Let $\{u,v\}$ be a linearly independent set in $\Bbb R^3$.  Show that the plane $\{su+tv\mid s,t \in\Bbb R\}$ through the origin in $\Bbb R^3$ is equal to the null space of some element of $(\Bbb R^3)^{*}$. (Where $(R^3)^*$ is the dual space of $R^3$).   Can we define a linear functional, say $L:R^3 \rightarrow R^2$?
122785
Let $\{u,v\}$ be a linearly independent set in $\Bbb R^3$.  Show that the plane $\{su+tv|s,t \in\Bbb R\}$ through the origin in $\Bbb R^3$ is equal to the null space of some element of $(\Bbb R^3)^{*}$.   Attempt: We are supposed to prove that $su+tv=ker(g)=0$ for some $g \in (R^3)^{*}$. And I am stuck now.   I want to argue algebraically. I know that $g= g(s) g_1+ g(t) g_2 $for $g_1, g_2$ being basis for g.  Then from definition of kernel we have   $g(s)=g(s) g_1 (s) +g(s) g_2 (s)=0 \rightarrow g(s)=0$  and $g(t)=g(s) g_1 (t) +g(s) g_2 (t)=0 \rightarrow g(t)=0$  So we have $su+tv=0=g(s)=g(t)$  I am not sure if that makes sense.
102650
Let $T$ be a linear operator on a finite dimensional space $V$, and let $p=p_{1}^{r_{1}} \cdots p_{k}^{r_{k}} $ be the minimal polynomial for $T$, and let $V= W_{1} \oplus\cdots\oplus W_{k}$ be the primary decomposition for $T$, i.e., $W_{j}$ is the null space of $p_{j}(T)^{r_{j}}$. Let $W$ be any subspace of $V$ which is invariant under $T$. Prove that  $W= (W \cap W_{1})\oplus (W \cap W_{2})\oplus \cdots \oplus (W \cap W_{k})$.
88237
I wrote a proof for the above question, but I am not sure whether it is right or not since I assumed linear independence.   Here's the proof:   Let $u$,$v$ be linearly independent vectors in $V$. $span(u)$, $span(v)$, $span(u+v)$ are all $T-invariant$.   $T(v)$ is an element in $span(v) \Longrightarrow T(v) = av$  $T(u)$ is an element in $span(u) \Longrightarrow T(u) = bu$  $T(u+v)$ is an element in $span(u+v) \Longrightarrow T(u+v) = T(u)+T(v)= bu+av = c(u+v)$  Hence, $bu+av-c(u+v) = 0 \Longrightarrow (b-c)u + (a-c)v = 0$ Since u and v are linearly independent, $(b-c)=(a-c)=0 \Longrightarrow b=a$   Hence, $T(w) = kw$, for all $w$ in $V$, and $k$ any scalar.   Is it right to assume linear independence? And is there any problem with my proof?
98076
Let $T: V \to V$ with the minimal polynomial $z^n$. Prove that there's a vector $v$ such that $v, Tv, T^2v, ..., T^{n-1}v$ are linearly independent.   The way I did it orginally was not allowed. No Jordan Forms are allowed because we did not cover them this semester. I asked my professor for a hint and all he said was we need to use induction. I wish I had something to work with, but I'm so  lost. It's my last problem. Thanks!
45876
Determine which of the following sets is a vector space.  $V$ is the line $y=x$, in the $xy$-plane: $V =\{[x,y]: \; y=x\}$  $W$ is the union of the first and second quadrants in the $xy$-plane: $W=\{[x,y]: \; y\geq0\}$  $U$ is the line $y = x+1$ in the $xy$-plane: $U=\{[x,y]: \; y=x+1\}$    I know that the answer is that "$V$" is the only vector space. However how do I come up with a quick proof to show that $V $is a vector space and that the other $2$ aren't?
79657
Let $V$ be the linear space of all real functions continuous on $[a, b]$. If $f\in V, g=T(f)$ means that $$g(x)=\int_a^b f(t)\sin(x-t)\,dt\hspace{1 cm} for\ a\le x\le b$$  We can see this as a linear transformation $T: V\to V$ the linearity of which is easy to show.$\\$ Now how do you compute the nullity and rank of this transformation?  I'm rather new to linear algebra so I'd appreciate it if you include lots of detail and explanation in your answer. Thank you in advance.  P.S: the final answer is infinite nullity and 2nd rank (according to the back of the book).  From calculus II by Tom M. Apostol - page 36
121665
Let $V$ and $W$ be vector spaces (say over the reals). There is a linear injection $V^* \otimes W^* \to (V \otimes W)^*$ which sends $\sum_i f_i \otimes g_i \in V^* \otimes W^*$ to the unique functional in $(V \otimes W)^*$ sending $v \otimes w \mapsto \sum_i f_i(v) \cdot g_i(w)$ for all $(v,w) \in V \times W$. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I'm aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the  on tensor products) but I must admit that I'm not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in $(V \otimes W)^*$ outside the range of this map. I'm having trouble cooking one up myself.  Added: Let me elaborate on my comment about cardinal arithmetic. Suppose that $X$ and $Y$ are infinite dimensional vector spaces over a field $k$. I'm confident that $\dim (X \otimes Y) = \dim X \cdot \dim Y$ holds. It is clear that $|X^*| = |k|^{\dim X}$ since we may identify a functional on $X$ with a function from a basis for $X$ to $k$. Also I think I've convinced myself that if $\dim X \geq |k|$ then in fact $|X| = \dim X$ ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that $|k| \leq \dim X \leq \dim Y$ it seems we have $$ \dim(X \otimes Y)^* = |k|^{ \dim X \cdot \dim Y} = |k|^{ \dim Y} = \dim Y^* = \dim X^* \cdot \dim Y^* = \dim (X^* \otimes Y^*)$$  which implies that in fact $(X \otimes Y)^*$ and $X^* \otimes Y^*$ are isomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about $|k|$) that there exists some isomorphism $(X \otimes Y)^* \to X^* \otimes Y^*$?
76871
I am trying to prove that $a_1$, $a_2$, $a_3$ are linearly independent.  I am asked to use vector product and prove that if $c_{1}a_{1} + c_{2}a_{2} + c_{3}a_{3} = 0$ then $c_1 = c_2 = c_3 = 0$  I am completely stuck on where to go with this problem. I would think that linearly independent then the null space of the space can only be equal to $0$. Would this prove it?
118588
Let $V$ and $W$ be two finite dimensional vector spaces over $\Bbb R$ and let $T_{1}:V\to V$ and $T_{2}:W\to W$ be two linear transformations whose minimal polynomials are given by $f_{1}(x)=x^{3}+x^{2}+x+1$ and $f_{2}(x)=x^{4}-x^{2}-2$.  Let, $T:V\oplus W\to V\oplus W$ be the linear transformation defined by   $T(v,w)=(T_{1}(v),T_{2}(w))$ for $(v,w)\in V\oplus W$ and let $f(x)$ be the minimal polynomial of $T$. Then which of the followings are correct?  (1) $\deg(f(x))=7$  (2) $\deg(f(x))=5$   (3) $\mathrm{nullity}(T)=1$    (4) $\mathrm{nullity}(T)=0$ .
115755
Let $E$ be $\mathbb C$-vector space of dimension $3$. Let $f$ be non zero endomorphism of $E$ such that $f^2=0$. show that there exists a basis $B=\{b_1,b_2,b_3\} $ of $E$ where the matrix of $f$ is given by :  $$\begin{array}{ccc} 0 &amp; 1 &amp; 0 \\  0 &amp; 0 &amp; 0 \\  0 &amp; 0 &amp; 0 \end{array}$$  My try : since $f$ is non zero, there exists $x\in E$ such that $f(x)\not =0$. Take $b_1=f(x)$ and $b_2=x$. Now $b_1,b_2$ can easily shown to be linearly independent. By the exchange lemma we can find $b_3$ in another generating system of $E$ to make $b_1,b_2,b_3$ into a basis of $E$. Now clearly $f(b_1)=f^2(x)=0$ and $f(b_2)=f(x)=b_1$. Now how to find  a $b_3$ such that $f(b_3)=0$ and $b_1,b_2,b_3$ is a basis of $E$? Thank you for your help.   Edit: For the linear independence of $b_1,b_2$, if $\alpha x+\beta f(x)=0$ then we apply $f$ to get $\alpha f(x)=0$ and since $f(x)\not =0$ then $\alpha=0$ it remains that $\beta f(x)=0$ and again this gives that $\beta=0$.
50117
This is my question:  Let P be a plane considered as a surface in 3-space. Show that its second fundamental form is zero. Conversely, show that if the second fundamental form of a surface is identically zero then the surface is a plane.  So far I have that the general equation of the plane is $$ax+by+cz+d=0.$$ Then $$M(u,v)=(u,v,\alpha u + \beta v + \gamma),$$ where $a,b,c,d,\alpha,\beta, \gamma$ are all constants and $c\not=0$. Then $M_u=(1,0,\alpha)$ and $M_v=(0,1,\beta)$. I know that $\vec{n} =\frac{M_u \times M_v}{|M_u \times M_v|}$ which gives $$ \vec{n}=\frac{(-\alpha,-\beta, 1)}{\sqrt{\alpha^2+\beta^2+1}}.$$ Hence, the unit norm is a constant. $$ M_{uu}=M_{uv}=M_{vv}=0$$ Hence $l=m=n=0$, so the second fundamental form is $$ ldu^2+mdudv+ndv^2=0.$$  Now I need to show the converse, but I don't know how. Please help :)
116303
Let $V$ be a subspace of $\mathbb R^3$ generated by the vectors $\{(1,1,1),(0,1,1)\}.$ There are $3$ vectors $u_1=(0,0,1),u_2=(1,1,0),u_3=(1,0,1)$.Then which of the following sets are connected?  $1.\left(\mathbb R^3\setminus V\right)\cup\{(0,0,0)\}$  $2.\left(\mathbb R^3\setminus V\right)\cup \{tu_1+(1-t)u_3:t\in [0,1]\}$  $3.\left(\mathbb R^3\setminus V\right)\cup\{tu_1+(1-t)u_2:t\in [0,1]\}.$  $4.\left(\mathbb R^3\setminus V\right)\cup \{(t,2t,2t):t\in \mathbb R\}$  Now $V$ is generated by two linearly independent vectors so is like $\mathbb R^2.$ So $\left(\mathbb R^3\setminus V\right)$ will be disconnected but giving the $(0,0,0)$ with them will make it connected.  No. $4$, I think will be connected also because $\{(t,2t,2t),t\in \mathbb R\}$ is in $V$ and contains the point $(0,0,0)$ Now, any neighbourhood of this point will have points from both $V$ and $\left(\mathbb R^3\setminus V\right)$ and so the union is connected.I'm not sure about no. $2$ and $3.$
89899
Hints only     Let $P = \text{Span } \{v_1, v_2\}$ be a plane in $\mathbb{R}^3$ with normal vector $n$. Show that $\{ v_1, v_2, n\}$ is a basis for $\mathbb{R}^3$. Hints only   Equation for $P$: $P = c_1v_1 + c_2v_2$. For real $c_1, c_2$. We have by definition, $n = v_1 \times v_2$.  To make sure $\{ v_1, v_2, n\}$ is a basis for $\mathbb{R}^3$. we must have   Span $\{ v_1, v_2, n\}$  =  $\mathbb{R}^3$ $\{ v_1, v_2, n\}$ Linearly independent.   I am having major trouble showing (1).  Can I get a SMALL hint?  I am not allowed to use dimension.  Showing set is LI  By definition of a plane, $\{v_1, v_2 \}$ is linearly independent thus $\overrightarrow{0} \not \in \{v_1, v_2 \}$  So $c_1v_1 + c_2v_2 + 0 \overrightarrow{n} = 0$ Thus $\{v_1, v_2, n \}$ is LI.   The span part is confusing
22458
[Theorem] Let $V$ be an inner product space, and let $S$ be an orthogonal subset of $V$ consisting of nonzero vectors. Then $S$ is linearly independent.     Also, orthogonal set and linearly independent set both generate the same subspace.  (Is that right?)    Then orthogonal $\rightarrow$ linearly independent but orthogonal $\nleftarrow$ linearly independent is that right?    One more question. For T/F, Every orthogonal set is linearly independent (F) Every orthonormal set is linearly independent (T) Why?
27467
I am trying to prove the following. I have seen it alluded to in other places of the internet (this site included) but without proof.  Let $L,L_1\ldots L_n$ be linear functionals on a vector space $X$. If $\bigcap_{i=1}^n ker(L_i) \subset ker(L)$ then there exists $t_i$ for  $i=1\ldots n \in \mathbb{R}$ such that $L = \sum_{i=1}^n t_i L_i$.   In other words, if the intersection of kernels of linear functionals is contained by the kernel of another linear functional then they are linearly dependent.  Related:
63325
May I please ask how to show the dual space of the vector space of all polynomials is isomorphic to the infinite-dimensional Euclidean vector space over the reals?  (i.e. Show that $(\Bbb{R}[X])^*$ and $\Bbb{R}^{∞}$ are isomorphic where $\Bbb{R}$ means the set of all reals)
81336
Let $V$ be a real inner product space and $S=\{v_1,v_2, \dots, v_n\}\subset V$. How am I to prove that $S$ is linearly independent if and only if the determinant of the matrix $$ a_{ij}=\pmatrix{\langle v_i , v_j \rangle}$$ is nonzero?    Just to be clear, the matrix we're talking about is this one:  $$\pmatrix{\langle v_1,v_1\rangle &amp; \langle v_1,v_2\rangle &amp;\langle v_1,v_3\rangle &amp; \cdots &amp; \langle v_1,v_{n-1}\rangle &amp; \langle v_1, v_n\rangle \\\langle v_2,v_1\rangle &amp; \langle v_2,v_2\rangle &amp;\langle v_2,v_3\rangle &amp; \cdots &amp; \langle v_2,v_{n-1} \rangle &amp; \langle v_2,v_n\rangle \\\langle v_3,v_1\rangle &amp; \langle v_3,v_2\rangle &amp;\langle v_3,v_3\rangle &amp; \cdots &amp; \langle v_3,v_{n-1}\rangle &amp; \langle v_3,v_n \rangle \\ \cdots&amp;\cdots&amp;\cdots&amp;\cdots&amp;\cdots&amp;\cdots\\\langle v_{n-1}, v_1\rangle &amp; \langle v_{n-1},v_2\rangle &amp;\langle v_{n-1},v_3\rangle &amp; \cdots &amp; \langle v_{n-1},v_{n-1}\rangle &amp; \langle v_{n-1},v_n\rangle \\\langle v_n,v_1\rangle &amp; \langle v_n,v_2\rangle &amp;\langle v_n,v_3\rangle &amp; \cdots &amp; \langle v_n,v_{n-1}\rangle &amp; \langle v_n,v_n\rangle \\ }$$    I highly doubt that anybody here has Roman's Advanced Linear Algebra, or maybe you do, but I think on page $261$ there is a small note on something which looks similar.    Should anybody need the code (C++) in their research, here is a gadget which streams LaTeX code to a file named "matrix.txt" for an $n \times n$ matrix such as this with some value for $n$:  ofstream fout; fout.open("matrix.txt");  int n; cout &lt;&lt; "Enter your desired n: "; cin &gt;&gt; n;  fout &lt;&lt; endl;  fout &lt;&lt; "$\\begin{pmatrix}" &lt;&lt; endl;  for( int j = 1 ; j &lt;= n ; j++ ) {     for( int i = 1 ; i&lt;= n ; i++ )     {         if( j == i &amp;&amp; i == n )         {             fout &lt;&lt; "\\langle v_" &lt;&lt; j &lt;&lt; "," &lt;&lt; "v_" &lt;&lt; i &lt;&lt; " \\rangle" &lt;&lt; endl;         }         else         {             if( i == n )             {                 fout &lt;&lt; "\\langle v_" &lt;&lt; j &lt;&lt; "," &lt;&lt; "v_" &lt;&lt; i &lt;&lt; " \\rangle\\\\" &lt;&lt; endl;             }             else             {                 fout &lt;&lt; "\\langle v_" &lt;&lt; j &lt;&lt; "," &lt;&lt; "v_" &lt;&lt; i &lt;&lt; " \\rangle&amp;" &lt;&lt; endl;             }         }     }     //fout &lt;&lt; endl; }  fout &lt;&lt; "\\end{pmatrix}$" &lt;&lt; endl &lt;&lt; endl;   For n equal to 5 you get this result:  $\begin{pmatrix} \langle v_1,v_1 \rangle&amp; \langle v_1,v_2 \rangle&amp; \langle v_1,v_3 \rangle&amp; \langle v_1,v_4 \rangle&amp; \langle v_1,v_5 \rangle\\ \langle v_2,v_1 \rangle&amp; \langle v_2,v_2 \rangle&amp; \langle v_2,v_3 \rangle&amp; \langle v_2,v_4 \rangle&amp; \langle v_2,v_5 \rangle\\ \langle v_3,v_1 \rangle&amp; \langle v_3,v_2 \rangle&amp; \langle v_3,v_3 \rangle&amp; \langle v_3,v_4 \rangle&amp; \langle v_3,v_5 \rangle\\ \langle v_4,v_1 \rangle&amp; \langle v_4,v_2 \rangle&amp; \langle v_4,v_3 \rangle&amp; \langle v_4,v_4 \rangle&amp; \langle v_4,v_5 \rangle\\ \langle v_5,v_1 \rangle&amp; \langle v_5,v_2 \rangle&amp; \langle v_5,v_3 \rangle&amp; \langle v_5,v_4 \rangle&amp; \langle v_5,v_5 \rangle \end{pmatrix}$  As is lengthily explained .