NewtonianAlch said:Homework Statement
A set of 6 vectors in R5 cannot be a basis for R5, true or false?
The Attempt at a Solution
I'm thinking true, because any set of 6 vectors in R5 is linearly dependent, even though some sets of 6 vectors in R5 span R5.
To be a basis it must be a linearly independent spanning set, so if it's linearly dependent, it cannot be a basis.
Am I correct?
A basis in linear algebra is a set of linearly independent vectors that span a vector space. This means that any vector in the vector space can be written as a linear combination of the basis vectors.
The number of vectors needed to form a basis for a vector space is equal to the dimension of the vector space. In the case of R5, which is a 5-dimensional vector space, a basis would consist of 5 linearly independent vectors.
No, a set of 6 vectors in R5 cannot be a basis for R5. This is because R5 is a 5-dimensional vector space, and a basis for a vector space must consist of a number of vectors equal to the dimension of the vector space.
Vectors are linearly independent if none of them can be written as a linear combination of the others. In other words, no vector in the set can be expressed as a combination of the other vectors using scalar multiplication and addition.
Yes, it is possible for a set of vectors to be linearly dependent in a vector space. This means that at least one of the vectors in the set can be written as a linear combination of the other vectors in the set. In the case of R5, if a set of 6 vectors can be written as a linear combination of the other 5 vectors, then it is not a basis for R5.