1. Given linear independence of n-dimensional vector a1a2... a (n-1), non-zero vector b is orthogonal to ai to prove a1,a2,a3...a (n-1), b is linear independence 2 The following vectors are organized into standard vector group a1=(1,-1,1) Ta2=(-1,1) Ta3=(1,1-1) T3 by Schmidt standard orthogonalization method Let a=(a1a

1. Given linear independence of n-dimensional vector a1a2... a (n-1), non-zero vector b is orthogonal to ai to prove a1,a2,a3...a (n-1), b is linear independence 2 The following vectors are organized into standard vector group a1=(1,-1,1) Ta2=(-1,1) Ta3=(1,1-1) T3 by Schmidt standard orthogonalization method Let a=(a1a

1. K1a1+k2a2+...+k (n-1) a (n-1)+knb=0, left multiplied by b transpose, b transpose is equal to 0, so kn=0, and k1=k2=...=kn-1=0 because a1, a2,...an-1=kn-1=0.2. b1=a1, b2=a2-{(a2, b1)/(b1, b1)}*b1, b3=a3-{(a3, b2)/(b...

Let n-dimensional vector a1,a2,...,as, the correct proposition is that if a1,a2,...,as is linearly independent, then a1+a2, a2+a3,...as-1+as, as+a1 are also linear Thank you for your explanation. Because it is the first time to learn, so, some knowledge points are not very flexible to understand. Also want to ask, I also know the same (A1+a2)-(a2+a3)+(a3+a4)-(a4+a1)=0. Linear correlation, but why can we get the correlation conclusion through this operation? This is also the case in exercises. Isn't it said that there is a linear correlation between k1 and k2.ks, which are not all 0? Isn't it 1,1,1,-1 now? I just can't bend a bend. Please advise me. Thank you

By using the vector a1, a2..as linear correlation proposed in the topic, we mean that there is a k1,k2...ks that is not completely zero, so that k1*a1+k2*a2+...+ks*as=0, where k1,k2...ks is a real number.

Why are pairs of orthogonal, non-zero vector groups linearly independent Why is pairs of orthogonal, non-zero vector groups linearly independent

Let η1,η2,...,ηs be a pairwise orthogonal nonzero vector.k1,k2,...,ks a combination coefficient such that k1 1+k2 2+...+ks s=0. The inner product of both sides of i=1,2,...,s, and ηi is k i·(ηi,ηi)=0(vector pairwise orthogonal, i.e. j=i, with (ηj,ηi)=0).

Why do pairs of orthogonal nonzero vector groups have to be linearly independent? Why do pairs of orthogonal nonzero vector groups be linearly independent?

Let x_1,..., x_n be n vectors and orthogonal to each other.
Suppose there is a constant k_1,..., kn, such that k_1x_1+.+k_nx_n=0, then we need to explain that k_1=...=k_n=0
In fact, the two sides of the equation are multiplied by x_i^T simultaneously
Then k_1x_i^Tx_1+k_2x_i^Tx_2+...+k_nx_i^Tx_n=0
Also x_i is orthogonal to other vectors, i.e. x_i^Tx_j=0(j\neqi)
Therefore, k_ix_i^Tx_i=0, i.e. k_i|x_i|^2=0(|x_i| represents the length of vector x_i)
So k_i=0.

The orthogonal vector group must be a linear independent vector group. Is it correct? The orthogonal vector set must be a linear independent vector set. Is it correct?

See how your textbook defines orthogonal vector groups
If there are all non-zero vector requirements, then correct

Prove: n-dimensional vector group α1,...,αm is linearly independent, vector β is orthogonal to each vector in α1,...,αm, then α1,...,αm,β is linearly independent Evidence: n-dimensional vector group α1,...,αm is linearly independent, vector β is orthogonal to each vector in α1,...,αm, then α1,...,αm,β is linearly independent

Let k1α1+...+kmαm+kβ=0(*) be the inner product of k1(α1,β)+...+km (αm,β)+k (β,β)=0 be orthogonal to every vector in β and α1,...,αm, so there is k (β,β)=0[ Note: here there must be β=0, otherwise the problem is incorrect] from β=0, so...