Riemann surface
matlab

In each case, determine whether $V$ is a vector space. If it is not a vector space, explain why not. If it is, find basis vectors for $V$.
(a) $V$ is the subset of $\mathbb{R}^3$ defined by
$$4 x-5 y+z=1$$
(b) Let the vector $\mathbf{w}=\left(w_1, w_2, \cdots, w_n\right)$ represent a portfolio’s holdings, where each component $w_i$ represents the fraction of the portfolio’s total market value in asset $i$. Let $V$ be the set of weight vectors that can represent market-neutral long/short portfolios. The weights $w_i$ satisfy $0<w_i \leq 1$ for long positions, $-1 \leq w_i<0$ for short positions, and
$$\sum_i w_i=0$$
(c) $V$ is the set of vectors in $\mathbb{R}^2$ for which $M \mathbf{v}=\mathbf{v}$, where
$$M=\left(\begin{array}{cc} 0 & -1 \ 2 & 3 \end{array}\right) .$$

Solution:
(a) $V$ is not a vector space since it does not contain the origin (i.e., the zero vector).
(b) $V$ is not a vector space because it is not closed under scalar multiplication or addition, which violate the inequality as well as the budget constraint.
(c) $M$ has eigenvalues of 1 and 2 , so the eigenvector
$$\mathbf{v}_1=\left(\begin{array}{c} 1 \ -1 \end{array}\right)$$
is a basis for the vector space $V$.

Calculate the trace and the determinant of the matrix. If the matrix is non-singular, compute its inverse. If the matrix is singular, determine its image and kernel.
(a)
$$\left(\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right)$$
(b)
$$\left(\begin{array}{ll} 3 & 4 \ 6 & 8 \end{array}\right)$$
(c)
$$\frac{1}{2}\left(\begin{array}{cc} 1 & -1 \ -1 & 3 \end{array}\right)$$
(d)
$$M=\left(\begin{array}{ll} 1 & \rho \ \rho & 1 \end{array}\right)$$
(e)
$$M=\left(\begin{array}{cc} x & x-x^2 \ 1 & 1-x \end{array}\right)$$

Solution:
(a)
$$M=\left(\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right), \operatorname{Tr} M=5, \operatorname{Det} M=-2, M^{-1}=\frac{1}{2}\left(\begin{array}{cc} -4 & 2 \ 3 & -1 \end{array}\right)$$
(b)
$$M=\left(\begin{array}{ll} 3 & 4 \ 6 & 8 \end{array}\right), \operatorname{Tr} M=11, \operatorname{Det} M=0, \operatorname{Im} M=\operatorname{span}\left{\left(\begin{array}{l} 1 \ 2 \end{array}\right)\right}, \operatorname{Ker} M=\operatorname{span}\left{\left(\begin{array}{c} 4 \ -3 \end{array}\right)\right}$$
(c)
$$M=\frac{1}{2}\left(\begin{array}{cc} 1 & -1 \ -1 & 3 \end{array}\right), \operatorname{Tr} M=2, \text { Det } M=1, M^{-1}=\frac{1}{2}\left(\begin{array}{cc} 3 & -1 \ -1 & 1 \end{array}\right)$$
(d) $M$ is non-singular provided $\rho^2 \neq 1$.
$$M=\left(\begin{array}{ll} 1 & \rho \ \rho & 1 \end{array}\right), \operatorname{Tr} M=2, \operatorname{Det} M=1-\rho^2, M^{-1}=\frac{1}{1-\rho^2}\left(\begin{array}{cc} 1 & -\rho \ -\rho & 1 \end{array}\right)$$
(e)
$$M=\left(\begin{array}{cc} x & x-x^2 \ 1 & 1-x \end{array}\right), \operatorname{Tr} M=1, \operatorname{Det} M=0, \operatorname{Im} M=\operatorname{span}\left{\left(\begin{array}{l} x \ 1 \end{array}\right)\right}, \operatorname{Ker} M=\operatorname{span}\left{\left(\begin{array}{c} 1-x \ -1 \end{array}\right)\right}$$

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