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Let $V$ be the set of all ordered pairs of real numbers, and consider the following addition and scalar multiplication operations on $\mathbf{u}=\left(u_{1}, u_{2}\right)$ and $\mathbf{v}=\left(v_{1}, v_{2}\right)$ $$ \mathbf{u}+\mathbf{v}=\left(u_{1}+v_{1}+1, u_{2}+v_{2}+1\right), \quad k \mathbf{u}=\left(k u_{1}, k u_{2}\right) $$ (a) Compute $\mathbf{u}+\mathbf{v}$ and $k \mathbf{u}$ for $\mathbf{u}=(0,4), \mathbf{v}=(1,-3),$ and $k=2$ (b) Show that $(0,0) \neq 0$ (c) Show that $(-1,-1)=0$ (d) Show that Axiom 5 holds by producing an ordered pair $-\mathbf{u}$ such that $\mathbf{u}+(-\mathbf{u})=\mathbf{0}$ for $\mathbf{u}=\left(u_{1}, u_{2}\right)$ (e) Find two vector space axioms that fail to hold.

In this video song, Call number 28 of Section four. Bone, which gives us two steps of a proven tells us what axioms. Uh, these two steps are possible by and the first one's asked him for, as it says, that any vector is added to the zero vector. So any vector you in, uh, vector space would be equal to you plus zero or vice versa. And you is also here. Zero is the vector that is being attitude. So you can also be zero in this. Um, in this case, so zero vector. It also applies for the zero vector. If you were to be zero vector. So that's why we're were able to weaken. Read Dr Step One and Step two is given to us by axiom seven, which states that, uh, any scaler value see times two vectors u plus V would equal to see you plus CV and they're interchangeable. So the scaler value just distributes to you to the Becker's

In this example, we have a set W that consists of vectors with four entries into kid here and all the requirement we have is that it solves these two equations for the four entries. What we want to do then, is determine if W. Is a subspace of our for To do that, let's begin working with the equations we see here on the inside so we can make sense of what this set W is really doing. So let me start with the first equation. I want to put it in a more standard form with zero on the right hand side. It's all right. A minus to B minus. Foresee equals zero for the first equation. Then for the second equation. This is a two way let's put in a minus C here minus three D and equal zero. So where two equations are linear, the right hand side Zehr zero. And if we like, we could write this in the form of a matrix equation, let's see how to do that. So we're going to write down first in Matrix. It's going to be of size to buy four because there's two rows and four variables Next we put in a vector containing the variables. So it'll be a, B, C and D and the right hand side Alright as the Zero Vector, which is a 21 by our two by one vector consisting entirely of zeros. Now we're ready to put in the Matrix. So what will do then is pull coefficients for very Blais. Coefficients are 12 for variable B. They are negative to zero. Then for see, we have negative four negative one. And for the variable d, we have zero and negative three. So let's call this a matrix A Here. I'll call this a vector X, and we have the zero vector indicated here. With that information, we can say that w is equal to first a B c D. We've indicated to be the vector X. It's all right and ex here. So it's a set of all vectors x such that first we'd have this equation to put in. This is our only restriction. But that restriction came into this form A X equals zero. Let's write that in. So we have a X equals zero vector here. So as soon as we get to this equality for this set. We recognize that this is the null space of the matrix A which has already indicated here. And that means we're in luck. Because now we have a theorem that tells us, since we have the null space, that w is a subspace of our four. And this is since W is equal to a null space of a so that completes our check for this particular set, W

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