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Actually we do not need quaternions, because we are working only with one rotation so we can assume that $R$ is rotation around z-axis. We can restrict ourselfs only to xy-plane. Rotations in 2d can be expressed by unit complex numbers and skew-symmetric matrices correspond to pure imaginary numbers. Cayley transformation for skew-symmetric matrices: $$S ... 0 If A is in Jordan Normal Form then the centralizer C(A) = \{X \in \mathbb M_n(k) \ | \ AX = XA\} of A isn't too hard to figure out. Then for invertible B we have C(BAB^{-1}) = BC(A)B^{-1} so this will let you figure out C(A) for any matrix A. Finally the set you're looking for is C(A_1) \cap \cdots \cap C(A_n). Of course this is a pain to ... 0 There is a link with the quaternion skew-field H=\{q=x+yi+zj+tk|x,y,z,t\in\mathbb{R}\} where ||q||^2=x^2+y^2+z^2+t^2. If ||q||=1, then q is a unit and q^{-1}==x-yi-zj-tk. Moreover e^q=e^x(\cos\sqrt{y^2+z^2+t^2}+\dfrac{\sin\sqrt{y^2+z^2+t^2}}{\sqrt{y^2+z^2+t^2}}(yi+zj+tk)). We consider a rotation Rot(\theta,u) where u=[a,b,c]^T is unitary. To ... 3 The statement$$ \operatorname{Col}(A) \subset \operatorname{Null}(B) $$means that for every v\in \operatorname{Col}(A) we have B v=0. But every element v\in \operatorname{Col}(A) is of the form v=Ax. Hence BAx=0 for all x. What does this say about BA? 1 The multivariable chain rule is not normally stated in terms of strange things like \frac{dv}{du}. It is stated in terms of total derivatives, i.e., linear transformations that approximate the map at a given point. In particular, it says that if T(u_0,v_0)=(x_0,y_0), then the derivative of the composition T^{-1}\circ T at (u_0,v_0) is the ... 0 We assume that the matrices are real, otherwise, it is more complicated. We study the function f(M)=-2trace(AB^TM)-x^TMy where M\in O(n). Since O(n) is a Lie group, the tangent space in M to O(n) is TS=\{MH|H\text{ is skew-symmetric}\}. Then D_Mf:K\rightarrow -2trace(AB^TK)-x^TKy=-trace((2AB^T+yx^T)K) and we seek M\in O(n) s.t., for every ... 0 The differential is a linear application Df(A)[\cdot] such that f(A+B)-f(A)=Df(A)[B]+o(\|B\|) for "small" matrices B. In your case L\in \mathcal L(M_n(\Bbb R)). i.e. it is not a matrix, but a linear application from the space of matrices to itself (it is a tensor of fourth order, if you want). We can write$$f(A+B)-f(A)= BA^{m-1} ...
Expanding on my comment ... Write $$S = \left[\begin{matrix} 0 & r & -q \\ -r & 0 & p \\ q & -p & 0 \end{matrix}\right] \qquad M = I + S = \left[\begin{matrix} 1 & r & - q \\ -r & 1 & p \\ q & -p & 1 \end{matrix}\right] = ( I - S )^\top = N^\top$$ Note that $M$ (and $N$) fix the unit vector $\mathbf{p} ... 1 Ok so I figured it out. Notice$w^\top X y = y^\top X^\top w$. So continuing the above expression:$p(w|X,y) \propto exp(-{1 \over 2}(2\sigma_n^{-2}y^\top X w -w^\top(\sigma_n^{-2}XX^\top -\Sigma_p^{-1} )w))= exp(-{1 \over 2}(w^\top A^{-1}w-2 \bar w^\top A^{-1}w))$where$A=(\sigma_n^{-2}XX^\top -\Sigma_p^{-1})^{-1} $and the answer follows. 0 Calculate $$\begin{bmatrix} S_1 & S_2 \\ S_3 & S_4 \end{bmatrix} · \begin{bmatrix} T_1 & T_2 \\ T_3 & T_4 \end{bmatrix} = \begin{bmatrix} S_1 T_1 + S_2 T_3 & S_1 T_2 + S_2 T_4 \\ S_3 T_1 + S_3 T_3 & S_2 T_2 + S_4 T_4 \end{bmatrix}.$$ This result holds as long as the dimensions involved fit together. This can be calculated ... 1 Let$A\oplus B$denote the block diagonal matrix with matrices$A$and$B$as diagonal entries. Assume$B\sim C$. Consider$A\oplus B$and$A\oplus C$. Because$B\sim C$, there exists a sequence of elementary row opperations that change$B$into$C$. Applying these same opperations in the same order to the lower section of$A\oplus B$(the$B$block), still ... 5 As$A$is normal, it is diagonalizable. So, there is an orthogonal matrix$P$such that: $$A=P^*DP,$$ where$D$is a diagonal matrix. Then there is a polynomial$p=p(x)=\sum_{k=0}^mc_kx^k$, such that:$p(D)=D^*$. If$D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$, then his polynomial takes$\lambda_j$to$\overline{\lambda}_j$. But $$p(A)=\sum_{k=0}^m ... 0 Solved. I found in paper formula which states it should be divided by some number which in case of first one is 2. See Update. 0 Another solution is to write (x,y) in shifted polar coordinates, that is:$$x = 5 + r \cdot \cos\theta \\ y = 6 + r \cdot \sin \theta$$where r is the distance of (x,y) to (5,6), and \theta is the angle from the line y = 6, to the point, anti-clockwise. You can write (x_\varphi, y_\varphi) = (5 + r \cdot \cos(\theta + \varphi), 6 + r \cdot ... 0 If \alpha is a path with \alpha(0) = g, then g^{-1}\alpha is a path through the identity. Now multiply that path with g. 1 We have$$ \left(\frac{1}{m}I+A\right)^{-1}=m(I+mA)^{-1}=m\sum_{n=0}^\infty (-1)^nm^nA^n=\sum_{n=0}^\infty (-1)^nm^{n+1}A^n. $$The series above (known as Newman series) converges whenever$$ \lvert m\rvert\cdot\|A\| <1. $$1 as far as I remember these type of sums are called Nuemann's series. \frac{1}{I-M}=\sum_0^\infty M^k under the "right" assumption on M.Operator M should be bounded ,you have that for free since you deal with matrices, and series conveges in norm. 2 Let us write G_{m} = G = mA (B + mA^{*}A)^{-1} in order to clarify that G_{m} depends on m. Ingredient 1. We show the following claim: Claim 1. There a uniform bound C > 0 satisfying \| G_{m} \| \leq C for all m. Indeed, for any vector u \in \Bbb{R}^{n} we get$$ ((B + mA^{*}A)u, u) = (Bu, u) + m (Au, Au)\geq (Bu, u) $$Since ... 1 The most straightforward way to see that your answer is incorrect is to calculate the rank of A, which is 1, since the columns (equivalently, the rows) span a space of dimension 1. Your answer has rank 2. Your argument is that a 2\times 2 Jordan matrix with two linearly independent eigenvectors having eigenvalue zero must be ... 1 Just because you have a certain multiplicity of an eigenvalue does not mean you only have one Jordan block containing that eigenvalue! For example if you had an eigenvalue of multiplicity 4 that could be 1 block of 4, it could be 1 block of 3 and 1 of 1, it could be blocks of size 2 and 2, it could be 2, 1, and 1, or finally it could ... 2 but I found it to be wrong Then you must have made a mistake somewhere. If you post your counterexample, we could find out which mistake. Let \beta be the symmetric bilinear form. Then there are two possibilities, \beta(v,v) = 0 for all v \in K^2, or \bigl(\exists v_1\in K^2\bigr)\bigl(\beta(v_1,v_1)\neq 0\bigr). In the first case, pick any ... 0 No. One may be tempted to expect \sqrt{1+x^2} to behave like x. However, since x^2 is not matrix monotone, it may cause \sqrt{1+x^2} to fail to be matrix monotone. I don't have a nice counterexample at hand, but if you have access to a numerical linear algebra package, you will see that when$$ A=\pmatrix{2.01&1\\ 1&3},\ B=\pmatrix{1\\ ... 1 $$\text{Let} \space A = U_1\Sigma_1V_1^* \space \text{and} \space B = U_2\Sigma_2V_2^* \space \text{(SVD of A and B)} \\ \text{Then} \space A^+ = V_1\Sigma_1^+U_1^* \space \text{and} \space B^+ = V_2\Sigma_2^+U_2^* \\$$ $$\text{Now} \space AB = 0 \\ U_1\Sigma_1V_1^*U_2\Sigma_2V_2^* = 0\\ U_1^*U_1\Sigma_1V_1^*U_2\Sigma_2V_2^*V_2 = 0\\ ... 3 NOTE: My original answer was about a different question. (I have considered only adding integer multiples of some row.) I have completely changed it, hopefully the new answer is correct. Other answers have already gave explanations about geometric meaning of \det(A)=1, so let me address the questions about row operations. Let us start by checking what ... 3 Usually when we talk about the determinant without any other information attached, the only really relevant information is whether or not it is zero. However, if we are interested in geometry, there is some significance to matrices with determinant 1. Namely, an important subset of them form the so-called special orthogonal group, which is just a fancy way ... 1 Simply, if the matrix, with determinant 1, represents a transformation, then it will preserves the volume. For example if the determinant of A is 5, then the volume of the image of a cube with volume 1 is 5 \times1=5. The identity matrix do the same thing. 0 Saying \det(A) = 1,means A is the determinant is$$ \begin{matrix} a_{11}&a_{12}&\cdots &a_{1n}\\ a_{21}&a_{22}&\cdots &a_{2n}\\ \vdots&\vdots&\ddots &\vdots\\ a_{n1}&a_{n2}&\cdots &a_{nn}\\ \end{matrix} $$and the determinant evalutes to one. Since \det(A) = \det(I), A = I_n where I_n is the identity ... 2 This is a pfaffian! See the wiki article for ways of computing it. 2 Let L_A be left multiplication by A. Then you can split F^{n\times n} into the direct sum of L_A-invariant n copies of F^n (the columns of M), on each of which L_A acts by the usual multiplication of a matrix by a column vector. Hence \det(L_A)=[\det(A)]^n. Similarly, \det(R_B)=[\det(B)]^n. Hence ... 0 You have the right idea. The k-dimensional subspaces of a diagonalizable linear operator can be found by taking the span of any k eigenvectors. Note that in this particular case, the eigenvectors are \pmatrix{1&0&0}^T,\pmatrix{0&1&0}^T, and \pmatrix{0&0&1}^T. 1 That depends on how you define vector derivative. There are generally two ways. One is applying abstract index notation, then$$\frac{d}{dx}x^T=\left(\frac{dx_i}{dx^j}\right)=(\delta_{ij})=(e_1\otimes\cdots\otimes e_n)^T$$where e_is are unit vector whose i th component is one and zero otherwise. Another way to look at it is to regard as directional ... 1 What sort of object can be the derivative of a vector-valued function whose values are row vectors and whose arguments are column vectors? Generally, what kind of object can be the derivative of a function whose values are members of one vector space W and whose arguments are members of another vector space V?$$ f: V\to W $$The answer is that the ... 0 Let A is a required 3x3 matrix and from the problem we have A [w1 w2 w3] = [f(w1) f(w2) f(w3)] Therefore A = [f(w1) f(w2) f(w3)]*[w1 w2 w3]^-1 find [f(w1) f(w2) f(w3)] = (\left[\begin{matrix}0&0&5\\1&-1&11\\0&0&2\end{matrix}\right]) and [w1 w2 w3]^- = ... 2 You can add or remove any row to any other (not the same) row, but in the first step you replace R3 by -R3 (+R1 but you can add). So by negating a row, you negate the determinant. 1 If you have an SVD representation of the matrix, you can use this to compute the product in linear time. I'm not sure how to efficiently compute this representation, though. You can also truncate the SVD of a higher-rank matrix to get a low-rank approximation. More specifically, if an m\times n matrix is of rank r, its (compact) SVD representation is ... 3 Note that any eigenvalue \lambda of A satisfies Av = \lambda v \tag{1} for some vector v \ne 0, so that A^2 v = A(Av) = A(\lambda v) = \lambda(Av) = \lambda(\lambda v) = \lambda^2 v, \tag{2} whence 0 = (A^2 + I)v = A^2v + v = (\lambda ^2 + 1)v, \tag{3} and since v \ne 0 this implies \lambda^2 + 1 = 0; \tag{4} but equation (4) has no ... 2 If Av=\lambda v then A^2v=\lambda^2v but on the other side -Iv=-v so we get \lambda^2=-1 which has no real solution. 1 The short SVD of uu^* is v\sigma v^* with v=\frac{u}{\|u\|} and σ=\|u\|^2. If you compute the bisector w=u+\frac{u_1+0}{|u_1|+0}\|u\|e_1 of u and e_1, then you can get a full unitary matrix V as the reflection matrix I-2\frac{ww^*}{\|w\|^2}, so that uu^*=V\Sigma V^* 0 Denote$$ A = \begin{bmatrix} 0 & \sqrt{2} & 0 & \sqrt{2} \\ \sqrt{2} & 0 & \sqrt{2} & 0 \end{bmatrix} $$Step 1: Find the spectral decomposition of AA^*. That is, find a diagonal matrix D and unitary matrix U for which$$ UDU^* = AA^* $$To get the diagonal entries of \Sigma, take the square root of the entries of D, so ... 1 We have$$\left[\array{A'\\B'}\right]\left[\array{A &B}\right]=\left[\array{A'A & I\\I &B'B}\right]\succeq 0.$$Thus for all u,v$$u'A'Au+v'B'Bv+2u'v\geq 0.$$With u=-\left(A'A\right)^{-1}v we get$$v'\left(A'A\right)^{-1}v+v'B'Bv-2v'\left(A'A\right)^{-1}v\geq 0\\ \iff v'\left(B'B-\left(A'A\right)^{-1}\right)v\geq 0\iff ... 1 Note that$a^3+sa+t=b^3+sb+t=c^3+sc+t=0$so that (adding the three expressions) $$(a^3+b^3+c^3)+s(a+b+c)+3t=0$$ Now equating coefficients in$x^3+sx+t=(x-a)(x-b)(x-c)$we get$a+b+c=0$(coefficient of$x^2$) and$t=-abc$Substituting these into the previous equation we get: $$a^3+b^3+c^3-3abc=0$$ NOTE: suppose our equation had been$x^3+rx^2+sx+t=0$. We ... 1$\det(A) = 3abc - a^3 - b^3 - c^3 = -(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$since$a + b + c = 0$by Viete's theorem. 0 Thanks for those kindly person answered or commented on my question. It's helpful. I find 2 ways to solve my problem. 1.The RV coefficient. Take each column of the matrix as an independent realization of a random vector. So, if I want to calculate matrix$A_1$and$A_2$, where$A_1 \in R^{n*k}$,$A_2 \in R^{m*k}$,$m,n \in N^+$, I turn this problem into ... 1 Your procedures are rather non-standard in terms of finding basis for columnspaces and rowspaces. In particular your second procedure for finding the rowspace is not only unorthodox, but appears to be incorrect. To fully answer all of your questions takes a bit of time, and I apologize in advance for the length. The fact is, you can find both basis in a ... 0 Yes: as you probably we'll know, the ideals of$R$and those of$M_n(R)$are in correspondence via the map$I\to M_n(I)$. Moreover this map preserves containment and intersections. If U is a right primitive ideal, say the annihilator of the simple R module S, then it's easy to show that$\oplus_{i=1}^n S$Is a simple right$M_n(R)$module with annihilator ... 2 Let's say more about this linear transformation. First $$A\in \ker t\iff t(A)=A+A^T=0\iff A^T=-A\iff A\in\mathcal{AS_n}(\Bbb R)$$ hence$0$is an eigenvalue of$t$with multiplicity equal to $$\dim\mathcal{AS_n}(\Bbb R)=\frac{n(n-1)}2:=\alpha_n$$ and if$A\in\mathcal{M_n}(\Bbb R)$then $$t(A)=A+A^T\in\mathcal{S_n}(\Bbb R)$$ and by the rank-nullity theorem ... 0 No, the kernel is not zero: There are matrices such that$A^T = -A$, called skew-symmetric. Try a matrix with zeros on the main diagonal. To show that the range of$t$is the collection of symmetric matrices, you need to check two things: For any$A$,$t(A) = A + A^T$is symmetric. For any matrix$Q$for which$Q^T = Q$, there is a matrix$A$such that ... 1 The set of anti-symmetric$2\times2$matrices, is the kernel of$t$:since if$A=-A^T$, then $$t(A)=0$$ 1 Consider the (4x2) matrix,$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \\ 2 & 4 \\ 2 & 5 \\ \end{bmatrix}$Augment$A$with the "resource" vector,$b = \begin{pmatrix} R1, R2, R3, R4 \end{pmatrix}^T$and then do elementary row operations on the augmented matrix to compute the row echelon form. I'll show the individual steps...so ... 0 You can always do it if the transfer function has non-repeated roots. For example, if you can put your transfer function into this form: $$G(s) = \frac{Y(s)}{U(s)} = \frac{c_1}{s-p_1} + \dots + \frac{c_n}{s-p_n}$$ then, it is equivalent to:$\$\begin{align} \begin{bmatrix}\dot{x}_1 \\ \dot{x}_2 \\ \vdots \\ \dot{x}_n\end{bmatrix} &= \begin{bmatrix}p_1 ...