# Change of basis on quotient of vector spaces

Problem statement Let's consider $\mathcal V = \mathbb R^n$ and $$\mathcal U = \{(x_1, \dots, x_n)\in \mathbb{R}^n: x_1+x_2+\ldots+x_n = 0\}$$

1. Calculate $\dim \mathcal V/\mathcal U$.
2. Let $a = (a_1,\ldots,a_n)$, $b=(b_1,\ldots,b_n)$ be two vectors of $\mathcal V$ so that $a,b\not\in \mathcal U$. Prove that $\mathcal{B}_a = \{a+\mathcal U \}$ and $\mathcal B_b = \{b+\mathcal U\}$ are basis of $\mathcal V / \mathcal U$.
3. Calculate the change of basis matrix from $\mathcal B_b$ to $\mathcal B_a$.

It's easy to see that $$\mathcal U \equiv \left\{\begin{array}{l} x_1 = -\lambda_2 -\cdots - \lambda_n\\x_2 = \lambda_2\\ \vdots \\x_n=\lambda_n \end{array} \right.$$ so $B = \{v_1=(-1,1,\ldots,0),\ldots,v_{n-1}=(-1,0,\ldots,1)\}$ is a basis of $\mathcal U$ and $\{(1,0,\ldots,0)+\mathcal U\}$ is a basis of $\mathcal V/\mathcal U$ since that vector $v_n = (1,0,\ldots,0)$ extends $B$ to a basis of $\mathcal V$. Now, for (2), since $a\not\in \mathcal U$, $a$ is a linear combination $a = \alpha_1 v_1 + \cdots + \alpha_n v_n$ with, at least, one $\alpha_i = 0$ where $1\leq i < n$ and the set $\{v_1,\ldots,v_{n-1},a\}$ is linearly independent, that is $a$ extends the base of $\mathcal U$, so $\mathcal{B}_a$ is basis of $\mathcal V/\mathcal U$. Analogously for $b$.

Is this correct? I'm stuck on (3), I don't know how to formalize that. I think that my way of solving (2) makes (3) harder . Any hint? Thanks in advance.

Edit: due to restrictions this should be solvable without using linear applications. Sorry for that.

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What is $\mathcal{U}$? I don't get it. Is it a sum ? An equation? It's probably supposed to be a set. I don't get what it is. – Git Gud Jan 26 '13 at 12:46
It is a vector space defined by the equation. In my algebra books it is used $\mathcal U$ instead of $U$. Sorry for the confusion. – V. Galerkin Jan 26 '13 at 12:55
The letter you use to represent it isn't an issue. So $$\mathcal{U}=\{(x_1,\dots,x_n)\in \mathbb{R}^n : x_1+\cdots +x_n=0\}\text{?}$$ – Git Gud Jan 26 '13 at 12:58
Yes, that's right. – V. Galerkin Jan 26 '13 at 12:59

(1) and (2) look good. In order to get the matrix in (3) you need to know how to express the basis vector $b+\mathcal{U}$ as a linear combination of the other basis vector $a+\mathcal{U}$. Note that, since $\mathcal{V}/\mathcal{U}$ is one-dimensional, "linear combination" here actually means scalar multiple.

In order to do that notice that $\mathcal{U}$ is the kernel of the linear form $$\varphi ~:~ \mathcal{V} \to \mathbb{R}, ~ (x_1,...,x_n)^T \mapsto \sum_{i=1}^n x_i$$ so that for any vector $v \in \mathcal{V}$ we have $v\in \mathcal{U} \Leftrightarrow \varphi(v)=0$.

Now we have $a,b \notin \mathcal{U}$, so that $\varphi$ has non-zero values on both $a$ and $b$.

Notice that $b-\frac{\varphi(b)}{\varphi(a)}a \in \mathcal{U}$ (this is not too difficult, just apply the map $\varphi$).

Therefore we get $b+\mathcal{U} = \frac{\varphi(b)}{\varphi(a)}a + \left( b-\frac{\varphi(b)}{\varphi(a)}a \right) + \mathcal{U} = \frac{\varphi(b)}{\varphi(a)}a +\mathcal{U}$ from which you obtain the desired base change matrix.

Edit: This can be formulated without using the word "linear form", although conceptually the idea will be the same. We have $a,b \notin \mathcal{U}$, which is equivalent to $\alpha:=\sum_{i=1}^n a_i \neq 0 \neq \sum_{i=1}^n b_i =: \beta$.

Now we basically use the same reasoning as before: $\beta - \frac{\beta}{\alpha}\alpha = 0$, from which we obtain that $b-\frac{\beta}{\alpha}a \in \mathcal{U}$, so that we now have $b+\mathcal{U} = \frac{\beta}{\alpha}a + \left(b - \frac{\beta}{\alpha}a \right) +\mathcal{U} = \frac{\beta}{\alpha}a + \mathcal{U}$, so the base change matrix is the $1\times 1$-matrix $\left( \frac{\beta}{\alpha} \right)$.

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Thank you very much, but I'm looking a way to calculate that matrix without using linear forms due to writing restrictions. I'll study your answer. – V. Galerkin Jan 26 '13 at 16:37
I've added an appropriate edit which avoids the term "linear form". The idea is still the same, however. – Oliver Braun Jan 26 '13 at 17:03
The $a_i$ are the entries of $a$, just like in (2) of the problem statement, where $a=(a_1,...,a_n)$. In (3) it says that neither $a$ nor $b$ lie in $\mathcal{U}$, so the sum of their coefficients, $\sum a_i$ and $\sum b_i$ that is, are necessarily non-zero. – Oliver Braun Jan 26 '13 at 17:19