Problem Is $\mathbb{R}^2$ a subspace of the complex vector space $\mathbb{C}^2$? Solution No, $\mathbb{R}^2$ is not a subspace of $\mathbb{C}^2$ - a subspace has the same addition and scalar multiplication as the space it’s a subspace of - see definition 1.32. But $\mathbb{C}^2$ is a vector space over $\mathbb{C}$, thus for any scalar $\lambda \in \mathbb{C}$ any vector $v$ of our subset $\mathbb{R}^2$ should satisfy $\lambda v \in \mathbb{R}^2$, but for $\lambda = i$, this is clearly not the case - take e....

Problem Suppose $b \in \mathbb{R}$. Show that the set of continuous real-valued functions $f$ on the interval $[0,1]$ such that $\int_0^1 f = b$ is a subspace of $\mathbb{R}^{[0,1]}$ if and only if $b = 0$. Solution $(\leftarrow)$ given that the zero-function $0(x) = 0$ has to be in the set for it to be a subspace $b = \int_0^1 0(x)dx = 0$ since $\int_0^1 f = b$ for all $f$ including when $f(x) = 0(x)$...

Problem Show that the set of differentiable real-valued functions $f$ on the interval $(-4,4)$ such that $f^\prime(-1) = 3f(2)$ is a subspace of $\mathbb{R}^{(-4,4)}$ Solution The zero-function $0(x) = 0$ is in the set, since $0 = \frac{d}{dt}0 = \frac{d}{dt}0(x) = 0^\prime(x)$ from which it follows that $0^\prime(-1) = 0 = 3 \cdot 0 = 3 \cdot 0(2)$ If $f$ and $g$ are in the subset, then $h(x) = f(x) + g(x)$ has the property that $h^\prime(x) = \frac{d}{dx}(f(x) + g(x)) = f^\prime(x) + g^\prime(x)$ and thus $h^\prime(-1) = f^\prime(-1) + g^\prime(-1) = 3f(2) + 3g(2) = 3(g + f)(2) = 3h(2)$ and thus $h(x)$ is in the subset....

Problem Verify all assertions in Example 1.35 Solution Subspace (a) $(\leftarrow)$ { $(x_1, x_2, x_3, x_4) \in \mathbb{F}^4 | x_3 = 5x_4 + b $ } is a subspace, then $(0,0,0,0) = 0$ is an element of this subspace by theorem 1.34, which implies $0 = x_3 = 5x_4 + b = 5\cdot 0 + b = b$. $(\rightarrow)$ $b = 0$, then: $0 = (0,0,0,0)$ satisfies the condition $x_3 = 5x_4 + b$ because $x_3 = 0 = 5 \cdot 0 + 0 = 5x_4 + b$, and is therefore in the subset....

Problem For each of the following subsets of $\mathbb{F}^3$, determine whether it is a subspace of $\mathbb{F}^3$: Solution We need to varify the 3 conditions for a subset of a vectorspace to be a subspace given in theorem 1.34 for each of the four given subsets. Subset (a) The additive identity $(x_1, x_2, x_3) = (0,0,0) = 0$ is in the subset because $0 + 2 \cdot 0 + 3 \cdot 0 = 0$....