Is there any internal relation between the rank of matrix and the rank of vector group?

Is there any internal relation between the rank of matrix and the rank of vector group?

Yes.
Some textbooks talk about the rank of vector group first, then the rank of matrix
In fact, the rank of the row vector group of the matrix = the rank of the column vector group = the rank of the matrix
This is called the triple - rank theorem of matrices.

Yes.
Some textbooks talk about the rank of vector group first, then the rank of matrix
In fact, the rank of the row vector group of the matrix = the rank of the column vector group = the rank of the matrix
This is called the three - rank theorem of a matrix.

The module of vector a is equal to 3 under the root sign, the module of vector b is equal to 2, and the angle between vector a and vector b is 30 degrees. The module of vector a is equal to 3 under the root number, the module of vector b is equal to 2, and the angle between vector a and vector b is 30 degrees.

Square of module of vector a plus vector b = square of vector a plus vector b
= Square of vector a + square of vector b + vector a * vector b
= Square of module of vector a + square of module of vector b + square of module of vector a * module of vector b * cos 30 degrees
=3+4+3=10
Module of vector a plus vector b = root number 10

Given that the modulus length of a is equal to the root number 2, the modulus length of b is equal to 3, and the angle between vector a and vector b is 45°, find the value range of x when the angle between vector a+x vector b and x vector a+ vector b is an acute angle... Given that the modulus length of a is equal to the root number 2, the modulus length of b is equal to 3, and the angle between vector a and vector b is 45°, find the value range of x when the angle between (vector a+x vector b) and (x vector a+ vector b) is an acute angle...

(A+xb)(b+xa)=xa2+xb2+(1+x2) ab=11x+3(1+x2)>0
X >(-11 85)/6 or x <(-11 85)/6, and x=1

The vectors a, b, c, d are known to be such that a modulus equals 1, b modulus equals the root number 2, b is projected on a by 1/2, and the vectors a-c are perpendicular to the vectors b-c The modulus of vector d-c is equal to 1. Then the maximum value of the modulus of vector d is equal to

The combination of numbers and shapes is suitable for this problem, as shown in Fig. OA=a, OB=b, OC=c, OD=d,

According to known conditions,|OB|=|AB|=√2,|OA|=1,
Because (a-c),(b-c), C is on a circle with AB diameter,
And |d-c|=1, so D is on a circle with C as the center and 1 as the radius,
When OC exceeds the midpoint E of AB and OD exceeds OC,|d|max,
At this time |OE|=√[(3/4)^2+(√7/4)^2]=1,|EC|=√2/2,|CD|=1,
Therefore |d|maximum value is 1+√2/2+1=2+√2/2.

Given that the module of vector a is equal to 1 and the module of vector b is equal to root number 2, if vectors a-b are perpendicular to vector a, the angle between vectors a and b is determined

Let vector a be the angle of vector b A
Vector a-b is perpendicular to vector a
∴(A-b).a=0
I.e. a2-a.b=0
1-A.b=0
A.b=1
CosA=(a.b)/(|a b|)=1/(1 2)=√2/2
The angle between vector a and vector b is 45 degrees.

Let vector a and vector b have the angle A
Vector a-b is perpendicular to vector a
∴(A-b).a=0
I.e. a2-a.b=0
1-A.b=0
A.b=1
CosA=(a.b)/(|a b|)=1/(1 2)=√2/2
The angle between vector a and vector b is 45 degrees.

Let vector a and vector b have the angle A
Vector a-b perpendicular to vector a
∴(A-b).a=0
I.e. a2-a.b=0
1-A.b=0
A.b=1
CosA=(a.b)/(|a b|)=1/(1 2)=√2/2
The angle between vector a and vector b is 45 degrees.

The vector a=(sinθ,-2) b=(1, cosθ), where (0,π/2)(1) is perpendicular to each other. (2) If sin (θ-Φ)=1/Root 10,0

(1) A·b=sinθ-2cosθ=0
So tanθ=2
Draw a right triangle with 1,2 sqr (5)
So sinθ=2/sqr (5), cosθ=1/sqr (5)
(2) Because θ,φ are all acute angles, sin (θ-φ)>0
So 0

(1) A·b=sinθ-2cosθ=0
So tanθ=2
Draw a right triangle with 1,2 sqr (5) on each side
So sinθ=2/sqr (5), cosθ=1/sqr (5)
(2) Because θ,φ are all acute angles, sin (θ-φ)>0
So 0