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Let $V$ be the first quadrant in the $x y$ -plane; that is, let$V=\left\{\left[\begin{array}{l}{x} \\ {y}\end{array}\right] : x \geq 0, y \geq 0\right\}$a. If $\mathbf{u}$ and $\mathbf{v}$ are in $V,$ is $\mathbf{u}+\mathbf{v}$ in $V ?$ Why?b. Find a specific vector $\mathbf{u}$ in $V$ and a specific scalar $c$ such that $c \mathbf{u}$ is $n o t$ in $V .$ (This is enough to show that $V$ is not a vector space.)

a.) Yes. b. $c=-1$ and $u=\left[\begin{array}{l}{1} \\ {1}\end{array}\right]$

Calculus 3

Chapter 4

Vector Spaces

Section 1

Vector Spaces and Subspaces

Vectors

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Hello. For this question, we need to do two things with this given subset of our where X and Y are just two constants that are positive. Um, we need to show that given any two arbitrary vectors in this space, uh, the vector edition of those two vectors is also in the space. Then we need to try to give an example of a vector and a constant so that when you multiply the two together, uh, it is not in this space, and this is checking to see if this is a subspace of our two. So for the addition, it's typically best just give yourself to arbitrary vectors in here and see what happens. So let's say we have one vector with two components, and we're gonna add it to another vector with two components where all four of these constants are positive so that these two vectors are in this space. Then when I add them together, even though this seems fairly trivial, you can see that X one plus x two is going to be greater than or equal to zero because x one and X two are greater than zero. And why one and why to are going to be a greater than zero because the individual parts of the addition are greater than zero. And so writing that out formally is the proof that if you take any two arbitrary vectors in the space and add them, you you're still end up in this space. Now, for the second part, we can pick any vector in here. I'm just going to keep it simple and pick the vector one one. And if we multiply, we need toe find a constant C such that this thing is not within that space. And there's a new infinite amount of constant you can pick. But if you pick c equals negative one, that implies that these two things multiply together, are negative one negative one just by basic laws of Vector Edition, which is not within the space, because these two things are both negative and that's it

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