True or False: Vector a=(1,1) and vector b=(t,-t) must be parallel to each other. True or False: vector a =(1,1) and vector b =(t,-t) must be parallel to each other.

True or False: Vector a=(1,1) and vector b=(t,-t) must be parallel to each other. True or False: vector a =(1,1) and vector b =(t,-t) must be parallel to each other.

Cuo

Given that vector a runs parallel to vector b and |a|>|b|>0, the direction of vector a+b A. Same as vector a direction B. Opposite to vector a direction C. Same as vector b direction D. Opposite to vector b direction Given that vector a runs parallel to vector b and |a|>|b|>0, then the direction of vector a+b A. Same as vector a direction B. Opposite to vector a direction C. Same as vector b direction D. Opposite to vector b direction

Option A
The two vectors are parallel and can be in the same direction (at this time, the result is in the same direction as a, b, excluding BD) or in the opposite direction. Because the modulus of a is greater than the modulus of b, the two vectors are in the same direction as a, excluding A for C.

Option A
The two vectors are parallel and can be in the same direction (the result is in the same direction as a, b, excluding BD) or in the opposite direction. Because the modulus of a is greater than the modulus of b, the two vectors are in the same direction as a, excluding A for C.

Option A
The two vectors are parallel, which can be in the same direction (the result is in the same direction as a, b, excluding BD), or in the opposite direction, because the modulus of a is greater than the modulus of b, so it is in the same direction as a, excluding A from C

Is it parallel to vector b when vector a = vector =0 vector, parallel to vector b?

Parallel,0 vector parallel to any vector

Given vector A (a,1), B=(b, b^2), if A+B is parallel to vector (0,1), then vector A+B=

Because A+B is parallel to vector (0,1) and A+B=(a+b,1+b^2)
So a+b=0,1+b^2=k
So: A+B =(0,1+b^2)
Because the condition is insufficient, the concrete b has no way to obtain!

Because A+B is parallel to vector (0,1) and A+B=(a+b,1+b^2)
So a+b=0,1+b^2=k
Therefore: A+B =(0,1+b^2)
Because the condition is insufficient, the concrete b does not have the method to obtain!

Because A+B is parallel to vector (0,1) and A+B=(a+b,1+b^2)
So a+b=0,1+b^2=k
Therefore: A+B =(0,1+b^2)
Because the condition is insufficient, the concrete b has no way to obtain!

Any n-dimensional vector can be linearized by n-dimensional vector group α1.α2....α n. Any n-dimensional vector can be linearized by n-dimensional vector group α1.α2....αn.

It's linear.
It is proved that any n-dimensional vector can be linearized by n-dimensional vector group α1,α2,...,αn.
So ε1,ε2,...,εn can be linearized by α1,α2,...,αn.
And any n-dimensional vector can be expressed linearly by ε1,ε2,...,εn
So vector groups ε1,ε2,...,εn are equivalent to α1,α2,...,αn.
So r (α1,α2,...,αn)= r (ε1,ε2,...,εn)= n.
So α1,α2,...,αn are linearly independent.

Let α,β be n-dimensional vector,β'α=0, then the eigenvalue of α'β' is? (Write down the steps, thank you)

αβ'α = α(β'α) = 0 = 0α
So 0 is the eigenvalue of αβ'.
Note:α is not a 0 vector.

αβ'α = α(β'α) = 0 = 0α
So 0 is the eigenvalue of αβ'.
Note:α is not 0 vector.