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Let $T:\mathbb{R}^3\ \rightarrow \mathbb{R}^2$ be a linear mapping defined by $T(x,y,z)=(a_1x+a_2y+a_3z,b_1x+b_2y+b_3z)$, such that $a_1,a_2,a_3,b_1,b_2,b_3\in \mathbb{R}$ (constants).

Let $\{v_1,v_2\}$ be the dual basis to the standard basis of $\mathbb{R}^2$ and $\{u_1,u_2,u_3\}$ be the dual basis to the standard basis of $\mathbb{R}^3$.

What is $T^*(v_1)$ and $T^*(v_2)$. Furthermore, what is $T^*(v_1)$ and $T^*(v_2)$ in terms of the basis $\{u_1,u_2,u_3\}$, where $T^*$ represents the transpose of $T$.

I am not really sure how to start, and I do apologize for the lack of work.

I do realize that if $A$ is the matrix representation of $T$, then its transpose $A^T$ will be the matrix representation of $T^*$. I am just unsure how to go about the problem, and am confused regarding how to evaluate $T^*$ of some vector. (i.e. $T^*(v_1)$).

Any help would be much appreciated. Thanks in advance!

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You have all the knlwledge need to solve the problem. Since the matrix of $T$ with respect to the standard bases is$$\begin{pmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\end{pmatrix},$$the matrix of $T^*$ with respect to the dual bases is its transpose$$\begin{pmatrix}a_1&b_1\\a_2&b_2\\a_3&b_3\end{pmatrix}.$$Therefore, $T^*(v_1)=a_1u_1+a_2u_2+a_3u_3$ and $T^*(v_2)=b_1u_1+b_2u_2+b_3u_3$.

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  • $\begingroup$ Oh wow, I feel stupid. I forgot how to take the matrix representation of that linear operator. That makes so much more sense now. Thank you very much! $\endgroup$ – The math god Mar 15 '18 at 10:52
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To pre-add to the answer of José Carlos:

The matrix representation of any linear map $T:V\to W$ consists of the column vectors $T(b_i)$ (coordinated with a fixed basis of $W$), where $b_i$ ranges over a fixed basis of $V$.

In case of $\Bbb R^n$'s we fix the standard basis: $\pmatrix{1\\0\\ \vdots},\, \pmatrix{0\\1\\ \vdots},\, \dots$
So that, to obtain the matrix representation of $T$, you simply need to apply it on the vectors $\pmatrix{1\\0\\0},\ \pmatrix{0\\1\\0},\ \pmatrix{0\\0\\1}$ to obtain the columns.

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