A is the nxm matrix, B is the mxn matrix, where n

A is the nxm matrix, B is the mxn matrix, where n

As known, r (AB)= r (E)= n.
Because r (AB)

Given, r (AB)= r (E)= n.
Because r (AB)

Let A be n*m type matrix, B be m*n type matrix, and I be nth order identity matrix. If AB=I, it is proved that the column vector group of B is linearly independent.

Because n = r (In)= r (AB)

Let A be n*m matrix, B be m*n, n

0

Let A and B be n*m and m*n matrices, respectively, and C=AB be a reversible matrix. It is proved that the column vector group of B is linearly independent.

It is proved that r (C)=n can be known reversibly from C.
So n = r (C)= r (AB)

It is proved that r (C)=n can be known reversibly by C
So n = r (C)= r (AB)

Let A, B be any two non-zero matrices satisfying AB=0, then (A) A's column vector group linear correlation, B's row vector group linear correlation Let A, B be any two non-zero matrices satisfying AB=0, then there must be (A) The column vector group of A is linearly correlated, and the row vector group of B is linearly correlated. (B) The column vector group of A is linearly correlated, and the column vector group of B is linearly correlated. (C) Linear correlation of row vector group of A and linear correlation of row vector group of B. (D) Row vector group linear correlation of A, column vector group linear correlation of B. My question is: D Why not? The steps in the book are: Let A be mXn matrix, B be nXs matrix, AB=0, and A and B are non-zero matrices, then R (A)+ r (B)≤n, r (A)≥1, r (B)≥1, so there must be r (A)< n and r (B)< n. Therefore, the column vector group of A is linearly related, and the row vector group of B is linearly related.

Give you a counterexample.
A =
1 0 1 2
0 1 3 4
B =
1 2
3 4
-1 0
0 -1

In triangle ABC, let vector BC=a, vector CA=b, vector AB=c (1) If the triangle is a regular triangle, prove that a*b=b*c=c*a,(2) if a*b=b*c=c*a, is the triangle ABC a regular triangle?

1
Let the side length of a regular triangle be k
Then: a·b=|BC CA|*cos (π-C)
=-K^2 cos (π/3)
=-K^2/2
B·c=|CA AB|* cos (π-A)
=-K^2 cos (π/3)
=-K^2/2
C·a=|AB BC|* cos (π-B)
=-K^2 cos (π/3)
=-K^2/2
Therefore: a·b=b·c=c·a
2
A·b=b·c
That is:|BC CA|* cos (π-C)=|CA AB|* cos (π-A)
I.e.|BC|cosC=|AB|cosA
I.e. sinAcosC = sinCcosA
I.e. sin (A-C)=0
A-C∈(-π,π), therefore: A-C=0
I.e. A=C
Similarly, from b·c=c·a, we can get: A=B
From c·a=a·b, we can get: B=C
I.e. A=B=C
The triangle ABC is a regular triangle.