It is proved that the rank of the product of matrix A and transposed A' is equal to the rank of A, i.e. r (AA')=r (A).

It is proved that the rank of the product of matrix A and transposed A' is equal to the rank of A, i.e. r (AA')=r (A).

It is proved that:(1) Let X1 be the solution of AX=0, then AX1=0, then A'AX1=A'(AX1)=A'0=0, then X1 is the solution of A' AX=0. So, the solution of Ax=0 is the solution of A'AX=0.(2) Let X2 be the solution of A' AX=0, then X2'A' AX2=0 is the solution of A'AX=0.

In what case is the rank of the matrix AB transposed less than the rank of the matrix A or B transposed? How to prove it?

Actually r (AB)

The rank of the product of matrices is less than or equal to the rank of any factor

This question does not need to use that conclusion can also prove ah, must use? Evidence: since K is a full-rank square matrix, it is invertible, and there is K inverse. If both sides of the equation are multiplied by K inverse, then K inverse ()=(). The first bracket is the vector group of beta, and the second bracket is the vector group of alpha. This means that the vector group of alpha can...

This question does not need to use that conclusion can also prove ah, must use? Evidence: because K is a full-rank square matrix, so reversible, there is K inverse, the two sides of the equation at the same time multiply K inverse, K inverse ()=(), the first bracket is the beta vector group, the second bracket is alpha vector group, so that the alpha vector group can...

This question does not need to use that conclusion can also prove ah, must use? Evidence: because K is a full-rank square matrix, so reversible, there is a K inverse, the two sides of the equation at the same time multiply K inverse, K inverse ()=(), the first bracket is the beta vector group, the second bracket is alpha vector group, so that the alpha vector group can...

Let A be a m*n matrix and prove that the rank of A is equal to the rank of its transposed matrix, that is, r (A)=r (A')

R (A) is equal to the rank of the row vector group of A, equal to the rank of the column vector group of A', equal to r (A')

R (A) equals the rank of the row vector group of A, equals the rank of the column vector group of A', equals r (A')

How to prove that the eigenvalue of the product of matrix A and its transposition is equal to the eigenvalue of the product of matrix A and its transposition.

Provided that A must be a matrix, otherwise there will be a difference of some zero eigenvalues
The more general conclusion for matrices is that the eigenvalues of AB and BA are exactly equal (counting algebraic multiplicity)
It's easy to prove, say, directly.
μIA
B μI
The determinant of is det (2I-AB) and is equal to det (2I-BA)

Why is the rank of the column vector less than or equal to 1?

If the column vector is not equal to 0, consider it as a matrix, and the matrix has only one non-zero column, so the column rank is 1, so that the rank of the original column vector is 1. If the column vector is zero, then the rank is O