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# Linear algebra: homework 1

Theorem: Prove that the set of all powers is a basis of the set of all polynomials.

Lemma: Two polynomials are equal if and only if their coefficients are equal. (prove it)

Proof: Powers are $1,x,x^2,\ldots,x^n,\ldots$. Let's name them: $v_0=1, v_1=x, v_2=x^2, \ldots, \in {\rm P}$. Then $B = \{v_0,v_1,\ldots,v_n,\ldots\}$, an infinite set.

Part 1: ${\rm span \hspace{3pt}} B={\bf P}$.

We need to prove that every element of ${\bf P}$ is a linear combination of the elements of $B$.

Given $p \in P$, find the coefficients of $v_0,\ldots,v_n,\ldots$, that give $p$. How?

Recall, $p$ is a polynomial $$p(x)=a_0+a_1x+a_2x^2+\ldots+a_nx^n,$$ so we rewrite: $$p=a_0v_0+a_1v_1+a_2v_2+\ldots+a_nv_n.$$

Then $a_0,\ldots,a-N$ are the coefficients.

Part 2: $B$ is linearly independent.

Suppose it is not, then there are $c_0,\ldots,c_n$ and $v_0,\ldots,v_n$ so that $c_0v_0+\ldots+c_nv_n=0$ ($0$ is a vector here) and not all $c_0,\ldots,c_n$ are zero. Rewrite

$$\begin{array}{} c_0\cdot 1 + c_1 \cdot x + \ldots + c_n x^n &= 0 \in {\bf P} \\ &= 0 \cdot 1 + 0 \cdot x + 0 \cdot x^2 + \ldots + 0 \cdot x^n = 0 \\ \end{array}$$

So, by lemma, $c_0=0, c_1=0, \ldots, c_n=0$. That's a contradiction!

Homework: Prove the set of all rotations of the $xy$-plane,around $0$, is closed under compositions.

$R_{\alpha}(A)=B$

How? Show that $R_{\beta}R_{\alpha}$ is $R_{\gamma}$ for some $\gamma$. Recognize from the picture $\gamma = \alpha + \beta$.

Next, find the matrix of $R_{\alpha},R_{\beta}$, multiply, show that this is $R_{\alpha+\beta}$. Use trigonometry, ${\rm sin}/{\rm cos}$ of sums.

How about without trigonometry? Recall: $A=(x,y) \in {\bf R}^2$ and $B=R_{\alpha} \left[ \begin{array}{} x \\ y \end{array} \right]$, then $R_{\alpha}$ is a $2 \times 2$ matrix?

What other representations? Polar coordinates!

$A=(r, \theta)$,

$$\begin{array}{} R_{\alpha}(r, \theta) &= (r,\theta+\alpha) \\ R_{\beta}(s, \varphi) &= (s, \varphi + \beta) \end{array}$$

Then $$\begin{array}{} R_{\beta}R_{\alpha}(r,\theta) &= R_{\beta}(r,\theta+\alpha) \\ &= (r,\theta+\alpha+\beta) \\ &= R_{\alpha+\beta}(r,\theta). \end{array}$$

Question: Why don't we use polar coordinates for linear algebra?

With respect to polar coordinates, ${\bf R}^2$ should have a basis:

maybe . No. Can't work, why?

Some of the axioms fail? Reason is $\theta$ is periodic.