How to judge whether two matrices are similar,

If and only if their characteristic matrices are equivalent
This conclusion is beyond the scope of linear algebra
The necessary conditions are the same determinant, the same eigenvalue and the same trace
When both matrices are diagonalizable, the similarity is only if and only if the eigenvalues are the same

A is a real matrix of order m * n, R (a) = n

It doesn't have to be that complicated
For any m-dimensional column vector x
X^T(AA^T)X
= (A^TX)^T(A^TX)
>=0 (nonnegativity of self inner product of real vector)
So AA ^ t is positive semidefinite#
To show that AA ^ t is not positive definite
Then R (a ^ t) = R (a) = n < m (number of columns of a ^ t)
So a ^ TX = 0 has a nonzero solution x0
In this case, x0 ^ t (AA ^ t) x0 = = (a ^ tx0) ^ t (a ^ tx0) = 0
So AA ^ t is not positive definite

Given that a is a real anti symmetric matrix, I-A ^ 2 is proved to be a positive definite matrix

This leads to the conclusion that the eigenvalues of real antisymmetric matrices are zero or pure imaginary numbers
So the eigenvalue of I-A ^ 2 is 1 or 1 - (KI) ^ 2 = 1 + K ^ 2 > 0
So I-A ^ 2 is a positive definite matrix

When is a matrix invertible and positive definite?

1. Under what conditions is a matrix invertible,
Let the matrix be m
Then M is a square matrix and | m | is not equal to 0
2. Let m be a symmetric matrix with real coefficients of order n if x = (x) for any nonzero vector_ 1,...x_ n) If x ′ MX > 0, it is called m positive definite. A positive definite matrix can be reduced to a standard form, i.e. identity matrix, under consistent transformation. All symmetric matrices (or Hermitian matrices) with eigenvalues greater than zero are also positive definite matrices
Another definition: a real symmetric matrix. Positive definite quadratic form f (x1, X2,...) The matrix A (a ′) with X ′, x n) = x ′ ax is called positive definite matrix. Theorem 1: the sufficient and necessary condition for a symmetric matrix A to be positive definite is that all eigenvalues of a are positive. Theorem 2: the sufficient and necessary condition for a symmetric matrix A to be positive definite is that all order principal subformulas of a are positive. Theorem 3: the sufficient and necessary condition for any matrix A to be positive definite is that a is congruent with the unit matrix

Let a and B be positive definite matrices of order n, then AB is: A. real symmetric matrix. B. positive definite matrix. C. invertible matrix. D. orthogonal matrix

This (c) is correct
Because a, B are positive definite
So | a | > 0, | B | > 0
So | ab | = | a | B | > 0
So AB is reversible

Excuse me: A, B are n-order real symmetric matrix, and are positive definite, then AB must be: a symmetric matrix B positive definite matrix C invertible matrix D orthogonal matrix
Why and why not

If it is positive definite, the main and minor expressions of the order are greater than 0
So | a ≠ 0, | B ≠ 0
So | ab | = | a | B | ≠ 0
So AB is reversible
So (c) is correct

If n-order matrices A and B are positive definite, then a and B must be () A. symmetric matrix B. orthogonal matrix C. positive definite matrix D. invertible matrix

Dear landlord
[positive solution]
This (d) is correct
Because a, B are positive definite
So | a | > 0, | B | > 0
So | ab | = | a | B | > 0
So AB is reversible
I wish you a happy New Year!
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