Coordinate vector with Respect to Basis (Matrices/Linear Algebra)

Question

I'm studying for a linear algebra test and I'm really stuck on changing basis or finding transformation vectors with respect to other basis and all of that. I've read so many guides and done so many problems, but the concepts won't stick. I feel like this link:

https://math.dartmouth.edu/archive/m24w07/public_html/Lecture12.pdf

is really good, but I don't understand the material. Everything makes sense up to slide five, where they write:

Suppose now that we choose as a basis for $V = \Bbb{R}^3$ the set $$B = \{ (1,0,0), (1,1,0), (1,1,1,)\}.$$

Then the coordinate vector of an element $(a_1, a_2, a_3)$ is: $$(a_1 - a_2,a_2 - a_3, a_{3})$$

How did they get that coordinate vector?

Answer 1

$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$

Edit: how you would actually figure this out is to solve $c_1(1,0,0)+c_2(1,1,0)+c_3(1,1,1)=(a_1,a_2,a_3)$

Edit 2: to answer your question, how they came up with the matrix for $T$. You have three things you know. First, $Tb_1=T[1,0,0]^t=(3,1,1)$. Second, $Tb_2=T[0,1,0]^t=(4,1,1)$. Third, $Tb_3=T[0,0,1]^t=(4,2,0)$. Then you have 9 equations, 9 unknowns. Simply checking here should be good enough.

Answer 2

There's a couple ways to do this:

You know that in the new basis a vector $v$ has coordinates:

$[b_1,b_2,b_3] \stackrel{\text{def}}{=} b_1(1,0,0) + b_2(1,1,0) + b_3(1,1,1)$.

So a matrix that turns "new to old" is:

$P = \begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$

This sends $[0,1,0] = 0(1,0,0) + 1(1,1,0) + 0(1,1,1)$ to $(1,1,0)$, for example.

But that's the opposite of what we want, we want "old to new", so we actually need to find $P^{-1}$.

We have some shortcuts we can take here:

$\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

is what we need to solve (convince yourself $P^{-1}$ HAS to be upper-triangular with $1$'s on the diagonal). This leads to:

$a+1 = 0\\b+c+1 = 0\\c+1 = 0$

which implies $a = -1,b = 0,c = -1$. Hence:

$P^{-1} = \begin{bmatrix}1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}$, and we see:

$P^{-1}(a_1,a_2,a_3)^t = [a_1-a_2,a_2-a_3,a_3]^t$, as your notes claim.

Answer 3

Using the following definition,

Let $V$ be a vector space finite dimensional over a field $F$ and $\beta=\{v_{1},v_{2},\ldots,v_{n}\}$ be an ordered basis for $V$. We will say that $$[v]_{\beta}=\begin{pmatrix}\alpha_{1}\\\alpha_{2}\\\vdots\\\alpha_{n}\end{pmatrix}\in F^{n}$$ denote the coordinates for a vector $v\in V$ respect to the ordered basis $\beta$ if, and only if, there exists $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\in F$ such that $$v=\alpha_{1}v_{1}+\alpha_{2}v_{2}+\cdots+\alpha_{n}v_{n}$$

Using that fact, we have $$\left[\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}\right]_{\beta}=\begin{pmatrix}\alpha_{1}\\\alpha_{2}\\\alpha_{3}\end{pmatrix} \iff \color{red}{\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}=\alpha_{1}\begin{pmatrix}1\\0\\0\end{pmatrix}+\alpha_{2}\begin{pmatrix}1\\1\\0\end{pmatrix}+\alpha_{3}\begin{pmatrix}1\\1\\1 \end{pmatrix}}$$ Solving the $\color{red}{\text{linear system}}$ to find $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$ we get $$\left[\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}\right]_{\beta}=\begin{pmatrix}a_{1}-a_{2}\\a_{2}-a_{3}\\a_{3}\end{pmatrix}$$ as desired.

Answer 4

The best way to understand the problem of changing bases would be to work in $\mathbb{R}^2$, where we can easily represent vectors.

What is a basis of $\mathbb{R}^3=\{(x,y,z):x\in\mathbb{R}, y\in\mathbb{R},z\in\mathbb{R}\}$? It is a set $\{v_1,v_2,v_3\}$ such that every element $v$ of $\mathbb{R}^3$ is written uniquely $v=\xi_1v_1+\xi_2v_2+\xi_3v_3$, with $\xi_1\in\mathbb{R},\xi_2\in\mathbb{R}$ and $\xi_3\in\mathbb{R}$. Here, $v_1=(1,0,0), v_2=(1,1,0)$ and $v_3=(1,1,1)$.

Let $a_1\in\mathbb{R},a_2\in\mathbb{R}$ and $a_3\in\mathbb{R}$. Let $w=(a_1,a_2,a_3)\in\mathbb{R}^3$. The "coordinate vector of w" is the triple $(\alpha_1,\alpha_2,\alpha_3)$ such that $w=(a_1,a_2,a_3)=\alpha_1v_1+\alpha_2v_2+\alpha_3v_3=\alpha_1(1,0,0)+\alpha_2(1,1,0)+\alpha_3(1,1,1)$. So, you have $\begin{cases} a_1=\alpha_1+\alpha_2+\alpha_3 \\ a_2=\alpha_2+\alpha_3\\ a_3=\alpha_3\end{cases}\iff \begin{cases} a_1=\alpha_1+\alpha_2+\alpha_3 \\ \alpha_2=a_2-a_3\\ \alpha_3=a_3\end{cases} \iff \begin{cases} \alpha_1=a_1-a_2 \\ \alpha_2=a_2-a_3\\ \alpha_3=a_3\end{cases}$