Chapter 2: Problem 5
Prove that \(\mathbf{F}^{\infty}\) is infinite dimensional.
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Chapter 2: Problem 5
Prove that \(\mathbf{F}^{\infty}\) is infinite dimensional.
These are the key concepts you need to understand to accurately answer the question.
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Suppose \(m\) is a positive integer. Is the set consisting of 0 and all polynomials with coefficients in \(\mathbf{F}\) and with degree equal to \(m\) a subspace of \(\mathcal{P}(\mathbf{F})\) ?
Suppose that \(U\) and \(W\) are both five-dimensional subspaces of \(\mathbf{R}^{9}\). Prove that \(U \cap W \neq\\{0\\}\).
Let \(U\) be the subspace of \(\mathbf{R}^{5}\) defined by $$ U=\left\\{\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right) \in \mathbf{R}^{5}: x_{1}=3 x_{2} \text { and } x_{3}=7 x_{4}\right\\} $$ Find a basis of \(U\).
Prove that if \(V\) is finite dimensional and \(U_{1}, \ldots, U_{m}\) are subspaces of \(V\), then $$ \operatorname{dim}\left(U_{1}+\cdots+U_{m}\right) \leq \operatorname{dim} U_{1}+\cdots+\operatorname{dim} U_{m} $$
You might guess, by analogy with the formula for the number of elements in the union of three subsets of a finite set, that if \(U_{1}, U_{2}, U_{3}\) are subspaces of a finite-dimensional vector space, then $$ \begin{array}{l} \operatorname{dim}\left(U_{1}+U_{2}+U_{3}\right) \\ =\operatorname{dim} U_{1}+\operatorname{dim} U_{2}+\operatorname{dim} U_{3} \\\ \quad-\operatorname{dim}\left(U_{1} \cap U_{2}\right)-\operatorname{dim}\left(U_{1} \cap U_{3}\right)-\operatorname{dim}\left(U_{2} \cap U_{3}\right) \\ +\operatorname{dim}\left(U_{1} \cap U_{2} \cap U_{3}\right) \end{array} $$ Prove this or give a counterexample.
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