Please explain why  If V� is a vector space and u� and v� are vectors in V�, then according to the axioms  6v⋅u is also a vector in V� is a false statement

The vector y is a numerical vector. Which ONE of the following best describes what the R code below will calculate?   sum(!is.na(y)) A logical vector of the same length as the vector y, which takes the value TRUE for values in y which are NOT missing and FALSE for values in y which ARE missing.   Replaces each value in y with TRUE if it is not missing and FALSE if it is missing   The number of non-missing values in the vector y.   The number of missing values in the vector y.   The propostion of non-missing values in the vector y.

Let U be a subspace of R", and let S = { ū, v, w } be a set of three non-zero vectors in R”. Fill in the blanks so that the following statements are true. If you believe you don't have enough information to conclude anything for any of the cases, choose the answer "???". Note that there is no intended relationship between the parts. In each case, assume only what the statement says to assume. If SCU, then dim(U) > (Enter the largest integer that makes the statement necessarily true.) If S is linearly independent and S C U, then dim(U) 3. If S' is a spanning set for U, then dim(U) 3. If span(S) = U then S If S is a basis for U, then S linearly independent. linearly independent.

2 3 Can vector w = 3 be written as a linear combination of v₁ = 1, V₂ = 4 4 vector w belong to span{v₁,v,? 1 3? In other words does -2