- #1

harvesl

- 9

- 0

Let [itex]V[/itex] be an [itex]n[/itex]-dimensional vector space over [itex]\mathbb{R}[/itex]

Prove that [itex]V[/itex] contains a subspace of dimension [itex]r[/itex] for each [itex]r[/itex] such that [itex]0 \leq r \leq n[/itex]

Since [itex]V[/itex] is n-dimensional, there exists [itex]v_1, v_2,...,v_n \in V[/itex] such that [itex]\left\{v_1,v_2,...,v_n\right\}[/itex] is a basis for [itex]V[/itex].

For an element [itex] x \in V[/itex], we can write [itex]x[/itex] as

[itex]x = \lambda_1 v_1 + \lambda_2 v_2 + ... + \lambda_n v_n[/itex] for [itex]\lambda_i \in \mathbb{R}[/itex]

Since [itex]Sp\left\{v_1,v_2,...,v_n\right\} = V[/itex]

Therefore we can also say that

[itex] V = \left\{\lambda_1 v_1 + \lambda_2 v_2 + ... + \lambda_n v_n\right\}[/itex]

Since [itex]V[/itex] is a vector space [itex]\left\{(0)\right\} \in V[/itex] and [itex]\left\{(0)\right\}[/itex] is clearly a 0-dimension subspace of [itex]V[/itex] as it is closed under both addition and multiplication.

Since any non-empty subset of a linearly independent set of vectors is also linearly independent we have that for [itex]q[/itex], where [itex] 1 \leq q \leq n[/itex]

[itex]\left\{v_1, v_2,..., v_q\right\} = \left\{\mu_1 v_1 + \mu_2 v_2 + ... + \mu_q v_q : \mu_i \in \mathbb{R}\right\}

\subseteq \left\{\lambda_1 v_1 + \lambda_2 v_2 + ... + \lambda_n v_n\right\} = V[/itex]

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Second question

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Let [itex]U,W[/itex] be subspaces of [itex]V[/itex] with [itex]U \subseteq W[/itex]. Show that there is some subspace [itex]W'[/itex] of [itex]V[/itex] such that [itex]W \cap W' = U[/itex] and [itex]W +W' = V[/itex]

Firstly, since [itex]W \cap W' = U[/itex],[itex]\ \ \ U \subseteq W'[/itex] because if [itex]x \in U[/itex] then [itex]x \in W'[/itex]

Therefore we can write that

[itex]W' = U \cup \left\{v_1, v_2, ... , v_m\right\}[/itex]

For some vectors [itex](v_1, v_2, ... , v_m) \in V \backslash U[/itex] but [itex](v_1, v_2, ... , v_m) \notin W[/itex] because if any [itex]v_i \in W[/itex] then [itex]v_i \in W \cap W'[/itex] and we would have that [itex]W \cap W' \neq U[/itex]

Since [itex]dim(V) = n[/itex], if we set [itex]dim(W) = x[/itex] we get that

[itex]dim(\left\{v_1, v_2, ... , v_m\right\}) = m = n-x[/itex]

Hence we can find [itex]m[/itex] linearly independent [itex]\left\{v_1, v_2, ... , v_m\right\} \notin Sp(W)[/itex] to give

[itex]W \cap W' = W \cap (U \cup \left\{v_1, v_2, ... , v_m\right\}) = U[/itex]

and

[itex] W + W' = \left\{w + w' : w \in W, w' \in W'\right\} = Sp(W \cup W') = V[/itex]

As [itex]dim(W \cup W') = x + n-x = n[/itex]

Again, thanks to anyone who helps!