Given the plane vector a =(2,-2), b =(3,4), a×b=a×c, then the minimum value of |c| is A2B root 2C1/2D root 2/2

Given the plane vector a =(2,-2), b =(3,4), a×b=a×c, then the minimum value of |c| is A2B root 2C1/2D root 2/2

A×b=a×c=-2, then -2=a*c|c|=(a*c)/|a|=-root 2/(2cosa), when cosa takes the minimum value -1,|c| gets the maximum value, select D

A×b=a×c=-2, then -2=a*c|c|=(a*c)/|a|=-root 2/(2cosa), when cosa takes the minimum value -1,|c| takes the maximum value D

It is proved that there exists n-dimensional vector x such that Ax=Bx It is proved that there is n-dimensional vector x such that Ax=Bx

The title should be a nonzero vector x.
By BC=0, r (B)+r (C)

Let A be a matrix of order n,α be a vector of order n, if A^n-1α=0, and A^nα=0, it is proved that α, Aα,..., A^n-1α are linearly independent.

Let k0 k1A...+k (n-1) A^(n-1)α=0
Simultaneous left multiplication A^(n-1)
Since A^nα=0, A^(i)α=0(i >=n)
Then k0A^(n-1)α=0 and A^n-1α=0, then k0=0
Then k1A...+k (n-1) A^(n-1)α=0
At the same time, multiply A^(n-2) by A^(n-2) in the same way to obtain k1=0.
By analogy, ki=0(i=0,1,2,3,...,n -1)
Then we get α, Aα,..., A^n-1α linearly independent

(Aα) Let A be a matrix of order n and α be a vector of dimension n. If the rank r (αT0)=r (A), then the system of linear equations () αT is the transposition of α A. Ax=α must have infinite solution B.Ax=α must have unique solution C.(Aα)(x) (αT0)(y)=0 Zero solution only D.(Aα)(x) (αT0)(y)=0 must have a non-zero solution (Aα) Let A be a square matrix of order n and α be a vector of dimension n. If the rank r (αT0)=r (A), then the system of linear equations () αT is the transposition of α A. Ax=α must have infinite solution B.Ax=α must have a unique solution C.(Aα)(x) (αT0)(y)=0 Zero solution only D.(Aα)(x) (αT0)(y)=0 must have a non-zero solution (Aα) Let A be a square matrix of order n and α be a vector of dimension n. If the rank r (αT0)=r (A), then the system of linear equations () αT is the transposition of α A. Ax=α must have infinite solution B.Ax=α must have unique solution C.(Aα)(x) (αT0)(y)=0 Zero solution only D.(Aα)(x) (αT0)(y)=0 must have a non-zero solution

R [A,α;α T,0]= r (A)

Let A, B be n-dimensional column vectors, then the rank of n-order matrix c=ab^t is r (a)=, why not equal to n, the answer is 0 or 1

Let A =(a1, a2,.an)^T, B =(b1, b2,.bn)^T
Then AB^T = a1b1 a1b2 a1b3.a1bn
a2b1 a2b2 a2b3 .a2bn
..
anb1 anb2 anb3 .anbn
Note any 2*2 submatrix aibj aibk
Asbj asbk
Its determinant is 0, so any k (or 2 or more) level of the sub-expression is equal to 0.
So rank of AB^T

A, b are all three-dimensional column vectors, matrix A = baT (b times (a transpose)), what is the rank of matrix A? Why? What if it's extended to n-dimension?

The rank of matrix A is less than or equal to 1.
Because r (A)= r (ba^T)