Here is the question I want to answer:
Let $x = (1,1) \in V = \mathbb R^2$ and $y = (1, 2, 1) \in W = \mathbb R^3.$ Calculate the coordinates of $x \otimes y \in V \otimes W$ with respect to the tensor product basis of the standard bases for $V$ and $W.$
Here is my solution to it:
Since the standard basis of $V = \mathbb R^2$ is $\{e_1^2 = \begin{pmatrix}
1\\
0
\end{pmatrix}, e_2^2 = \begin{pmatrix}
0\\
1
\end{pmatrix}\}$ and the standard basis of $V = \mathbb R^3$ is $\{e_1^3 = \begin{pmatrix}
1\\
0\\
0
\end{pmatrix}, e_2^3 = \begin{pmatrix}
0\\
1\\
0
\end{pmatrix}\}, e_3^3 = \begin{pmatrix}
0\\
0\\
1
\end{pmatrix}\}$ then a basis of $V \otimes W$ is $$\{ e_1^2 \otimes e_1^3 , e_1^2 \otimes e_2^3, e_1^2 \otimes e_3^3, e_2^2 \otimes e_1^3, e_2^2 \otimes e_2^3, e_2^2 \otimes e_3^3\} = \{ \begin{pmatrix}
1\\
0\\
0\\
0\\
0\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
1\\
0\\
0\\
0\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
0\\
1\\
0\\
0\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
0\\
0\\
1\\
0\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
0\\
0\\
0\\
1\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
0\\
0\\
0\\
0\\
1
\end{pmatrix}\}$$ Therefore, $\begin{pmatrix}
1\\
1
\end{pmatrix} \otimes \begin{pmatrix}
1\\
2\\
1
\end{pmatrix} = \begin{pmatrix}
1\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}\\
1 \begin{pmatrix}
1\\
2\\
1
\end{pmatrix}
\end{pmatrix} = \begin{pmatrix}
1\\
2\\
1\\
1\\
2\\
1
\end{pmatrix} = 1\begin{pmatrix}
1\\
0\\
0\\
0\\
0\\
0
\end{pmatrix}+ 2 \begin{pmatrix}
0\\
1\\
0\\
0\\
0\\
0
\end{pmatrix}+ 1 \begin{pmatrix}
0\\
0\\
1\\
0\\
0\\
0
\end{pmatrix}+ 1 \begin{pmatrix}
0\\
0\\
0\\
1\\
0\\
0
\end{pmatrix}+ 2 \begin{pmatrix}
0\\
0\\
0\\
0\\
1\\
0
\end{pmatrix}+ 1 \begin{pmatrix}
0\\
0\\
0\\
0\\
0\\
1
\end{pmatrix}\}.$
Then the coordinates of $x \otimes y$ is $(1,2,1,1,2,1).$
Is my solution correct?
Best Answer
I’ll write $a,b$ for the basis of $V$ and $u,v,w$ for the basis of $W$. Then $x=a+b$ and $y=u+2v+w$. Hence $$ \newcommand{\o}{\otimes} x\o y= (a+b)\o (u+2v+w) = a\o u + b\o u + 2 a \o v + 2 b \o v + a\o w + b\o w $$ Which is the same as your answer but easier to write.