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Back to the class

Section 3.3 #3: Find the inverse of $\begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}$, if it exists.
Solution: The inverse does not exist. Note that by Theorem 3.8 we could compute the inverse using the formula $$\begin{bmatrix} a&b \\ c&d \end{bmatrix}^{-1} = \dfrac{1}{ad-bc} \begin{bmatrix} d& -b \\ -c & a \end{bmatrix},$$ provided that $ad-bc \neq 0$. From this we see that there is no inverse, because $3(8)-6(4)=0$.

Section 3.3 #54: Find the inverse (if it exists): $\begin{bmatrix} 1&1&0 \\ 1&0&1 \\ 0&1&1 \end{bmatrix}$.
Solution: We use the row reduction method: compute $$\begin{array}{ll} [A \hspace{8pt} I] &= \begin{bmatrix} 1&1&0&1&0&0 \\ 1&0&1&0&1&0 \\ 0&1&1&0&0&1 \end{bmatrix} \\ &\sim \vdots \\ &\sim \begin{bmatrix} 1&0&0& \frac{1}{2}& \frac{1}{2} & -\frac{1}{2} \\ 0&1&0&\frac{1}{2}&-\frac{1}{2}&\frac{1}{2} \\ 0&0&1&-\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \end{bmatrix} \\ &= [I \hspace{8pt} A^{-1}]. \end{array}$$ Therefore we see that $$A^{-1} = \begin{bmatrix} \frac{1}{2}& \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2}&-\frac{1}{2}&\frac{1}{2} \\ -\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \end{bmatrix}.$$

For the next set of problems, recall the 10 axioms of vector spaces on pg.429: for all $\vec{u}, \vec{v}, \vec{w} \in V$ and $\alpha, \beta, \gamma \in \mathbb{F}$,
Axiom 1. $\vec{u}+\vec{v} \in V$
Axiom 2. $\vec{u}+\vec{v}=\vec{v}+\vec{u}$
Axiom 3. $(\vec{u}+\vec{v})+\vec{w} = \vec{u}+(\vec{v}+\vec{w})$
Axiom 4. There exists some $\vec{0} \in V$ such that $\vec{u}+\vec{0}=\vec{u}$
Axiom 5. There exists some $-\vec{u} \in V$ such that $\vec{u} + (-\vec{u}) = \vec{0}$
Axiom 6. $\alpha \vec{u} \in V$
Axiom 7. $\alpha (\vec{u}+\vec{v}) = \alpha \vec{u} + \alpha \vec{v}$
Axiom 8. $(\alpha + \beta) \vec{u} = \alpha \vec{u} + \beta \vec{u}$
Axiom 9. $\alpha (\beta \vec{u}) = (\alpha \beta)\vec{u}$
Axiom 10. $1 \vec{u} = \vec{u}$

Section 6.1 #2: Vector space or not? If so, prove it, if not, explain why: $(V,\mathbb{F})=\left( \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \colon x \geq 0, y \geq 0 \right\},\mathbb{R} \right)$ with the usual vector addition and scalar multiplication.
Solution: It is not a vector space. Axiom 6 fails because, for instance, if $u=\begin{bmatrix} 1 \\ 1 \end{bmatrix} \in V$, then $-u=\begin{bmatrix} -1 \\ -1 \end{bmatrix} \not\in V$.

Section 6.1 #10: Vector space or not? If so, prove it, if not, explain why: $(V,\mathbb{F}) = \left( \left\{ \begin{bmatrix} a & b \\ c&d \end{bmatrix} \colon ad = 0 \right\}, \mathbb{R} \right)$ with usual matrix addition and scalar multiplication.
Solution: It is not a vector space. Axiom 1 fails: consider the matrices $A = \begin{bmatrix} 0 &0 \\ 0&1 \end{bmatrix} \in V$ and $B = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \in V$ while $A+B=\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \not\in V$.

Section 6.1 #16: Vector space or not? If so, prove it, if not, explain why: $(V,\mathbb{F}) = (\mathbb{C}^{2 \times 1},\mathbb{C})$ with usual vector addition but scalar multiplcation defined by $$c \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} \overline{c} z_1 \\ \overline{c} z_2 \end{bmatrix}.$$ Solution: It is a vector space. Let us see why by to verifying the axioms: let $\vec{u} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}, \vec{v}=\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}, \vec{w}=\begin{bmatrix} w_1 \\ w_2 \end{bmatrix} \in V$ and $\alpha, \beta, \gamma \in \mathbb{C}$
Axiom 1. $\vec{u}+\vec{v} = \begin{bmatrix} u_1+v_1 \\ u_2+v_2 \end{bmatrix} \in V$
Axiom 2. $\vec{u}+\vec{v} = \begin{bmatrix} u_1+v_1 \\ u_2+v_2 \end{bmatrix} = \begin{bmatrix} v_1+u_1 \\ v_2+u_2 \end{bmatrix} = \vec{v}+\vec{u}$
Axiom 3. $(\vec{u}+\vec{v})+\vec{w}= \begin{bmatrix} (u_1+v_1)+w_1 \\ (u_2+v_2)+w_2 \end{bmatrix} = \begin{bmatrix} u_1 + (v_1+w_1) \\ u_2+(v_2+w_2) \end{bmatrix} = \vec{u}+(\vec{v}+\vec{w})$
Axiom 4. $\vec{0} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ and $\vec{u}+\vec{0} = \begin{bmatrix} u_1 + 0 \\ u_2 + 0 \end{bmatrix} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \vec{u}$
Axiom 5. $-\vec{u}=\begin{bmatrix} -u_1 \\ -u_2 \end{bmatrix}$ and $\vec{u}+ (-\vec{u})= \begin{bmatrix} u_1 + (-u_1) \\ u_2 + (-u_2) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \vec{0}$
Axiom 6. $\alpha \vec{u}=\begin{bmatrix} \overline{\alpha}u_1 \\ \overline{\alpha} u_2 \end{bmatrix} \in V$
Axiom 7. $\alpha (\vec{u}+\vec{v}) = \begin{bmatrix} \overline{\alpha}(u_1+v_1) \\ \overline{\alpha}(u_2+v_2) \end{bmatrix} = \begin{bmatrix} \overline{\alpha} u_1 + \overline{\alpha}v_1 \\ \overline{\alpha} u_2 + \overline{\alpha} v_2 \end{bmatrix} = \alpha \vec{u} + \alpha \vec{v}$
Axiom 8. $(\alpha + \beta) \vec{u} = \begin{bmatrix} (\overline{\alpha+\beta})u_1 \\ (\overline{\alpha + \beta})u_2 \end{bmatrix} = \begin{bmatrix} (\overline{\alpha}+\overline{\beta})u_1 \\ (\overline{\alpha}+\overline{\beta})u_2 \end{bmatrix}= \begin{bmatrix} \overline{\alpha}u_1 + \overline{\beta}u_1 \\ \overline{\alpha}u_2 + \overline{\beta} u_2 \end{bmatrix} = \alpha \vec{u} + \beta \vec{u}$
Axiom 9. It is well-known that $\overline{\alpha \beta}=\overline{\alpha} \overline{\beta}$. Therefore compute $$\alpha (\beta \vec{u}) = \alpha \begin{bmatrix} \overline{\beta}u_1 \\ \overline{\beta} u_2 \end{bmatrix} = \begin{bmatrix} \overline{\alpha}\overline{\beta} u_1 \\ \overline{\alpha}\overline{\beta} u_2 \end{bmatrix} = (\alpha \beta) \vec{u}.$$ Axiom 10. Calculate $$1 \vec{u} = \begin{bmatrix} 1 (u_1) \\ 1(u_2) \end{bmatrix} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \vec{u}$$ Since all 10 axioms of a vector space hold for this structure, we conclude that it is a vector space. $\blacksquare$

For the following problems recall Theorem 6.2: let $(V,\mathbb{F})$ be a vector space and $W \subset V$. Then $(W, \mathbb{F})$ is a subspace of $(V,\mathbb{F})$ if and only if both
a.) If $\vec{u},\vec{v} \in W$, then $\vec{u}+\vec{v} \in W$.
b.) If $\vec{u} \in V$ and $\alpha \in \mathbb{F}$, then $\alpha \vec{u} \in W$.

Section 6.1 #24: Is $W = \left\{ \begin{bmatrix} a \\ 0 \\ a \end{bmatrix} \right\}$ a subspace of $(V,\mathbb{F})=(\mathbb{R}^3,\mathbb{R})$?
Solution: It is a subspace. Let $\vec{u} = \begin{bmatrix} u_1 \\ 0 \\ u_1 \end{bmatrix}, \vec{v} = \begin{bmatrix} v_1 \\ 0 \\ v_1 \end{bmatrix} \in W$. First verify a.):
$$\vec{u}+\vec{v} = \begin{bmatrix} u_1+v_1 \\ 0 \\ u_1+v_1 \end{bmatrix} \in W,$$ so a.) holds. Now let $\alpha \in \mathbb{F}$ and verify b.):
$$\alpha \vec{u} = \begin{bmatrix} \alpha u_1 \\ 0 \\ \alpha u_1 \end{bmatrix} \in W;$$ now an application of Theorem 6.2 completes the proof.

Section 6.1 #34: Is $W=\{bx + cx^2 \colon b,c \in \mathbb{R}\}$ a subspace of $(V,\mathbb{F})=(\mathcal{P}_2,\mathbb{R})$?
Solution: It is a subspace. Let $\vec{u}=u_1 x + u_2 x^2 \in W$ and $\vec{v}=v_1 x + v_2 x^2 \in W$. Now verify a.):
$$\vec{u}+\vec{v} = (u_1 x + u_2 x^2) + (v_1 x + v_2 x^2) = (u_1+v_1)x + (u_2+v_2)x^2 \in W.$$ Now let $\alpha \in \mathbb{F}$ and verify b.):
$$\alpha \vec{u} = \alpha(u_1x + u_2 x^2) = (\alpha u_1)x + (\alpha u_2) x^2 \in W.$$ Now an application of Theorem 6.2 completes the proof.

Section 6.1 #36: Is $W=\{a+bx+cx^2 \colon a,b,c \in \mathbb{R}, abc=0\}$ a subspace of $(V,\mathbb{F})=(\mathcal{P}_2,\mathbb{R})$?
Solution: It is not a subspace. To see why, consider $\vec{u}=1+x \in W$ and $\vec{v}=x^2 \in W$. However, $$\vec{u}+\vec{v} = 1+x+x^2 \not\in W,$$ which tells us that a.) fails. However Theorem 6.2 is an "if and only if", and so the failure of a.) also implies the failure of the sentence "$(W,\mathbb{F})$ is a subspace". Therefore $(W,\mathbb{F})$ is not a subspace of $(\mathcal{P}_2,\mathbb{F})$.