Problem 14 Show that if $S_{1}$ and \(S_{... [FREE SOLUTION]

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Linear Algebra

Stephen H. Friedberg, Arnold J. Insel, Lawrence E. Spense

$Math Studyset 91影视 Explanations$ Math

5 Edition

Chapter 1: Problem 14

Show that if $S_{1}$ and $S_{2}$ are arbitrary subsets of a vector space V, then $\operatorname{span}\left(S_{1} \cup S_{2}\right)=\operatorname{span}\left(S_{1}\right)+\operatorname{span}\left(S_{2}\right)$. (The sum of two subsets is defined in the exercises of Section 1.3.)

Short Answer

Expert verified

We have proven that $\operatorname{span}(S_1 \cup S_2) = \operatorname{span}(S_1) + \operatorname{span}(S_2)$ by showing that each subspace is a subset of the other. We first showed any vector in $\operatorname{span}(S_1 \cup S_2)$ can be expressed as a linear combination of vectors in $S_1 \cup S_2$ and separated into sums from $\operatorname{span}(S_1)$ and $\operatorname{span}(S_2)$, making it an element of $\operatorname{span}(S_1) + \operatorname{span}(S_2)$. Then, we showed any vector in $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ can be written as the sum of two vectors from $\operatorname{span}(S_1)$ and $\operatorname{span}(S_2)$ and expressed as a linear combination of vectors in $S_1 \cup S_2$, making it an element of $\operatorname{span}(S_1 \cup S_2)$. Thus, we concluded that these subspaces are equal: $\operatorname{span}(S_1 \cup S_2) = \operatorname{span}(S_1) + \operatorname{span}(S_2)$.

Step by step solution

Show that $\operatorname{span}(S_1 \cup S_2)$ is a subspace of $\operatorname{span}(S_1) + \operatorname{span}(S_2)$

We want to show that for any vector in the span of the union, say $v \in \operatorname{span}(S_1 \cup S_2)$, we also have that $v \in \operatorname{span}(S_1) + \operatorname{span}(S_2)$. Since $v \in \operatorname{span}(S_1 \cup S_2)$, we can express $v$ as a linear combination of vectors in $S_1 \cup S_2$: $$v = a_1v_1 + a_2v_2 + \dots + a_nv_n$$ where $v_i \in S_1 \cup S_2$ and $a_i$ are scalars for all $i$. For each $v_i$, we have two possibilities: $v_i \in S_1$ or $v_i \in S_2$. So we can re-write this as: $$v = \sum_{i: v_i \in S_1} a_iv_i + \sum_{i: v_i \in S_2}a_iv_i$$ Notice that the summation on the left contains only vectors in $S_1$, so it belongs to $\operatorname{span}(S_1)$. Similarly, the summation on the right contains only vectors in $S_2$, so it belongs to $\operatorname{span}(S_2)$. Thus, we conclude that $v \in \operatorname{span}(S_1) + \operatorname{span}(S_2)$.

Show that $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ is a subspace of $\operatorname{span}(S_1 \cup S_2)$

For this step, we want to show that if a vector is in the sum of the spans, it must also be in the span of the union. In other words, we want to show that $\forall v \in \operatorname{span}(S_1) + \operatorname{span}(S_2), v \in \operatorname{span}(S_1 \cup S_2)$. Any vector $v$ can be written as the sum of two vectors: $$v = a_1v_1 + a_2v_2 + \dots + a_mv_m + b_1w_1 + b_2w_2 + \dots + b_nw_n$$ where $v_i \in S_1$, $w_j \in S_2$ and $a_i, b_j$ are scalars. Notice that for each term in the sum, the corresponding vector is in $S_1 \cup S_2$, so we can rewrite $v$ as a linear combination of the vectors in $S_1 \cup S_2$: $$v = a_1v_1 + a_2v_2 + \dots + a_mv_m + b_1w_1 + b_2w_2 + \dots + b_nw_n$$ This means that $v \in \operatorname{span}(S_1 \cup S_2)$.

Conclude that $\operatorname{span}(S_1 \cup S_2) = \operatorname{span}(S_1) + \operatorname{span}(S_2)$

We have shown in the previous steps that $\operatorname{span}(S_1 \cup S_2)$ is a subspace of $\operatorname{span}(S_1) + \operatorname{span}(S_2)$ and vice versa. Since they are both subspaces of each other, we can now conclude that they are equal subspaces: $$\operatorname{span}\left(S_{1} \cup S_{2}\right)=\operatorname{span}\left(S_{1}\right)+\operatorname{span}\left(S_{2}\right)$$

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91影视

Show that if \(S_{1}\) and \(S_{2}\) are arbitrary subsets of a vector space V, then \(\operatorname{span}\left(S_{1} \cup S_{2}\right)=\operatorname{span}\left(S_{1}\right)+\operatorname{span}\left(S_{2}\right)\). (The sum of two subsets is defined in the exercises of Section 1.3.)

Short Answer

Step by step solution

Show that \(\operatorname{span}(S_1 \cup S_2)\) is a subspace of \(\operatorname{span}(S_1) + \operatorname{span}(S_2)\)

Show that \(\operatorname{span}(S_1) + \operatorname{span}(S_2)\) is a subspace of \(\operatorname{span}(S_1 \cup S_2)\)

Conclude that \(\operatorname{span}(S_1 \cup S_2) = \operatorname{span}(S_1) + \operatorname{span}(S_2)\)

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