Tag Info

Hot answers tagged

2

One of the ingredients of a vector space is a definition of scalar multiplication; but you need to know what field the scalars are in! A vector space $V$ over $\mathbb{F}$ has as part of its data a map (scalar multiplication) $\mathbb{F}\times V\to V$. The same set $V$ could be given a different vector space structure with a multiplication map ...


2

$$\langle v| A^{\dagger} = \sum _n e^{-i\alpha_n} \langle n |$$ I don't think that your version is true. Consider $$A= \left[ \matrix { 0&i\\ 1&0 } \right]$$ $$A^{\dagger}= \left[ \matrix { 0&1\\ -i&0 } \right]$$ and apply them to the vector $\left[\matrix{1\\0}\right]$ then express in the standard basis. $$A ...


2

Simply write $$\left\{ \begin{array}{rcl} 4+t&=&6+2s\\ 5+t&=&11+4s\\ -1+2t&=&-3+s \end{array} \right.$$ And then solve the system formed by the two fisrt equations. If the solution satisfies the third one, then the lines intersect and its intersection can be found substituting the found value for $t$ in the first line, or the value ...


2

For (b) first note that $f_{V^0}+f_{W^0}\in V^0+W^0$ implies $$ (f_{V^0}+f_{W^0})(s)=f_{V^0}(s)+f_{W^0}(s)=0+0=0 $$ whenever $s\in W\cap V$ since $W\cap V$ is a subspace of both $V$ and $W$. This proves that $V^0+W^0\subset(V\cap W)^0$ Now, note that \begin{align*} \dim\bigl((V\cap W)^0\bigr) &= \dim(U)-\dim(V\cap W) \\ &= ...


2

Let $\vec{x}, \vec{y}, \vec{z} \in V$, where $V$ is a vector space. We are given that \begin{align}\vec{x}+\vec{y}=\vec{x}+\vec{z}\end{align} Now since, $\vec{x}, \vec{y}, \vec{z} \in V$, we know they satifsy the vector space axioms. From these axioms, consider the vector $-\vec{x}.$ Adding this vector to both sides of the above equation and simplifying ...


2

Hint: The zero vector must be in every subspace.


1

You can also take one of the elements and place it on the right side. In this equation, let's place $-\frac{1}{2x-1}$ to the right side. You will get the following equation: $\Large \frac{5}{x+1} = \frac{1}{2x-1}$. We can now apply the following rule: $\frac{a}{b}=\frac{c}{d} \rightarrow ad=bc$, so we get this the following equation: $5(2x-1) = 1(x+1) \iff ...


1

Yes: Your minimal polynomial tells you that since A has two different eigenvalues (2 and -1), A is diagonalizable. Find P such as : $P^{-1}*AP = \begin{pmatrix}2&0\\0&-1\end{pmatrix}$ That can be done by calculating A-2*I, and find a combination of columns that gives 0. Such a combination will give an eigenvector of the eigenvalue 2. A-2*I = ...


1

Yes; it's true since $$\left\langle vv^Tx,x\right\rangle=\left\langle v^Tx,v^Tx\right\rangle=\left|\left|v^T x\right|\right|^2\ge0 .$$


1

Let $0\ne v\in V$. Then $kv$ is $T$-invariant, hence $Tv=a_v v$ for some $a_v\in k$. Then for any multiple $w\in kv$, clearly $Tw=a_vv$ as well. Likewise, if $w\in V\setminus kv$ we obtain $Tw = c_ww$ and $T(v+w)=c_{v+w}(v+w)$ with $c_w,c_{v+w}\in k$. From $T(v+w)=Tv+Tw$ we infer $$c_{v+w}(v+w)=c_vv+c_ww $$ i.e. $$(c_{v+w}-c_v)v + (c_{v+w}-c_w)w=0.$$ Now ...


1

Note that $\{v_1,v_2,v_3\}$ is a linearly dependent set if and only if the matrix $$ A= \begin{bmatrix} h & -1/2 & -1/2 \\ -1/2 & h & -1/2 \\ -1/2 & -1/2 & h \end{bmatrix} $$ satisfies $\det A=0$. Since $$ \det A=\frac{1}{4}(h-1)(2h+1)^2 $$ it follows that $\{v_1,v_2,v_3\}$ is a linearly dependent set if and only if $h=1$ or ...


1

As Emanuele suggests, you will have a system of three linear equations for two unknowns. You can create an augmented matrix and then row reduce. We have the equations $t-2s=2,t-4s=6,2t-s=-2$ which implies the following augmented matrix: $$\begin{pmatrix} 1 & -2 & 2 \\ 1 & -4 & 6 \\ 2 & -1 & -2 \end{pmatrix}\Rightarrow ...


1

The answer for your first question is yes. The answer for your second is no. The next proposition is well known. It was proved here. Proposition: Let $V,W$ be vector spaces over $k$. Let $w=\sum_{i=1}^ra_i\otimes b_i=\sum_{i=1}^sv_i\otimes w_i\in V\otimes W$. If $\{a_1,\ldots, a_r\}$ is linear independent then $\text{span ...


1

Suppose $a_{k+1}f(v_{k+1}) \ +\ a_{k+2}f(v_{k+2})\ +...+\ a_nf(v_n)$ = $0$ $\Rightarrow$ $f(a_{k+1}v_{k+1}+...+a_nv_v)=0$ $\Rightarrow$ $a_{k+1}v_{k+1}+...+a_nv_n\in Ker\ (f)$ $\Rightarrow$ $a_{k+1}v_{k+1}+...a_nv_n=b_1v_1+...b_kv_k$ (Why ?) $\Rightarrow$ $(-b_1v_1)+(-b_2v_2)+...+(-b_kv_k)+a_{k+1}v_{k+1}+...+a_nv_n=0$. Now $\{v_1,..,v_n\}$ is a basis for ...


1

Since $H\simeq M\oplus M^{\perp}$, any vector $v\in H$ can be uniquely written as a sum $v=m+m'$ with $m\in M,m'\in M^{\perp}$. Then apply $p:=p_M+p_{M^{\perp}}$. By definition of projection, $p_M(m)=m$ and $p_M(m')=0$, and similarly $p_{M^{\perp}}(m)=0$ and $p_{M^{\perp}}(m')=m'$. Taking the sum gives you $p(v)=p_M(m+m')+p_{M^{\perp}}(m+m')=m+m'=v$ for any ...


1

Let $w_1,\ldots,w_k$ be an orthonormal basis of $M$. Then, $p_Mv=\sum_i\left<v,w_i\right>w_i$. Then for any $\forall j\in\{1,\ldots, k\}$, we have $$ \left<v-p_Mv,w_j\right>=\left<v,w_j\right>+\sum_i\left<v,w_i\right>\left<w_i,w_j\right>\\ ...


1

What you want to do is write an arbitrary $(x,y)$ as a linear combination of $(1,0)$ and $(2,1)$ and then discard the $(1,0)$ component. $$ (x,y) = y(2,1)+(x-2y)(1,0). $$ So the projection you want is $p(x,y)=y(2,1)$, which can be represented by a matrix acting on column vectors instead: $$ p\left[\begin{array}{c}x \\ ...


1

Well, what is the definition of span of a list? Let $V$ be a vector space. Let $v_1,...,v_p \in V$, I want to show that $$ Span( v_1,...,v_p) = \{ \lambda_1 v_1 + ... + \lambda_p v_p : \lambda_k \in \mathbb{K} \} $$ is a subspace of $V$. Notice if you add two vectors in the span, you obtain again a linear combination of these vectors and therefore lying ...


1

A linear map $T$ is an isomorphism if and only if it both injective (one-to-one) and surjective (onto). A linear map $T$ is injective if and only if $\dim(\ker(T)) = 0$. Try to show that this is the case.


1

The underlying theorem is: http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem In your special case, if $\dim(\mathrm{Ker} T)>0$ you know that there is $v\neq 0$ such that $Tv=0$. So $Tv=T0$ and the map is not injective.


1

It respects operation (obvious). It is surjective (easy to show). Assume $w = (v_1, \dots, v_n)$ that $f(w) = 0$. Okay, let's write $f(w) = v_1 + \dots v_n = 0$. Then $v_i = 0, \forall i \in \{1, \dots, n\}$ because $V_i \cap (V_1 + \dots + V_{i-1} + V_{i+1} + \dots + V_n) = \{0\}$. From this we got that $w = 0$ and $f$ injective.


1

You should redefine $f$ as $$f(A)=\frac{1}{2}(A-A^t).$$ This is the map that is usually called anti-symmetrization, and solving the exercise should tell you why!


1

Notice that $$f(A)=A-A^t\implies f^2(A)=f(A)-f(A)^t=A-A^t-A^t+A=2A+2(f(A)-A)$$ hence we get $$f(A)^2=2f(A)$$ hence the polynomial $x(x-2)$ with simple roots annihilates $f$ and then $f$ is diagonalizable. Moreover, $A$ is an eigenvector associated to $0$ iff $A=A^t$ iff $A\in S_2(\Bbb R)$ which has the dimension $\frac{2\times 3}{2}=3$ and $A$ is an ...


1

Let $A$ have entries $A[i,j] = a_{ij}$. Since $A$ is upper triangular, $a_{ij} = 0$ when $i>j$. We note that $$ (A^TA)[i,j] = \sum_{k=1}^n a_{ki} a_{kj}\\ (AA^T)[i,j] = \sum_{k=1}^n a_{ik}a_{jk} $$ So that the diagonal entries are given by $$ (A^TA)[i,i] = \sum_{k=1}^n (a_{ki})^2\\ (AA^T)[i,i] = \sum_{k=1}^n (a_{ik})^2 $$ Setting the first diagonal ...


1

Suppose $A$ is $(n+1)\times(n+1)$ upper triangular and write it in block form $$ A=\begin{bmatrix} a_{11} & x^T \\ 0 & Y \end{bmatrix} $$ where $x$ is a column vector with $n$ rows and $Y$ is $n\times n$ and upper triangular. Then $$ AA^T= \begin{bmatrix} a_{11} & x^T \\ 0 & Y \end{bmatrix} \begin{bmatrix} a_{11} & 0^T \\ x & Y^T ...



Only top voted, non community-wiki answers of a minimum length are eligible