Let \(V\) be a vector space over the field \(\mathbb{F}\) having dimension \(n\).

Let \(\{v_1,\ldots,v_n\}\) be a basis for \(V\).
We call \((v_1,\ldots,v_n)\) an **ordered basis** for \(V\).

By fixing an order on the basis elements, we can now represent every vector in \(V\) as an \(n\)-tuple.

More precisely, let \(v \in V\). Recall that there exist scalars \(\lambda_1,\ldots,\lambda_n\in \mathbb{F}\) such that \(v = \lambda_1 v_1 + \cdots + \lambda_n v_n.\)

Define \([v]_{(v_1,\ldots,v_n)}\) to be the \(n\)-tuple \(\begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{bmatrix}\).

We call \([v]_{(v_1,\ldots,v_n)}\) a **tuple representation**
(or **coordinate representation** of
\(v\) with respect to the ordered basis \((v_1,\ldots,v_n)\).

Clearly, if we choose a different basis or a different ordering of the basis vectors, we get a different representation for the same vector. That is why it is so important to specify the ordered basis when working with tuple representations.

Note that \( \Gamma = \left ( \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 3\end{bmatrix}\right )\) is an ordered basis for \(\mathbb{R}^2\). Let \(u = \begin{bmatrix} 3 \\ 2 \end{bmatrix}\). What is \([u]_\Gamma\)?

**Solution.**We first write \(u\) as a linear combination of the ordered basis elements. That is, we want to find real numbers \(\alpha\) and \(\beta\) such that \(u = \alpha \begin{bmatrix} 1 \\ -1 \end{bmatrix} + \beta \begin{bmatrix} 2 \\ 3\end{bmatrix}\). Thus, we need to solve the system \begin{eqnarray*}3 = \alpha + 2\beta \\ 2 = -\alpha + 3 \beta \end{eqnarray*} Solving gives \(\alpha = 1\) and \(\beta = 1\). Hence, \([u]_\Gamma = \begin{bmatrix} 1 \\ 1\end{bmatrix}\).-
Let \(V\) denote the vector space of polynomials in \(x\) with real coefficients having degree at most \(2\). Let \(\Gamma\) denote the ordered basis \((1, x+1, x^2+1)\) and let \(\Omega\) denote the ordered basis \((x-1, x^2+1, x^2-1)\). (Check that these are indeed ordered bases.) We want to find the tuple representations of \(x\) with respect to these ordered bases.

To determine \([x]_\Gamma\), we need to find real numbers \(\alpha, \beta,\gamma\) such that \[x = \alpha (1) + \beta (x+1) + \gamma (x^2+1).\] Simplifying the right-hand side, we obtain \[x = \gamma x^2 + \beta x + (\alpha + \beta + \gamma). \] Since the left-hand side has no \(x^2\) term, we must have \(\gamma = 0\). Now, comparing the coefficients of \(x\) on both sides, we get \(\beta = 1\). Since the left-hand side has no constant term, we must have \(\alpha = -1\).

In other words, \(x = (-1)1 + 1(x+1) + 0x^2\), giving \([x]_\Gamma = \begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix}\).

For \([x]_\Omega\), note that \(x = 1(x-1) + \frac{1}{2}(x^2+1) + \left(-\frac{1}{2}\right)(x^2)\). Hence, \([x]_\Omega = \begin{bmatrix} 1 \\ \frac{1}{2} \\ -\frac{1}{2}\end{bmatrix}\).

The beauty of these tuple representations of vectors is that we can now work with tuples instead. For instance, if \(u,v\in V\) and \(\Gamma\) is an ordered basis for \(V\), then one can easily check that \([u+v]_{\Gamma} = [u]_{\Gamma} + [v]_{\Gamma}.\) In other words, adding two vectors in \(V\) simply requires adding the corresponding tuple representations in \(\mathbb{F}^n\).

And if \(\lambda\) is a scalar, then \([\lambda u]_{\Gamma} = \lambda [u]_{\Gamma}\). Thus, multiplying a vector by a scalar simply involves multiplying the corresponding tuple representation by the same scalar.

- Let
\(\Gamma = \left ( \begin{bmatrix} 1 \\ 2\end{bmatrix},
\begin{bmatrix} 2 \\ 3 \end{bmatrix}\right )\) and
\(\Omega = \left ( \begin{bmatrix} 1 \\ -1\end{bmatrix},
\begin{bmatrix} 1 \\ -2 \end{bmatrix}\right)\) be two
ordered basis for \(\mathbb{R}^2\).
Let \(u = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\) and \(v = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\). Find the following:

\([u]_{\Gamma}\)

\([u]_{\Omega}\)

\([v]_{\Gamma}\)

\([v]_{\Omega}\)

Give a \(2 \times 2\) matrix \(A\) such that \([x]_\Omega = A [x]_\Gamma\) for all \(x \in \mathbb{R}^2\).