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Example 157 Let $V=\mathbb{R}[x]$ denote the vector space of polynomials in $x$ with real coefficients, and let $W$ denote the vector space of real sequences $\left(a_{n}\right)_{n=0}^{\infty}$. Then $V$ and $W$ are both infinite dimensional but are not isomorphic.
Solution. The set $B=\left\{1, x, x^{2}, x^{3}, \ldots\right\}$ is a basis for $V$. That it is linearly independent shows that $V$ is not finite-dimensional. The set of sequences $\left\{\left(\delta_{i n}\right)_{n=0}^∞ \mid i \geqslant 0\right\}$ is linearly independent and so $W$ is also infinite-dimensional.
However the set $S=\left\{\left(t^{n}\right)_{n=0}^{\infty} \mid t \in \mathbb{R}\right\}$ is an uncountable linearly independent subset of $W$ and hence $W$ does not have a countable basis. We prove that $S$ is linearly independent below. Suppose that
$$
\alpha_{1}\left(t_{1}^{n}\right)_{n=0}^∞+\dots+\alpha_{k}\left(t_{k}^{n}\right)_{n=0}^∞=(0)
$$
for real numbers $\alpha_{1}, \ldots, \alpha_{k}, t_{1}, \ldots, t_{k}$ with the $t_{i}$ distinct. Then for all $n \geqslant 0$ we have
$$
\alpha_{1} t_{1}^{n}+\cdots+\alpha_{k} t_{k}^{n}=0 .
$$
These equations for $0 \leqslant n< k$ can be rewritten as the single matrix equation
$$
\left(\begin{array}{ccccc}
1 & 1 & 1 & \cdots & 1 \\
t_{1} & t_{2} & t_{3} & \cdots & t_{k} \\
t_{1}^{2} & t_{2}^{2} & t_{3}^{2} & \cdots & t_{k}^{2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
t_{1}^{k-1} & t_{2}^{k-1} & t_{3}^{k-1} & \cdots & t_{k}^{k}
\end{array}\right)\left(\begin{array}{c}
\alpha_{1} \\
\alpha_{2} \\
\alpha_{3} \\
\vdots \\
\alpha_{k}
\end{array}\right)=\left(\begin{array}{c}
0 \\
0 \\
0 \\
\vdots \\
0
\end{array}\right) .
$$
As the $t_{i}$ are distinct, then the above $k \times k$ matrix is invertible (Vandermonde matrix). Hence $α_1=\dots=α_k=0$, $S$ is an uncountable linearly independent set. No such set exists in $V$ and hence $W$ is not isomorphic to $V$. | 
Czhang271828
Posted at 2023-5-21 15:05:01
