If two vectors parallel to the same vector are collinear vectors, the answer is right. If the same vector is zero, how to explain?

The zero vector has multiple directions, right?

A vector whose modulus is equal to zero is called a zero vector, and is denoted as 0. Note that the direction of the zero vector is arbitrary. But we stipulate that the direction of the zero vector is parallel to, but not perpendicular to, any vector. Because the direction is arbitrary, the direction of the zero vector is any direction in space. There are multiple directions, which can be correct, and preferably any direction.

How can the rank of a matrix equal to 1 be decomposed into the product of column vector and row vector When can a matrix be decomposed into the product of a column vector and a row vector?

Let A be n*n matrix, rank (A)=1
Let A=(a1,..., an), ak, k=1,..., n be the n-dimensional column vector
Let a1 be not a zero vector, then rank (A)=1
Ak=bk*a1, number of bk
Then A =(a1, b2*a1,..., bn*a1)= a1*(1, b2,..., bn)
If A=uv, u is the column vector, v is the row vector, and u and v are not zero vectors, record v=(v1,..., vn)
Then rank (A)=rank (uv)=rank (u (v1,..., vn))
=Rank (uv1,..., uvn)=1

It is known that the unit vector A (0,3) B (2,0) C (-1,3) opposite AB+2AC is How to change from (0,3) to (0,1)? It is known that the unit vector A (0,3) B (2,0) C (-1,3) opposite AB+2AC is How did you change from (0,3) to (0,1)?

It is known that the unit vector A (0,3) B (2,0) C (-1,3) opposite AB+2AC is
Vector AB=(2,0)-(0,3)=(2,3)
Vector AC=(-1,3)-(0,3)=(-1,0),
Vector AB+2AC =(0,-3),
The module of vector AB+2AC is 3,
AB+2AC The opposite unit vector is =(0,-3)/-3=(0,1)
Note: The unit vector in the same direction of vector a is the modulus of a divided by a, and the unit vector in the opposite direction of vector a is the inverse number of the modulus of a divided by a.

Known vector A, B, and AB = A+2 B, BC=-5 A+6 B, CD =7 A-2 B, must be collinear () A.A, B, D B.A, B, C C.B, C, D D.A, C, D Known vector A, B, and AB = A+2 B, BC=-5 A+6 B, CD =7 A-2 B, then certain collinear () A.A, B, D B.A, B, C C.B, C, D D. A, C, D

∵

BD =

BC+

CD=-5

A+6

B+7

A-2

B=2

A+4

B=2

AB
∴

AB∥

BD
Because straight line AB and BD have common point B
So points A, B, D are in the same straight line.
Therefore, 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). A linear algebraic problem.

Let A be a matrix of m×n.
It can be proved that r (A'A)=r (A) by proving that Ax=0 and A' Ax=0 are two homogeneous equations of n-element.
1. Ax=0 must be the solution of A'Ax=0.
2. A'Ax=0→x' A'Ax=0→(Ax)' Ax=0→Ax=0
Therefore, the two equations are the same solution.
Similarly r (AA')= r (A')
In addition, r (A)=r (A')
So r (A)=r (A')=r (AA')=r (A' A)

If two vectors parallel to the same vector are collinear vectors, the answer is right. If the same vector is zero, how to explain?