VS.M11 :
Let \(V\) be the set \(\mathbb{C}^2\) with the usual vector addition, but with scalar multiplication definded by:
\(\alpha\begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} \alpha y \\ \alpha x\end{bmatrix}\)
Determine whether or not \(V\) is vector space with these operations.
JB:
Relying heavily on the given definition of VS (vector space) in the text (pg 271), I’ll check if any of the 10 condtions may not be met. Upon careful inspection, it should be noted that the \(x\) and \(y\) get flipped upon being multiplied by scalar \(\alpha\).
Looking at the last condition, \(O\) (pg 271), we see that if \(\vec{\textbf{u}}\in\textit{V}\), then \(1\vec{\textbf{u}}=\vec{\textbf{u}}\).
Applying this condition to the provided, we get:
\(1\begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 1 y \\ 1 x\end{bmatrix}\) which would only provide a valid result if \(x=y\). This can be more easily seen by way of example where \(x\not=y\):
\(1\begin{bmatrix} 2 \\ 6\end{bmatrix} \not= \begin{bmatrix} 6 \\ 2 \end{bmatrix}\).
\(\therefore{}\) \(V\) is not a vector space because \(1\vec{\textbf{u}}\not=\vec{\textbf{u}}\).