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$n>2$ let $G$ be a subgroup of $GL(n)$ such that $O(n)\subset G$ and $dim(G)=dim(O(n))+1$. We denote by $G_0$ the connected component of $G$. We know that $SO(n)$ is simple, this implies that $SO(n)\subset [G_0,G_0]$. Thus $dim([G_0,G_0]\geq dim(G_0)-1$. We can't have $dim([G_0,G_0])=dim(G_0)$ since this would implies that $G$ is simple a fact which is ...

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Assuming that you meant $T:V\to W$, you know that $\operatorname{Rng}(T)\subseteq W$, but the only $n$-dimensional subspace of $W$ is itself, so equality holds.

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Orthogonal transformations preserve the norm of a vector. You can take the product of that group with the one-dimensional multiplicative group that scales the vectors by nonzero real factors (a copy of $\Bbb{R} \backslash \{0\}$), in other words, the nonzero multiples of the unit matrix.

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$$ABXA^{−1}B^{−1} = I + A$$1st step :$\times A^{-1}$from left $$A^{-1} ABXA^{−1}B^{−1} = A^{-1}(I + A)\\BXA^{−1}B^{−1} = A^{-1}+I$$2nd step $\times B^{-1}$from left $$B^{-1}(BXA^{−1}B^{−1} = A^{-1}+I)\\XA^{−1}B^{−1} = B^{−1}A^{-1}+B^{−1}$$3rd step $\times B$from right side $$(XA^{−1}B^{−1} = B^{−1}A^{-1}+B^{−1})B\\XA^{−1} = B^{−1}A^{-1}B+I$$4th step ...

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Hint: $A^{-1} A = AA^{-1} = I$ So, for example, just like $\frac{1}{4} x = 7 + y$ becomes $x = 28+4y$ multiplying both sides by $4$... $AX=I+B$ becomes $X = A^{-1} + A^{-1}B \quad$ left-multiplying both sides by $A^{-1}$, and $XA=I+B$ becomes $X = A^{-1} + BA^{-1} \quad$ right-multiplying both sides by $A^{-1}$. Remember that, in fact, $AX \neq XA$ for ...

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By definition $$B_{ij}=(w_j)_i,\qquad B^T_{ij}=(w_i)_j$$ Then \begin{align} (Aw_k)_i&=A_{ij}(w_k)_j=B_{ih}B^T_{hj}(w_k)_j+d\delta_{ij}(w_k)_j\\ &=(w_h)_i(w_h)_j(w_k)_j+d(w_k)_i\\ &=(w_h)_id_k^2\delta_{hk}+d(w_k)_i\\ &=d_k^2(w_k)_i +d(w_k)_i\\ &=(d_k^2+d)(w_k)_i=c_k(w_k)_i \end{align}

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Permuting rows two and three, as well as columns two and three yields the matrix $$A_2:= \begin{pmatrix} 1275 & 0 & -169 & -208 \\ 0 & 1275 &-208 & -256 \\ -169 & -208 & 1531& -208 \\ -208 & -256& -208 & 1444\\ \end{pmatrix}.$$ Denoting $B:=\pmatrix{-169 ... 0 The matrix formed by the product$\mathbf v\mathbf w^T$represents a linear transformation that maps the entire space onto the line containing the vector$\mathbf v$. If$\|\mathbf v\|=1$, then$\mathbf v\mathbf v^T$represents orthogonal projection onto$\mathbf v$(if$\|\mathbf v\|\ne1$, you can of course get orthogonal projection by dividing by ... 0 The definition of determinant through permutation will give the answer. For a matrix$C=(c_{i,j})_{m \times m}$$$\mathrm{det}(C)=\sum \limits_{\sigma \in S_{m}} \mathrm{sgn}~\sigma\cdot c_{1,\sigma(1)}c_{2,\sigma(2)} \cdots c_{m,\sigma(m)}$$ If we write,$J=(x_{i,j})_{2n \times 2n}then, $$\mathrm{det}(J)=\sum \limits_{\sigma \in S_{2n}} ... 1 What you looking for is a Thomas algorithm which is a simplified form of a Gaussian Elimination or if you want LU decomposition. Matlab looks fast because Matlab identifying such special cases and call in such case to a very effective solver for banded-matrices. This solver is little bit more general then Thomas algorithm. The very rough idea is that you ... 3 Just to expand on the comment of @Myself above... Adjoining algebraics as matrix maths Sometimes in mathematics or computation you can get away with adjoining an algebraic number \alpha to some simpler ring R of numbers like the integers or rationals \mathbb Q, and these characterize all of your solutions. If this number obeys the algebraic equation ... 1 This should help you out. It is a pretty somplete list. https://en.wikipedia.org/wiki/Invertible_matrix 2 One has:$$J:=\begin{pmatrix}0 & A\\B & 0\end{pmatrix}=\begin{pmatrix}A& 0\\0 & B\end{pmatrix}\times\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}.$$You only have to compute:$$\varepsilon:=\det\left(\begin{pmatrix}0 & I_n\\I_n & 0\end{pmatrix}\right).$$Indeed, using the first equality, one has: ... 1 There are block matrix determinant identities that will give you this, but if you want to solve using basic determinant formulas consider the full expansion formula for the determinant using permutations, giving a sum of n! terms for an n \times n matrix, where each term is either 1 or -1 multiplied by n entries in the matrix. If a term includes an ... 0 First note that the obvious approach -- inverting L using a sparse direct method like sparse QR (found in the excellent SuiteSparse package, for instance) -- should take on the order of seconds for a matrix of this size, not 10 hours. If performance is critical, it seems that the topic of numerically extracting the diagonal of the inverse of a matrix is ... 1 That A(h) is invertible for small enough h is clearly true, as you said, because A and \det are continuous. Now, for the boundedness, I assume that the matrix norm you use, is the standard spectral norm. Since A is continuous, B(h)=A(h)A(h)^T is also continuous and in particular the spectral radius \rho(B(h)) of B(h) is continuous in h. ... 2 So \forall\ \textbf{x}\in\mathbb{R}^{2n} we have \textbf{x}^T\begin{pmatrix}A & 0\\ 0 & A\end{pmatrix}\textbf{x}\geq0. Let \textbf{y}\in\mathbb{R}^n arbitrary, we want to show \textbf{y}^TA\textbf{y}\geq0. To this end, let \textbf{0}\in\mathbb{R}^n and define \textbf{x}=\left(\begin{array}{c}\textbf{y}\\ \textbf{0}\end{array}\right). Then ... 3 The answer is positive, assume A \in \mathbb{R}^{n \times n}, then for any x \in \mathbb{R}^n, take (x^T, x^T)^T to test the diagonal matrix:$$\begin{pmatrix} x^T & x^T \end{pmatrix} \begin{pmatrix}A & 0 \\ 0 & A \end{pmatrix} \begin{pmatrix}x \\ x\end{pmatrix} = 2x^TAx \geq 0$$implies that x^TAx \geq 0, i.e., A is positive ... 2 It is both. It is also diagonal. But it isn't "strictly upper triangular" or "strictly lower triangular" (those require 0s on the diagonal). 0 Yes, you're right with that. Note that for matrix multiplication you always have \underbrace{A}_{\in\mathbb{K}^{m\times n}} \cdot \underbrace{B}_{\in\mathbb{K}^{n\times k}} = C\in\mathbb{K}^{m\times k}. Your vector-vector multiplication (the vector is in \mathbb{K}^{3\times 1}) in the form vv^T where v=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix} ... 3 Hint What can you say about the traces of the given matrices? (Alternatively, for three of the choices, one can find a suitable matrix B for which the equation holds for all A.) 1 You can eliminate answers A and D by noting that the zero matrix B always satisfies that identity. 1 Note that -x^3+6x^2+9x-14=-(x-1)(x+2)(x-7), so we may assume that M is the diagonal matrix with entries 1,-2,7 on the diagonal. Then it's easy to see that the characteristic polynomial of M^{-3} is given by (x-1)(x+(1/2)^3)(x-(1/7)^3). 1 As stated, the result is false. For example, let x=\begin{pmatrix}1&-1 \end{pmatrix} and y=\begin{pmatrix}1&1\end{pmatrix} then$$x^Ty=\begin{pmatrix}1&1\\-1&-1 \end{pmatrix}.$$That matrix has 0 as only eigenvalue and it is clearly not diagonalisable. Actually, the statement should be Let x,y be two non zero row vectors, ... 3 That's wrong. For x = (0,1) and y = (1,0),$$ x^t y = \begin{pmatrix} 0\\ 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix} $$is not diagonizable. 0 Let \mu=E(X). Then cov(X)=E(XX^T)-\mu\mu^T and cov(AX)=E((AX)(AX)^T)-(A\mu)(A\mu)^T=A(E(XX^T)-\mu\mu^T)A^T=Acov(X)A^T. 0 While the estimate is true with C=1 for the identity matrix and the all-one matrix, it is not true in general. Take the following matrix:$$ A = \pmatrix{ 1 & 1 & \dots & 1 & 1\\ 1 & 1 & & 0 & 0\\ 1 & 0 & \ddots & 0 & 0\\ 1 & 0 & & 1 & 0\\ 1 & 0 & \dots & 0 & 1\\ }. $$That ... 0$$\nabla_x\|x^HAx - b\|_2^2 = \nabla_x(x^HAx - b)^2=4(x^HAx - b)Ax$$0 In which ring do you work, in K or in K[A] \subset M_n(K) the commutative subring generated by the matrix A. If you work in K you certainly cannot consider A as an element of the ring. If you work in K[A], what is the meaning of A in det(A-\lambda I_n), is it a scalar ? There is however, a trick which can make the proof correct. We know ... 0 As Gerry Myerson points out, you divide by the norm to produce a unit vector. \|w_1\|=\sqrt2, not 1/\sqrt2. When you divide w_1 by this you get \frac1{\sqrt2}(1,-1) for the first vector as before. \|w_2\|=1/\sqrt2. Again, you divide w_2 by this value, giving \sqrt2(-1/2,-1/2)=\frac1{\sqrt2}(-1,-1) for the second. 0 A nice trick for finding eigenvectors for 2 and 3 dimensional matrices is to exploit the Cayley-Hamilton theorem (it works for higher dimensions too, but the cost of multiplying the matrices makes it less attractive). The Cayley Hamilton theorem says that if p(x) is the characteristic polynomial of a square matrix A, then p(A) = 0 (scalars in this ... 0 dealing with the more difficult direction - i will leave the details for you to complete, since this is a useful exercise to go through, in terms of familiarizing yourself with use of the index notation, and gaining practice in thinking at the required level of abstraction. but there may be a fairly intuitive and straightforward solution along the following ... 1 Hints: A=\mathbf v\mathbf w^T\implies\operatorname{rank}A=1 should be pretty easy to prove directly. Multiply a vector in \mathbb R^m by A and see what you get. For the other direction, think about what A does to the basis vectors of \mathbb R^m and what this means about the columns of A. Solution Suppose A=\mathbf v\mathbf w^T. ... 0 Let's define some intermediate variables$$\eqalign{ P &= V^{-1} \cr M &= X^TPX \cr F &= M^{-1} \cr }$$whose differentials are$$\eqalign{ dP &= -V^{-1}\,dV\,V^{-1} \cr dM &= X^T\,dP\,X \cr dF &= -M^{-1}\,dM\,M^{-1} \cr }$$That last differential is the one we're interested in, so let's successively substitute variables until ... 0 We'll think the matrices as endomorphism of the space V. And so we'll think the product of AB as composition of endomorphism.$$V\underset{A}{\longrightarrow}V\underset{B}{\longrightarrow}V$$First of all we have to note that linearity of A and B implies that the image of the vector 0 is always 0 and so we have the following relation on the kernel: ... 1 Up to a minor bug, your ideas are correct. You eigenvector is v_1=\begin{pmatrix}1\\0\\0\end{pmatrix}. You want v_2 such that (A-2I)v_2=v_1, or, as you said, \begin{pmatrix}a\\1\\0\end{pmatrix} for arbitrary a (because first generalisd eigenvector is defined up to an eigenvector). For the v_3 you want (A-2I)v_3=v_2 ... 0 Hint: the matrices of W have the form$$ W =\begin{bmatrix} a&b\\ c&-b \end{bmatrix} $$so$$ \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \in W $$now check for$$ \begin{bmatrix} a&b\\ c&-b \end{bmatrix} + \begin{bmatrix} x&y\\ z&-y \end{bmatrix} \quad and \quad k \begin{bmatrix} a&b\\ c&-b \end{bmatrix} $$have the ... 1 It is. You simply have to check the three properties: 0\in W, A,B\in W implies A+B\in W and A\in W implies \lambda A \in W. 1 The description of all three homogeneous spaces can be derived from the standard linear action of U(3) on \mathbb C^3. Since unitary maps preserve the lenght of vectors, this restricts to a transitive action on the Unit sphere in \mathbb C^3 which is S^5. The stabilizer of a point in S^5 is easily seen to be isomorphic to U(2), whence ... 1 Invertible:$$ \begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ \begin{bmatrix}1&0\\0&-1\end{bmatrix},\ \ \begin{bmatrix}0&1\\1&0\end{bmatrix},\ \ \begin{bmatrix}0&1\\-1&0\end{bmatrix}. $$Not invertible: the canonical basis, i.e.$$ \begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ ... 0 It’s not going to be a particularly simple expression, but you can generateA$from$a$with a sequence of matrix operations. I’ll describe the building blocks here and let you work out the specific operations for what you’re trying to do on your own. Left-multiplying a matrix by the$k$th row of the identity matrix picks out the$k$th row of that ... 0 If you just want four linearly independent vectors in the end (you don't care about orthogonality and such), then picking from the standard basis vectors also works. For instance, it is pretty clear that$(0,1,0,0)$is linearly independent from the two vectors that you have. Now you just have to check the other three (odds are any of them would work, but you ... 1 Yes and yes. For the invertible, think about the identity matrix, and try to change the diagonal values. Alternatively, think about the matrices of the form $$\begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}$$ For the un-invertible, think about the simplest basis for the set of matrices you can possibly think of. 2 The method is certainly sound. The determinant is indeed $$\det A(a)=3\det \begin{bmatrix} 1 & -1 & a \\ 0 & a & 0 \\ 5 & 1 & 5 \end{bmatrix} =3a\det\begin{bmatrix} 1 & a \\ 5 & 5 \end{bmatrix} =3a(5-5a)$$ which is different from zero if and only if$a\ne 0$and$a\ne 1$. 2 Hint : For which$a$do we have$15a-15a^2=15a(1-a)=0$? 1 The non-symmetric algebraic Riccati equation is (1)$XAX+B_1X+XB_2+C=0_2$where$A,B_1,B_2,C$are given in$M_2(\mathbb{C})$and the unknown is$X\in M_2(\mathbb{C})$. If$A,B_1,B_2,C$are generic (to get an idea, consider random choices) matrices, then this equation can be rewritten (2)$X^2+AX+B=0$. Now, if$A,B$are generic, then (2) admits exactly$6$... 0 This would be an example of an adjacency matrix$A$of a graph that is not simple. Each entry$A_{ij}$in the matrix corresponds to the number of edges between vertices$i$and$j$. For example, there are 5 edges between$a$and$d$, since$A_{a,d}=5$. If$A_{ij}$is a blank entry, then there are no edges between vertices$i$and$j$. From this, it should ... 1 By diagonalization, you can find that any diagonal matrix$A$can be represented as $$A = P D P^{-1},$$ where$D$is a diagonal matrix in which each element is an eigenvalue, and then$P$is a nonsingular matrix (i.e. its columns are linearly independent). Now, the power of a diagonalization is that you can easily figure out the power of a matrix (pun ... 0 We will use the following notation: For matrix$T\in F^{n\times n}$, we have$L_T: F^{n\times 1} \to F^{n\times 1}$is the linear transformation defined by $$v \mapsto Tv$$ Now note that$\text{Im } L_{AB} = \text{Im } (L_A \restriction_{\text{Im} B})$. Thus the image of$L_{AB}$is a subset of the image of$L_{A}$. Hence$\text{rank} (AB) \le ...

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Hint: a rotation that is neither $180^\circ$ nor $0^\circ$ has no real eigenvalues.

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