Why can the matrix rank of row vector multiply column vector be less than or equal to 1?

Why can the matrix rank of row vector multiply column vector be less than or equal to 1?

The properties according to rank are r (AB)<=min (r (A), r (B))
The rank of both row and column vectors is 1, so r (AB)<=1, i.e. the product is less than or equal to 1

Why is the rank of A less than or equal to the rank of B when the column vector group of A can be represented by the column vector group of B?

When the column vector group of matrix A can be represented by the column vector group of matrix B
There must be a C with A=BC.(You write each expression and combine them to get this expression.)
R (A)= R (AB)<= min {R (A), R (B)}<= R (B)

When the column vector group of matrix A can be represented by the column vector group of matrix B
There must be a C with A=BC,(You write each expression and combine it to get this expression)
R (A)= R (AB)<= min {R (A), R (B)}<= R (B)

Given that the absolute value of vector a is equal to 1, a·b =1/2,(a-b)·(a+b)=1/2, the angle between a+b and a-b is α, then the value of cosα is?

Given |a|=1, a·b=1/2,
(A-b)·(a+b)=1/2
A^2-b^2=1/2, then b^2=1-1/2=1/2, so |b|=under the root (1/2)
According to a·b=1/2,|a||b|cosα=1/2
Then cosα=(1/2)/[| a|| b|]=(2 below the root)/2.

Given |a|=1, a·b=1/2,
From (a-b)·(a+b)=1/2
A^2-b^2=1/2, then b^2=1-1/2=1/2, so |b|=under the root (1/2)
According to a·b=1/2,|a||b|cosα=1/2
Then cosα=(1/2)/[| a|| b|]=(2 below the root)/2.

If the angle of vectors a, b is θ, then cosθ is equal to the product of the quantities of vectors a, b divided by the product of their moduli

This proposition is correct if vector a, b is not zero vector, but is incorrect if vector a, b has zero vector. So it is incorrect to say that the proposition "if the angle of vector a, b is θ, then cosθ is equal to the quantity product of vector a, b divided by the product of their modules ".

The proposition is correct if the vectors a, b are not zero, but is incorrect if there are zero vectors in the vectors a, b. Therefore, it is incorrect to say that the proposition "if the angle of vectors a, b is θ, then cosθ is equal to the quantity product of vectors a, b divided by the product of their moduli ".

What is cos equal to if vector a =(2.1-2), b =(-1.2.2)?

Vector a=(2,1,-2), b=(-1,2,2),
A·b=2×(-1)+1×2+(-2)×2=-4
|A|=3,|b|=3
Then Cos (a, b)=a·b/(|a||b|)
=-4/9

For vectors a (a1, a2, a3) and b (b1, b2, b3) → → How to deduce the formula of vector product a·b=a1b1+a2b2+a3b3?

A =(a1,0,0)+(0, a2,0)+(0,0, a3)
And then just ride it up... Will you?