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Problem: Let $V$ be a $n$-dimensional vectorspace and let $\beta = \left\{v_1, v_2, \ldots, v_n\right\}$ be a basis for $V$. Prove that the coordinate map $\phi_{\beta} : V \rightarrow \mathbb{R}^n$ is linear and determine the matrix $[\phi_{\beta}]_{\beta}^{\gamma}$ with $\gamma$ the standard basis for $ \mathbb{R}^n $.

Attempt at solution: Take vectors $v, w \in V$. Then we have \begin{align*} v = \sum_{i=1}^n \lambda_i v_i \qquad \text{and} \qquad w= \sum_{i=1}^n \zeta_i v_i \end{align*} for unique scalars $\lambda_i$ and $\zeta_i \in \mathbb{R}$. It follows that \begin{align*} v+ w &= \sum_{i=1}^n \lambda_i v_i + \sum_{i=1}^n \zeta_i v_i \\ &= \sum_{i=1}^n (\lambda_i + \zeta_i) v_i. \end{align*} Hence we get \begin{align*} \phi_{\beta}(v+w) &= (\lambda_1 + \zeta_1, \lambda_2 + \zeta_2, \ldots, \lambda_n + \zeta_n) \\ &= (\lambda_1, \lambda_2, \ldots, \lambda_n) + (\zeta_1, \zeta_2, \ldots, \zeta_n) \\ &= \phi_{\beta} (v) + \phi_{\beta} (w). \end{align*} Now take a scalar $\mu \in \mathbb{R}$. Then we get \begin{align*} \mu v = (\mu \lambda_1) v_1 + (\mu \lambda_2) v_2 + \ldots + (\mu \lambda_n) v_n, \end{align*} from which it follows that \begin{align*} \phi_{\beta}(\mu v) &= (\mu \lambda_1, \mu \lambda_2, \ldots, \mu \lambda_n) \\ &= \mu (\lambda_1, \lambda_2, \ldots, \lambda_n) \\ &= \mu \phi_{\beta} (v). \end{align*} So $\phi_{\beta} : V \rightarrow \mathbb{R}^n$ is linear.

Then for the matrixrepresentation, I'm not sure what I need to write in the columns of the matrix. If I take a vector $v \in V$, then $\phi_{\beta} (v) = (\lambda_1, \lambda_2, \ldots, \lambda_n)$. And if $\gamma$ is the standard basis for $\mathbb{R}^n$, then this gives $\lambda_1 (1,0,0, \ldots, 0) + \lambda_2 (0,1,0, \ldots, 0) + \ldots + \lambda_n (0,0, \ldots, 1)$. So In the first column just come the coordinates $\lambda_i$? I'm confused about this.

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  • $\begingroup$ Could you clarify what "coordinate map" means? $\endgroup$ – Rory Daulton Jul 3 '15 at 19:12
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    $\begingroup$ @RoryDaulton I think it's pretty clear from his attempted solution. If for $v\in V$ I can write $v=\lambda_1v_1+\dots+\lambda_nv_n$ with respec to basis $\beta$, then the coordinate map would be $\phi_{\beta}(v)=(\lambda_1,\dots,\lambda_n)$ $\endgroup$ – Alex Mathers Jul 3 '15 at 19:22
  • $\begingroup$ Note that that $[v]_\beta = \phi(v) = [\phi(v)]_\gamma$ $\endgroup$ – Omnomnomnom Jul 3 '15 at 19:56
  • $\begingroup$ @Omnomnomnom, so what does that mean? Was I right in stating that in the columns are just the coordinates of $v$? $\endgroup$ – Kamil Jul 3 '15 at 20:00
  • $\begingroup$ @Kamil it means that the $i$th column is the coordinates of $v_i$ with respect to $\beta$. $\endgroup$ – Omnomnomnom Jul 3 '15 at 20:24
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If $T$ is any linear transformation, the matrix $[T]^{\gamma}_{\beta}$ can be written $\begin{bmatrix} [T(v_1)]_{\gamma} & \cdots & [T(v_n)]_{\gamma}\end{bmatrix}$. In other words, the $i$th column of $[T]^{\gamma}_{\beta}$ is $T(v_i)$ written in $\gamma$ coordinates.

For $\phi_{\beta}$, the vector $\phi_{\beta}(v_i)$ is $e_i = (0,\ldots0 ,1,0, \ldots,0)$, so the $i$th column of $[\phi_{\beta}]_{\beta}^{\gamma}$ is $e_i$. This is the same as saying $[\phi_{\beta}]_{\beta}^{\gamma}$ is the identity matrix.

Another way to see this is the equation: $$[\phi_{\beta}]_{\beta}^{\gamma}[v]_{\beta} = [\phi_{\beta}(v)]_{\gamma}$$ But if $v= \lambda_1v_1 + \cdots + \lambda_nv_n$, then we have $[v]_{\beta} = [\phi_{\beta}(v)]_{\gamma} = (\lambda_1,\ldots,\lambda_n)$, and so the matrix $[\phi_{\beta}]_{\beta}^{\gamma}$ is the identity.

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Your proof that $\phi_\beta$ is linear is correct.

You're on the right track to compute $[\phi_\beta]_\beta^\gamma$. For a complete computation, write $$ \phi_\beta(v_j) = \left(\lambda_{1j},\dotsc,\lambda_{nj}\right) = \lambda_{1j}e_1+\dotsb+\lambda_{nj}e_n $$ for $1\leq j\leq n$. Hence $$ [\phi_\beta]_\beta^\gamma = \begin{bmatrix} \lambda_{11} & \cdots & \lambda_{1n} \\ \vdots & \ddots & \vdots \\ \lambda_{n1} & \cdots & \lambda_{nn} \end{bmatrix} $$

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Would this not just be the identity matrix? For example, if we write $v=\sum_{i=1}^{n}\lambda_iv_i$ then can equivalently write $v=(\lambda_1,\dots,\lambda_n)_{\beta}$ with respect to the given basis. But if we have $\phi_{\beta}(v)=(\lambda_1,\dots,\lambda_n)\in\mathbb{R}^n$, then we haven't changed it, we've just taken the scalar values. So the matrix would just be the identity.

Unless of course, it wants the "coordinate map" with respect to the standard basis.

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  • $\begingroup$ I don't understand your post. What you mean with 'then we haven't changed it'. I don't understand why it is the identity matrix. $\endgroup$ – Kamil Jul 3 '15 at 19:47

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