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--- math121b:ex3 [2020/02/08 14:13]
pzhou created
+++ math121b:ex3 [2020/02/20 15:02] (current)
pzhou
@@ Line 6: / Line 6: @@
 . True or False
-  * All vector spaces come equipped with a preferred inner product.
+  * Each vector space comes equipped with a preferred inner product.
-  * All vector spaces come equipped with a preferred basis.
+  * Each vector spaces comes equipped with a preferred basis.
   * If  $v \in V$  is an element in a vector space  $V$ , then it determines a dual element  $v^* \in V^*$ .
   * Let  $P$  be the vector space of smooth  $\R$ -valued function on  $[0,1]$ . For example,  $f(x) = x^2-2$ , is an element in  $P$ , or  $f(x) = \frac{1}{x+1}$ . Then, the function  $\Phi: P \to \R$ , defined by sending  $f(x) \in P$  to  $\int_0^1 x^2 f(x) dx$  is a linear function on  $P$ .
 . Area
-  * Consider the vector space  $\R^2$ . Let  $\vec v_1 = (1,1), \vec v_2 = (0,2)$ . Let  $P(\vec v_1, \vec v_2)$  denote the area of the parallelogram (skewed rectangle) generated by  $\vec v_1, \vec v_2$ .    P(\vec v_1, \vec v_2) = ?
+  * Consider the vector space  $\R^2$ . Let  $\vec v_1 = (1,1), \vec v_2 = (0,2)$ . Let  $P(\vec v_1, \vec v_2)$  denote the **signed** area ((Signed area means $P(v,w) = - P(w,v)$)) of the parallelogram (skewed rectangle) generated by  $\vec v_1, \vec v_2$ .    P(\vec v_1, \vec v_2) = ?
   * From the above computation, can you deduce  $P(\vec v_1 + 3 \vec v_2, \vec v_2) = ?$  which formula did you use?
   * How about  $P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=?$
@@ Line 24: / Line 24: @@
   *  $\| e_1 - 2 e_2 \| = ?$
   *  $P(e_1, e_2) =$ ?
-  * Can you find two vectors  $v_1, v_2 \in \R^2$ .
+  * Can you find two vectors  $v_1, v_2 \in \R^2$ , such that  $v_1, v_2$  has the same properties as  $e_1, e_2$ ?
 . Let  $V=\R^2$  be the Euclidean vector space of 2-dim, and  $v, w$  be two vectors in it. Suppose we know that
@@ Line 31: / Line 31: @@
    * What if  $V$  is  $n$ -dimensional, does the above conclusion still holds?
+==== Tangent Vectors ====
+. Tangent vector of a subspace in  $\R^2$ . Let  $S^1$  denote the unit circle in  $\R^2$ , i.e
+ $S^1 = \{(x,y) \in \R^2 \mid x^2 + y^2 = 1 \}.$
+Let  $v = (0,1)$ , then for which point  $p \in S^1$ , is the vector  $(p,v)$  a tangent vector of  $S^1$  at  $p$ ?
+. Let  $f(x,y) = x^2 - y^2$ , and let  $\Gamma_f = \{(x,y, z) \mid z = f(x,y) \}$  the graph of  $f$  in  $\R^3$ .
+Then for the point  $p=(2, 3, -5)$  on  $\Gamma_f$ , find two linearly independent tangent vectors in  $T_p \Gamma_f$ .
+===== Solution =====
+==== Metric Tensor. Length, Area and Volume element ====
+. True of False
+  * False. Although, you can equip any finite dimensional vector space with an inner product, there is no one holier than the other.
+  * False. The whole point of introducing abstract linear vector space is to do away with basis (at least on the level of definition, even though in practice, it is useful to pick a basis).
+  * False.
+  * True.
+.
+  *  $P(\vec v_1, \vec v_2) = \det \begin{pmatrix} 1 & 0 \cr 1 & 2 \end{pmatrix} = 2$ .
+  *  $P(\vec v_1 + 3 \vec v_2, \vec v_2) = P(\vec v_1, \vec v_2)  = 2$ . You can use either the property of determinant modify a column by a multiple of another column does not change the value. Or, you can use the fact that  $\vec v_1 \wedge \vec v_2) = (\vec v_1 + 3 \vec v_2) \wedge \vec v_2$ . Same thing.
+  *
+ $P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2) = \det \begin{pmatrix} a & b \cr c & d \end{pmatrix} P(\vec v_1, \vec v_2) =  2 (ad-bc)$
+.
+  *  $\| e_1 \| = \sqrt{3}$ ,  $\| e_2 \| = \sqrt{2}$
+  *  $\| e_1 + e_2 \|^2 = g(e_1, e_1) + g(e_2, e_2) + 2 g(e_1, e_2) = 3 + 2 + 2 = 7$ , hence   $\| e_1 + e_2 \| = \sqrt{7}$ . Another way to get  $\| e_1 + e_2 \|^2$  is by
+ $\begin{pmatrix} 1 & 1  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr 1 \end{pmatrix} = 7$
+  *  $\| e_1 - 2 e_2 \|^2 =  \begin{pmatrix} 1 & -2  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr -2 \end{pmatrix} = 7$
+  *  $P(e_1, e_2) = \sqrt{\det(g)} = \sqrt{5}$
+  * Can you find two vectors  $v_1, v_2 \in \R^2$ , such that  $v_1, v_2$  has the same properties as  $e_1, e_2$ ?  $v_1 = (\sqrt{3}, 0), v_2 = (\sqrt{2} \cos\theta, \sqrt{2} \sin \theta)$  where  $\cos\theta = \frac{1}{\sqrt{2} \sqrt{3}}$ . We are using the formula
+ $v \cdot w = \| v \| \|w \| \cos \theta$
+there  $\theta$  is the angle between  $v$  and  $w$ .
+. Try using the angle formula above to show that the two vector are parallel. It does not depends on whether you are in 2-dim or higher dim.
+==== Tangent Vectors ====
+. For point  $p=(1,0)$  and  $p=(-1,0)$ , the vector  $v = (0, 1)$  is a tangent vector in  $T_p S^1$ . Just draw the picture to see.
+. The defining equation for  $\Gamma_f$  is  $z - f(x,y) =0$ , then tangent vectors  $\vec v_p$  to  $\Gamma_f$  at point  $p$  satisfies
+ $\langle dz -d f|_p , v_p \rangle = 0$
+In concrete form, we have at $p=(2,3,-5)$,
+ $dz -d f|_p = dz - 2x dx + 2y dy|_p = d z - 4 dx + 6 dy$
+thus, if  $v_p = v_x \d_x + v_y \d_y + v_z \d_z$ , then the only requirement we have is
+ $-4 v_x + 6 v_y + v_z = 0$
+The solution space is spanned by  $v_p = (1, 0, 4)$  and  $(0, 1, -6)$ . More correctly, we have
+ $\d_x + 4 \d_z, \quad \d_y - 6 \d_z$
+are two linearly independent vectors on  $T_p \Gamma_f$ .

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