# Tag Info

1

Rewriting the equations in the linearized form, you can use a block decomposition $$\left[ \begin{array}{ccc} A & B\\ C& D \end{array} \right] \left[ \begin{array}{ccc} x_0\\ x_1\end{array} \right]=\left[ \begin{array}{ccc} y_0\\ y_1\end{array} \right]$$ and solve for the vector sections that you want ...

0

If you want only vector definition for skew symmetric matrix assume $r$ is a vector $= [r_x r_y r_z]^T$ in the frame described by versors $i, j, k$ Then skew-symetric matrix $S(r)$ assigned to the vector $r$ is constructed this way $$S(r) = \begin{pmatrix} r \times i \ \ r \times j \ \ r \times k \end{pmatrix}$$ As you see in the form of matrix you ...

0

This is only a partial answer. (I assume you want $N_1$ and $N_2$ invertible, otherwise $N_1=0$ solves) Given $M$ its matrix of cofactors $Cof(M)$ (se for instance here) is such that $Cof(M)=M^{-1}\det M$. As $Cof(M)$ is made by minors of $M$ if $M$ is integer then $Cof(M)$ is. If $M_2N_2=M_1N_1$ we have $$N_2=\frac{Cof(M_2)M_1}{\det M_2}N_1$$ This is ...

0

The formula $\dim\operatorname{Ker} T+\dim\operatorname{Im} T=\dim V=n$ should be enough for this. If $T$ is injective, then $\dim( \operatorname{Ker} T) =0$ so $\dim(\operatorname{Im} T) =n$, therefore since $\operatorname{Im} T$ is a subspace of $V$ you get $\operatorname{Im} T=V$. The inverse is easier: a bijection is always, by definition, injective. ...

0

I think it is essentially correct, but I would write things a bit differently. You are right on using rank-nullity. But then I would say that the only vector $z$ with $T(z)=0$ is the zero vector. Then the kernel is 0-dimensional, i.e., nullity is $0$, so rank is $n$, and then $T(V)$ is an n-dimensional subspace of $V$, and so it must be the whole of $V$ ...

1

The first part is okay, but the second one (finding $T(1)$) isn't. Note that linear transformations give you $$T(1) = -T(-2x^2 + (2x^2 - 1)) = -T(-2x^2) - T(2x^2 - 1) \\ = -(3x^2 - 3x) - (-3x +2) = -3x^2 - 2$$ With your approach, you'd have to find that $$1 = -1\cdot (-2x^2) + 0\cdot (\frac12 x + 4) - 1\cdot(2x^2 - 1) = \pmatrix{-1\\0\\-1}_{\mathcal B}$$ So ...

0

Hint: Every hyperplane can be associated with a couple of antipodal points on $S^{n-1}$ and every point with a Voronoi cell.

2

Yes; a matrix is uniquely determined by the linear transformation it defines, and vice versa. One explicit proof is: $$A=AI = A[e_1 \dots e_n] = [Ae_1 \dots Ae_n] \\ B=BI=B[e_1 \dots e_n] = [Be_1 \dots Be_n]$$ so because $Ae_i=Be_i$ we get $A=B$. Here $e_i$ is the $i$th unit coordinate vector.

1

Some comments: That $V$ has a $4$-element basis doesn't mean that $V=\mathbb{R}^4$, just that $V\cong \mathbb{R}^4$. I suggest that you right in the beginning something like "Consider the isomorphism $V\to \mathbb{R}^4, v_i\mapsto e_i$." Then you can transfer every question to $\mathbb{R}^4$ and solve it there instead. Depending on how strict your teacher ...

2

You seem a little bit confused about your definitions, so I'll try and make some things a bit clearer. Definition 1 If $K$ is a field, an absolute value on $K$ is a map $|.|\colon K\to \mathbb R_{\ge 0}$ such that the following axioms apply: $|x|=0$ if and only if $x=0$. For all $x,y\in K$, $|xy|=|x||y|$ For all $x,y\in K$, $|x+y|\le |x|+|y|$. We ...

1

Since $\Bbb{C}$ is algebraically closed, you can always find eigenvalues of every endomorphism. You have to use induction on $n$. For $n=1$, the thesis is obvious. Now, suppose $n \ge 2$ and that you proved the thesis for $0, \dots, n-1$. Let $W$ be an invariant subspace of dimension $1$ (this is spanned by any eigenvector $v_1$). Then $T$ induces an ...

0

The characteristic polynomial of this matrix is $$\det(\lambda I - A) = \lambda^2 + \theta_2 \lambda + \theta_1$$ Since similarity transforms leave the characteristic polynomial alone, the matrix is unique. It remains to be shown that a $2\times 2$ matrix with this characteristic polynomial is indeed similar to $A$. For this we use the JNF. A $2\times 2$ ...

1

As OP was able to work out from my comment, the matrix, $M$, of $T$ with respect to the basis $B$ is diagonal, its diagonal entries being the eigenvalues associated with the eigenvectors $v_1,\dots,v_n$. The matrix for $S$ is $aI$, no matter what basis is used, so the matrix for $T-S$ is $M-aI$, which is also diagonal.

1

A positive definite matrix can be diagonalized as $A= U D U^H$. Then, $A^H A = A A^H = U D^2 U^H$, so the eigenvalues of $A^H A$ are the squared eigenvalues of $A$. But the eigenvalues of $A^H A$ are the squared singular values of $A$. So we have eigenvalues squared = singular values squared. Since $A$ is positive definite, eigenvalues are non-negative and ...

0

$T(1,1)$ and $T(1,2)$ should be expressed in basis $\mathcal B'$, not $\mathcal B$. Let $\{e_1, e_2\}$ be the canonical basis. Set $u_1=\begin{bmatrix}1\\1 \end{bmatrix}=e_1+e_2$, $\,u_2=\begin{bmatrix} 1\\2\end{bmatrix}=e_1+2e_2$. From these, you deduce: $$e_1=2u_1-u_2,\quad e_2=u_2-u_1.$$ You've proved $\,T(u_1)= 3e_1-2e_2$, $\,T(u_2)=4e_1-5e_2$, ...

0

If a finite number (the first few) of terms of a sequence are given and if that sequence is:- (1) in Arithmetic Progression; or (2) in Geometric Progression; or (3) in some nicely behaved pattern (like a Harmonic Progression or a Fibonacci Sequence ), then answer is YES – a formula for the general term can be found. Otherwise is NO, because the values ...

0

Characteristic zero is really necessary for Weyl's theorem, because there are counterexamples over fields of characteristic $p>0$. Algebraically closed is not really necessary, even if we first need to assume it for, say, applying Schur's lemma. Afterwards one can apply an argument, that the statement is already true for $K$ if it was true for the ...

3

Note that ${\rm dim}\ im(T) + {\rm dim}\ ker(T)=n$. So we have a claim : $$im(T)\cap ker(T)=\{ 0\}$$ Assume that $Tx=v\neq 0\in im(T)\cap ker(T)$. Then $$0=( x,Tv)=(Tx,v)=(v,v)$$ Contradiction.

0

The diagonal of a matrix product (AB) is the dot product of row's of the first matrix (A) with the corresponding column of the second matrix (B).

2

If $rank(A) < n$, the strictness of $S$ may be lost, since it is now possible that $x \ne y$ but $S(x) \cap S(y) \ne \emptyset$. That is because there is $w \ne 0$ such that $Aw = 0$. You should be able to construct a counterexample to strict monotonicity using that information.

2

You have found a basis. Let these vectors be the columns of a matrix $A$, and solve $Ax=(1,1,1)^T$. Let $A = \begin{pmatrix} 2 & 1 & -2 \\ 5 & 2 & 0 \\ 4 & -3 & 1 \end{pmatrix}$. Compute $$Ax = \begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix}.$$ Solving, we get $x = \begin{pmatrix} \frac{11}{45} \\ -\frac{1}{9} \\ ... 2 These two equations should be enough: 1) The sum$w+v=(1,1,1)$2)$u.v=0$, where$.$is the dot product. Equivalently,$v=\frac {<(1,1,1),(1,2,-3)>}{||(1,2,-3)||}(1,2,-3)+\frac{<(1,1,1),(-2,0,1)>}{||(-2,0,1)||}(-2,0,1))$, where$<,>$is the standard dot product in$\mathbb R^3$and$||v||$is the standard (square of the) length given ... 2 The way you have things defined, it looks like you can read off the column vectors you seek from that matrix with respect to the linear transformation. The matrix of$L$with respect to$\mathcal{S}$and$\mathcal{T}$is one that is precisely $$\begin{bmatrix}[L\mathbf{v_1}]_{\mathcal{T}} & [L\mathbf{v_2}]_{\mathcal{T}} & ... 0 A parallelogram with side lengths a and b, with angle \theta between them always has area$$ab\sin\theta,$$regardless of the dimension in which the parallelogram lives. Here, a = \lVert {\bf a} \rVert, b is defined analogously, and you can find \theta using the standard technique involving the dot product. As you're seeing, there are many ... 0 Yes, C=AB if B=\begin{bmatrix}v_1&0&\cdots & 0 \\ 0& v_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0&0&\cdots & v_n \end{bmatrix}. 0 Form a matrix A whose columns are the given vectors. Then \det (A^TA) is the square of the area of the parallelogram. This generalizes easily to k vectors in {\Bbb R}^n. 0 Take the cross product. Or, more generally for vectors in \Bbb R^n, take the wedge product. http://en.wikipedia.org/wiki/Exterior_algebra 0 Hint: An orthogonal projection is what we call idempotent, meaning that applying the map twice to a vector is the same as applying the map once. Thus if we have an eigenvector \vec{v} with$$ P\vec{v}=\lambda\vec{v}, $$then we have both$$ P^2\vec{v}=P\lambda\vec{v}=\lambda^2\vec{v} $$and$$ P^2\vec{v}=P\vec{v}=\lambda\vec{v}. $$Thus ... 0 We have V=E\oplus E^\perp, so every vector x in V can we written uniquely as x=y+z with y\in E and z\in E^\perp. Letting T be the orthogonal projection, we have that:$$T(x)=\lambda x \iff x=\lambda x + 0 \iff (\lambda -1)x=0$$Handling the cases \lambda =0 and \lambda \neq 0 separately might be useful. 0 The daily returns is an element-by-element multiplication. In MATLAB this is R=H.*D | 100 100 110 | | 0 0 5 | | 0 0 550 | | 0 200 200 | | 3 0 0 | = | 0 0 0 | The sum by stock is given by the diagonals of$$T = H D^\top$$| 100 100 110 | | 0 0 5 |T | 550 300 | | 0 200 200 | | 3 0 0 | = | 1000 0 | Going down the ... 1 Note that x = y if and only if -x = -y. Also, x = -y if and only if -x = y. So there are really only two possibilities:$$x = y \qquad \mbox{or} \qquad x = -y.$$In other words, once you know that x^2 = y^2, then you know that x and y have the same magnitude (the same absolute value); you also know that either x and y are exactly the ... 0$$\sqrt {x^2}= \sqrt {y^2} \implies |x|=|y| \implies \pm y=\pm x$$https://www.desmos.com/calculator/debdx80itq 1 If x^2=y^2, then y=\pm x, which is the same as x=\pm y. But \sqrt{x^2}=|x|. 1 First understand the following \mathbf{Thereom:} Let \{v_1,…,v_n\} be any basis of an inner product space V. Then there exists an orthonormal basis \{u_1,…,u_n\} of V such that the change of basis matrix from \{v_i\} to \{u_i\} is triangular i.e. for k=1,2.., n, u_k= a_{k1}v_1+a_{k2}v_2+..+a_{kk}v_{k} The proof comes from applying the Gram ... 0 You need to prove that$$ V = E + E^\perp $$and$$ E \cap E^\perp = \{0\} $$then it follows from an elementary theorem. 0 If the matrix (in general, let's say n \times n) is A and the initial population vector {\bf v}, after k generations the population vector is A^k {\bf v}. If there are n linearly independent eigenvectors {\bf u}_j and \lambda_j, we have {\bf v} = \sum_j c_j {\bf u}_j, and A^k {\bf v} = \sum_j c_j \lambda^k {\bf u}_j. If all ... 0 Since you say that elements f,g,h of V are scalar valued functions, and f can be applied to x\in\Bbb R^1, I conclude that V is the vector space of functions \Bbb R^1\to\Bbb R. Then f(x)\in\Bbb R is a scalar, from which you can indeed make a vector [f(x)]\in\Bbb R^1. But you cannot apply that to f, which is a function \Bbb R^1\to\Bbb R. On ... 1 Your matrix is the sum between an identity matrix and a circulant matrix, so the characteristic polynomial is given by:$$ p(\lambda)=(1-\lambda)^n-(-1)^n \tag{1}$$and the determinant is given by (-1)^n p(0), so it is 2 if n is odd and zero otherwise. 0 You can approach this with a diagrammatic map$$ \mathbb{R}^3\xrightarrow{T}\mathbb{R}^2\xrightarrow{S}\mathbb{R}^3. $$Now consider the composition of the two maps. 0 In \mathbb{Z}_2 the system matrix has determinant 0, which means that the solution cannot be unique. Therefore, let us look at only the first n-1 equations, with x_n considered not as an unknown, but as a parameter:then you solve the system backwards and get the solution x_1 = p_n + x_n (remeber that in \mathbb{Z}_2 + and - coincide), x_2 = ... 1 The general matrix is given by the sum between the identity matrix and a circulant matrix, hence its characteristic polynomial over \mathbb{Q} is given by:$$ p(\lambda)=(\lambda-1)^n-1.$$Over \mathbb{F}_2 such a matrix cannot be invertible since the sum of the elements in every row/column is zero, hence (1,1,\ldots,1) is an eigenvector associated ... 0 Gram-Schmidt finds, given a basis of a subspace, an orthonormal basis of this subspace. Its main tool is the following formula, which defines the orthogonal projection of a vector \vec v onto another vector \vec u:$$p_{\vec u}(\vec v)=\frac{(\vec u,\vec v)}{(\vec u,\vec u)}\vec u.$$What you should do: Find an orthonormal basis of your subspace and ... 0 Hint; your final vectors are not correct. The point of GS it to get an orthogonal set of vectors. Are yours orthogonal? You are starting off with two non orthogonal vectors , that is v_1=( 1 , 1 , 1) and v_2= ( 1 , 2 ,1) The GS algorithm proceeds as follows; let w_1=(1,1,1) then we define$$w_2= v_2- \frac{\langle v_1 , w_1 \rangle}{\langle w_1 , ... 0 Are you sure about what is$<f,f>$? Try an example, let say$f(x)=\cos x$. Observe that $$<f,f>=\int_0^1 [f(x)]^2\, dx$$ and$[f(x)]^2\geqslant 0$for all$x$. Now you can use the fact that "the integral is the area under the curve" to argue that$<f,f>$is always nonnegative. 0 All you have to do is to find the$A^{\infty} = \lim\limits_{n \rightarrow +\infty} A^n$and see what happens when you apply it to arbitrary "distribution" of pie and cake lovers. However, check your eigenvalues: this matrix clearly has one eigenvalue equal to 1 (I vaguely recall, but I'm not sure that all matrices which columns sum to 1 have this ... 0 For convenience, define$F=(I\otimes X^T)AX+BX+C$You want to find the derivative of$\|F\|^2_F = F:FStart with the differential \eqalign{ d(F:F) &= 2\,F:dF \cr &= 2\,F:\Big[(I\otimes dX^T)AX+(I\otimes X^T)A\,dX+B\,dX\Big] \cr &= 2\,F:(I\otimes dX^T)AX+2\,F:(I\otimes X^T)A\,dX+2\,F:B\,dX \cr &= 2\,AXF^T:(I\otimes ... 0 Alamos is right. Yet, using \chi_A, \chi_B can be calculated without knowing the (\lambda_i). For instance, let \chi_A(x)=x^4+3x^3-6x^2+2x-1 and B=P(A)=A^3-7A^2+3A-I (note that we may assume that degree(P)\leq 3). Then I,B,B^2,B^3,B^4 are known polynomials in A of degree \leq 3. Consequently they are linearly dependent and we deduce ... 0 If \lambda=1 is an eigenvalue, then there is a vector r such that Ar=r Let r=xi+yj+zk Ar=r would mean that 3x-2y+3z=x or 2x-2y+3z=0 -y+3z=y or 3z=2y -x+2y-2z=z or -x+2y-3z=0 Are these equations consistent? 2A-I=\begin{bmatrix} 3 & -2 & 3\\ 0 & -2 & 3\\ -1 & 2 & -3 \end{bmatrix}$And you can see the two last columns are proportionate so the matrix is not invertible and$\lambda=1$is an eigenvalue My computation of$\det(A-\lambda I)=-\lambda^3+\lambda^2+13\lambda-13$1 This is an awfully complicated way of deriving the derivative or gradient. It is simpler to show that$\|J(\theta+h)-J(\theta) - (X \theta-y)^T X h\|$is bounded above by$\|X^TX\| \|h\|^2$from which we see that$DJ(\theta)(h) = (X \theta-y)^T X h = \langle X^T (X\theta -y), h \rangle$, where$\langle \cdot,\cdot \rangle \$ is the usual inner product in ...

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