Does a vector span a matrix?

To find a basis for the span of a set of vectors, write the vectors as rows of a matrix and then row reduce the matrix. The span of the rows of a matrix is called the row space of the matrix. The dimension of the row space is the rank of the matrix. The row space and the column space always have the same dimension.

Is a basis for any subspace of a vector space always a basis for the vector space?

Let V be a subspace of Rn for some n. If either one of these criterial is not satisfied, then the collection is not a basis for V. If a collection of vectors spans V, then it contains enough vectors so that every vector in V can be written as a linear combination of those in the collection.

Does Q the set of all rational numbers in R form a vector space give reason for your answer?

By the Subspace Criteria, D is a subspace of the vector space of all functions on R and, thus, is a vector space in its own right. multiplication with scalars drawn from R. Hence, the set of all rational numbers is not a vector space over R.

Could a set of three vectors in R4 span all of R4 explain what about n vectors in Rm when n is less than M?

Solution: A set of three vectors can not span R4. To see this, let A be the 4 × 3 matrix whose columns are the three vectors. By the same reasoning, the echelon form of an m × n matrix B whose columns are n vectors in Rm, where n

Can a vector space have more than one basis?

A vector space can have several bases; however all the bases have the same number of elements, called the dimension of the vector space. This article deals mainly with finite-dimensional vector spaces. However, many of the principles are also valid for infinite-dimensional vector spaces.

Why is Q not a subspace of R?

It simply fails the test: a \in R and q \in Q —- > a q \in Q. The reason is that if one takes a to be an irrational number, then for any nonzero rational number q, we have a q \notin Q.

Why Q is not a vector space over R?

We’ve just noted that R as a vector space over Q contains a set of linearly independent vectors of size n + 1, for any positive integer n. Hence R cannot have finite dimension as a vector space over Q. That is, R has infinite dimension as a vector space over Q.