I WILL GIVE THE BRAINLIST V2
2 answers:
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Answer:
A), B) and D) are true
Step-by-step explanation:
A) We can prove it as follows:
B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then
.
C) Consider . This set is orthogonal because
, but S is not orthonormal because the norm of (0,2) is 2≠1.
D) Let A be an orthogonal matrix in . Then the columns of A form an orthonormal set. We have that
. To see this, note than the component
of the product
is the dot product of the i-th row of
and the jth row of
. But the i-th row of
is equal to the i-th column of
. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then
E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.
In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set and suppose that there are coefficients a_i such that
. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then
then
.