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In Exercises 29 and $30, V$ is a nonzero finite-dimensional vector space, and the vectors listed belong to $V$ . Mark each statement True or False. Justify each answer. (These questions are more difficult than those in Exercises 19 and $20 .$ )a. If there exists a set $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\}$ that spans $V,$ then $\operatorname{dim} V \leq p .$b. If there exists a linearly independent set $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\}$ in$V,$ then $\operatorname{dim} V \geq p$c. If $\operatorname{dim} V=p,$ then there exists a spanning set of $p+1$ vectors in $V .$

a) True b) Truec) True.

Calculus 3

Chapter 4

Vector Spaces

Section 5

The Dimension of a Vector Space

Vectors

Johns Hopkins University

Harvey Mudd College

University of Nottingham

Idaho State University

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problem. 29 has a set of givens and gives us three statements and asks whether or not these statements are true or false. The givens Are you a vector Space V, which is non zero in finite dimensional, and all of the vectors listed below belong to the Now let's look at a A gives us the vector space of you. One to V. P tells us that this vector space spans me, then says then the dimension a v Slesin or equal to P. Now, in order to solve this, we're gonna consider two types of spanning sets. First, we're gonna consider a linearly independent, Spanning said for a linearly independent, spanning set. That means none of the actors in this set can be written as a linear combination of any of the out other vectors. Oh, linearly independent spinning set is also known as a basis, and by definition, if we have a basis, the dimension of the equals P. Now the other type of set that we are going to consider is a linear Lee depended spanning set. If we have a linearly dependent spinning. So that would mean that some of the vectors in the set V one to V. P could be written as linear combinations of other vectors in the set. Therefore, those vectors would not contribute to the dimension at all. Therefore, the dimension of the is less than p. So a is true Now we're gonna consider be Once I erase all of this, B is gonna be solved fairly similarly, by considering two different types of sets for B, we have the set V one to V. P, and this set is linearly independent. Then we're given the dimension of the is greater than or equal to P and we want to know whether or not this is true or false. So let's consider our two types of linearly independent sets. There's a spanning set which we already know from A That would be a basis which would make the dimension of the equal to P. Now, if we have a not spanning set, we have a not spanning set. That would mean some of the vectors in the set don't contribute to the dimension. All therefore to mention of the is greater then p so be is also true. Next we're gonna consider, see? All right, So for C the statement. ISS If the dimension of the equals p then there exists a spanning set of P plus one vectors. After thinking through and be, we know that a linearly depended spanning set exists, and in linearly dependent, spanning sets, the dimension of the is less than P. But so if the dimension of the is less than P, there could exist expanding set of P plus one vectors. Now, for this one, we can consider a short example say that we have the equals, our swear that would make the dimension of the equal to the dimension of R squared, which equals two equals p. And then we could also have a set of 30 for one vector 10 for another vector 01 for another factor.

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