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Chapter 5: Problem 10

Suppose \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) is defined by \(F(x, y, z)=(x+y+z, 2 x-3 y+4 z) .\) Show that \(F\) is linear.

Short Answer

Expert verified
To show that \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(x+y+z, 2x-3y+4z)\) is linear, we must demonstrate that it satisfies additivity and homogeneity. By computing \(F(\boldsymbol{u}+\boldsymbol{v})\) and \(F(\boldsymbol{u})+F(\boldsymbol{v})\) for arbitrary vectors \(\boldsymbol{u}\) and \(\boldsymbol{v}\), we find that \(F(\boldsymbol{u}+\boldsymbol{v}) = F(\boldsymbol{u})+F(\boldsymbol{v})\), fulfilling the additivity condition. Additionally, by computing \(F(c\boldsymbol{u})\) and \(cF(\boldsymbol{u})\) for a scalar value c, we find that \(F(c\boldsymbol{u}) = cF(\boldsymbol{u})\), fulfilling the homogeneity condition. Therefore, F is a linear function.

Step by step solution

01

Define vectors u and v in R^3

Let \(\boldsymbol{u}=(u_1,u_2,u_3)\) and \(\boldsymbol{v}=(v_1,v_2,v_3)\) be two arbitrary vectors in R^3. We will show that F is additive by finding F(u + v) and comparing it to F(u) + F(v).
02

Compute F(u + v)

We need to compute \(F(\boldsymbol{u}+\boldsymbol{v})\): \(\boldsymbol{u}+\boldsymbol{v}=(u_1+v_1,u_2+v_2,u_3+v_3)\) Now, let's find \(F(\boldsymbol{u}+\boldsymbol{v})\): \(F(u_1+v_1, u_2+v_2, u_3+v_3)= ( (u_1+v_1) + (u_2+v_2) + (u_3+v_3),\ 2(u_1+v_1) - 3(u_2+v_2) + 4(u_3+v_3)) \)
03

Compute F(u) + F(v)

Now, let's compute \(F(\boldsymbol{u})+F(\boldsymbol{v})\): \(F(\boldsymbol{u})=(u_1+u_2+u_3, 2u_1-3u_2+4u_3) \) and \(F(\boldsymbol{v})=(v_1+v_2+v_3, 2v_1-3v_2+4v_3)\) So, \(F(\boldsymbol{u})+F(\boldsymbol{v})= (u_1+u_2+u_3 + v_1+v_2+v_3, 2u_1-3u_2+4u_3 + 2v_1-3v_2+4v_3)\)
04

Check if F(u + v) = F(u) + F(v)

Comparing the results from Step 2 and Step 3, we can see that they are the same: \(F(\boldsymbol{u}+\boldsymbol{v}) = F(\boldsymbol{u})+F(\boldsymbol{v})\) Thus, F satisfies the additivity condition.
05

Compute F(cu) for a scalar c

Now we need to check if F is homogeneous. Let c be any real number, and compute \(F(c\boldsymbol{u})\): \(F(cu_1,cu_2,cu_3) = (c(u_1+u_2+u_3), 2(cu_1)-3(cu_2)+4(cu_3))\)
06

Check if F(cu) = cF(u)

Now, we compute \(cF(\boldsymbol{u})\): \(cF(\boldsymbol{u}) = c(u_1+u_2+u_3, 2u_1-3u_2+4u_3) = (c(u_1+u_2+u_3), 2(cu_1)-3(cu_2)+4(cu_3))\) Comparing the results from Step 5 and Step 6, we can see that they are the same: \(F(c\boldsymbol{u}) = cF(\boldsymbol{u})\) Thus, F satisfies the homogeneity condition.
07

Conclusion

Since the function F satisfies both additivity and homogeneity, we can conclude that F is a linear function.

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Most popular questions from this chapter

Suppose \(F: V \rightarrow U\) and \(G: U \rightarrow W\) are linear. Prove (a) \(\operatorname{rank}(G \circ F) \leq \operatorname{rank}(G)\) (b) \(\operatorname{rank}(G \circ F) \leq \operatorname{rank}(F)\)

Suppose \(F: V \rightarrow U\) is linear. Show that \(F(-v)=-F(v).\)

For each of the following linear operators \(T\) on a vector space \(V\) and ordered bases \(\beta\), compute \([T]_{\beta}\), and determine whether \(\beta\) is a basis consisting of eigenvectors of \(T\). (a) $\mathrm{V}=\mathrm{R}^{2}, \mathrm{~T}\left(\begin{array}{l}a \\\ b\end{array}\right)=\left(\begin{array}{l}10 a-6 b \\ 17 a-10 b\end{array}\right)\(, and \)\beta=\left\\{\left(\begin{array}{l}1 \\\ 2\end{array}\right),\left(\begin{array}{l}2 \\ 3\end{array}\right)\right\\}$ (b) $\quad \mathrm{V}=\mathrm{P}_{1}(R), \mathrm{T}(a+b x)=(6 a-6 b)+(12 a-11 b) x$, and $$ \beta=\\{3+4 x, 2+3 x\\} $$ (c) $\mathrm{V}=\mathrm{R}^{3}, \mathrm{~T}\left(\begin{array}{l}a \\ b \\\ c\end{array}\right)=\left(\begin{array}{r}3 a+2 b-2 c \\ -4 a-3 b+2 c \\\ -c\end{array}\right)$, and $$ \beta=\left\\{\left(\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right),\left(\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right),\left(\begin{array}{l} 1 \\ 0 \\ 2 \end{array}\right)\right\\} $$ (d) \(\mathrm{V}=\mathrm{P}_{2}(R), \mathrm{T}\left(a+b x+c x^{2}\right)=\) $$ (-4 a+2 b-2 c)-(7 a+3 b+7 c) x+(7 a+b+5 c) x^{2}, $$ and \(\beta=\left\\{x-x^{2},-1+x^{2},-1-x+x^{2}\right\\}\) (e) $\mathrm{V}=\mathrm{P}_{3}(R), \mathrm{T}\left(a+b x+c x^{2}+d x^{3}\right)=$ $$ -d+(-c+d) x+(a+b-2 c) x^{2}+(-b+c-2 d) x^{3}, $$ and \(\beta=\left\\{1-x+x^{3}, 1+x^{2}, 1, x+x^{2}\right\\}\) (f) $\mathrm{V}=\mathrm{M}_{2 \times 2}(R), \mathrm{T}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{ll}-7 a-4 b+4 c-4 d & b \\\ -8 a-4 b+5 c-4 d & d\end{array}\right)$, and $$ \beta=\left\\{\left(\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right),\left(\begin{array}{rr} -1 & 2 \\ 0 & 0 \end{array}\right),\left(\begin{array}{ll} 1 & 0 \\ 2 & 0 \end{array}\right),\left(\begin{array}{rr} -1 & 0 \\ 0 & 2 \end{array}\right)\right\\} $$

5.90. Let \(F: V \rightarrow U\) be linear and let \(W\) be a subspace of \(V\). The restriction of \(F\) to \(W\) is the \(\operatorname{map} F | W: W \rightarrow U\) defined by \(F | W(v)=F(v)\) for every \(v\) in \(W\). Prove the following: (a) \(F | W\) is linear; (b) \(\operatorname{Ker}(F | W)=(\operatorname{Ker} F) \cap W\) (c) \(\operatorname{Im}(F | W)=F(W)\)

Prove Theorem 5.9: Suppose \(V\) has finite dimension and \(\operatorname{dim} V=\operatorname{dim} U .\) Suppose \(F: V \rightarrow U\) is linear. Then \(F\) is an isomorphism if and only if \(F\) is nonsingular.
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