Chapter 3A HW Questions

Review questions:

They are just their own objects with their own definitions. A vector space is something over a field with properties of addition, scalar multiplication, and distributive

True, let $U$ have a basis $u_1,u_2,....,u_n$ and $X$ have basis $x_1,x_2,....,x_n$. Then, by the proof of 2.33 in LADR, we know $V$ has basis $u_1,u_2,....,u_n,x_1,x_2,....,x_n$.

For this, first lets observe the standard basis for ${\mathcal{P}}_n(F)$ as $\{1,x,x^2,x^3,...,x^n\}$ Now, as the dimension of both is $n+1$ we just have to prove linear independence or that it is a spanning list We can easily prove independence as the only way to get the $0$ function is by setting all coefficients in each individual $f$ to 0. This is only possible if $0$ is written as the following linear combination: $0=0f+0f^{\prime }+0f^{\prime \prime }+....+0f^{(n)}$ Thus, it is linearly independent and a basis of ${\mathcal{P}}_n(F)$

False, unless the map is surjective.

Through scaler multiplicative closure, each ${\lambda }_j{v}_j$ is in each subspace containing $v_j$. With additive closure, the sum of all such ${\lambda }_j{v}_j$ with another scaler times a multiple is also within every subspace containing $v_i,v_j$. thus, this must be in the intersection as scaler multiplicative and additive closure says the linear combination of $v_i,...,v_n$ is in every subspace containing $v_i,...,v_n$.

Yes, because otherwise all scaler multiplies including the missing vectors would be in $U$. As $\text{dim}U<\text{dim}V⟹U\ne V$ we know this is not the case.

Homework 3A questions

$f(x,y)=\text{floor}(x+y)$

This is the comp of functions. So starting slow, if we take any polinomial, and square it, then the map does not work. Take the two polynomials X and 1, letting $p(x)=x^2$. Then if we look at $T(x+1)\ne x^2+1^2.$

Proof:

1. $T$ is a scalar multiple of the identity $⟺$ $ST=TS$ for every $S\in \mathcal{L}(V)$
2. $\Rightarrow$
   1. Assume $T$ is a scalar multiple of the identity. Then, $T=\lambda 1$ for some $\lambda \in F$.
   2. Then we have it so that $ST=S(\lambda 1)=(\lambda S)1=\lambda (S1)=\lambda (1S)=(\lambda 1)S=TS$
   3. This holds due to definition 3.5
3. $\Leftarrow$
   1. Assume $ST=TS$.
   2. Then, $ST=TS=1TS$