# Tag Info

0

Like in the definition of Wikipedia you are correct, $k$ cannot be equal $0$. In fact elementary matrices represent the steps that you do with the gaussian algorithm, and you are not allowed to multiply a row/column by $0$ to get the Row Echelon Form because when you multiply a row/column by $0$ "you loose the information" of the row/column and that is ...

2

Yes it's wrong. Notice that you should write the image of the polynomials in the given basis i.e. $$D(1+x)=1$$ and $$D(1+x+x^2)=1+2x=2(1+x)-1$$ so the second column of the matrix is $(1,0,0)^T$ and the third column is $(-1,2,0)^T$.

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Matrices consisting of function entries are called matrix valued functions, they are studied in matrix calculus for example. Vectors with function entries are called vector valued functions. In general, if $V$ is a vector space one can define spaces of functions on $[a,b]$ with values in $V$, e.g. $L^2([a,b],V)$ with $V=\mathbb{R}^n$ has the inner product ...

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Assuming that the functions are in $L^1$ is not enough for the integral $\int_a^b f_i(x)g_i(x)\,dx$ to converge. If you want your space to have an inner product, assume the functions come from $L^2$. In which case you have a vector-valued $L^2$ space, or $L^2$ space of vector-valued functions. E.g., Cauchy–Schwarz inequality on vector-valued L2 space ...

0

Every linear map $f:\mathbb R^n\to\mathbb R^m$ is represented by a $(m\times n)$ matrix, say $A$. The operator norm of $A$ is equal to the Lipschitz constant of $f$, directly from definitions. Same holds for the linear map defined by the pseudoinverse matrix $A^\dagger$. The map is Lipschitz, with the Lipschitz constant $\|A^\dagger \|$. You should be ...

2

Here is a computational contribution that treats the case of a square matrix. As pointed out this problem can be solved using the Polya Enumeration Theorem. We just need to compute the cycle index of the group acting on the slots of the matrix, subsitute $1+z$ into the cycle index to get the generating function and finally set $z=1$ to ...

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You should give Sequalator a try. It is capable of solving thousands of Linear Simultaneous Equations with great precision and speed. On my mediocre laptop it is able to solve 1000 equations just below a second. I am getting the same answer as yours in Sequalator with 0% error. To verify that the answer is correct you can check the error i.e. ...

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Denote scalar product of vectors $v,u$ by $(v,u)$, norm of vector $v$ by $\|v\|=\sqrt{(v,v)}$. Lemma 1. Let $A$ be a symmetric positive operator on $\mathbb{R}^n$, $f\in \mathbb{R}^n$ be a vector. Then $(Af,f)\cdot (A^{-1} f,f)\geq \|f\|^4$. Proof. Let $A=B^2$, where $B=\sqrt{A}$ is positive. Then $(Af,f)=(B^2f,f)=(Bf,Bf)=\| Bf\|^2$, ...

1

Two commuting matrices can be diagonalized by the same matrix. The positive semi definite follows immediately.

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$A$ is in the form $A=sU+V$ where the $u_{i,j}$ are $1$ and the $v_{i,j}\in\{0,1\}$. Then $\det(A)=as+b$ where $a=trace(Uadj(V))$ and $b=\det(V)$. If $\det(V)=0$, then we are done ; thus we assume that $V$ is invertible. Note that $ab$ has the same signum as $trace(UV^{-1})$ and it remains to show that $trace(UV^{-1})>0$ (the sum of the entries of ...

3

The set of $n \times n$ matrices that are "invertible in practice" is exactly the set of $n \times n$ matrices that are invertible, as long as $n < 10^{10}$. For $n \geq 10^{10}$, invertible matrices are invertible in practice, but not the other way around. For a counterexample, consider the matrix given by $$A_{ij} = \begin{cases} 1 - 1/n & i=j\\ ... 1 Technically, yes, but practically, no. Technical answer: Consider the n\times n matrix$$M = \left[\begin{array}{ccccc}1 & x & x & \ldots & x\\x & 1 & 0 & \ldots & 0\\x & 0 & 1 & \ldots & 0\\ \vdots & & & \ddots & \\ x & 0 & 0 &\ldots & 1\end{array}\right].$$By ... 2 Even when one assume A(t) is symmetric, the alternate form is still incorrect. For a counter-example, consider$$A(t) = \begin{bmatrix}0 & 1 & 0\\1 & 0 & t\\0 & t & 0\end{bmatrix}$$We have$$\frac{d}{dt} S(t) = \begin{bmatrix}0 & 0 & 1\\0 & 2t & 0\\1 & 0 & 2t\end{bmatrix} \ne \begin{bmatrix}0 & 0 ...

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Let $\text{can} = \{1, x, x^2\}$ be the standard basis for $\mathcal{P}_2(\Bbb R)$. You claim that you got the matrix $[A]_{\text{can}}$. We want $[A]_{\text{B}}$, if I understood you right. If $[I]_{B, \text{can}}$ is the matrix obtained putting the coordinates of $B$'s polynomials, in the $\text{can}$ basis, in columns, then we have: [A]_B = [I]_{B, ... 0 Denote the three elements of B by p_1, p_2, p_3. We have \begin{align*} Ap_1(x) &= (x^2 - 2)p_1(1) - xp_1'(x)\\ &= 2x^2 - 4 - x\\ &= -4(1+x) + 3x + 2x^2\\ &= -4(1+x) + 3(x+x^2) - x^2\\ &= -4p_1(x) + 3p_2(x) - 1p_3(x) \end{align*} I'm sure, you can do the other two along the same lines. 2 Think about the ij entry of A^T A; it's s_{ij} = \sum_k a_{ki} a_{kj} $$Take the derivative with respect to t (using primes to denote that) to get$$ s'_{ij} = \sum_k a_{ki}' a_{kj} + \sum_k a_{ki} a_{kj}' $$The claim is that this is just 2 \sum_k a_{ki}' a_{kj}, after some index-shuffling, and that's true if the matrix is symmetric, and not ... 6 Well, the set of all eigenvectors of an eigenvalue forms a subspace. I.e. if u,v \in Eig(A,\lambda) then a_1 u+a_2v\in Eig(A,\lambda): For Av=\lambda v and Au=\lambda u we have: A(a_1u+a_2v)=a_1Au+a_2Av=a_1\lambda u+a_2\lambda v=\lambda(a_1u+a_2v) Note: 0 is not an eigenvector. So the set above is only a space if we add 0 to the space. 2 Continuing JimmyK4542's answer, we could write the following two relationships using the Woodbury Matrix identity:$$(A+D)^{-1} = A^{-1}-A^{-1}(D^{-1}+A^{-1})^{-1}A^{-1} (D^{-1}+A^{-1})^{-1} = D - D(A+D)^{-1}D $$In the second, we've treated D^{-1} as large and A^{-1} as the perturbation, which is sensical if D was small to begin with. Plugging ... 0 Q1) Any true norm will do here. The Frobenius norm is a bit easier to work with, because of course the squared norm is easily differentiated. I don't believe the spectral norm is differentiable. Q2) A more general class of functions that can be used instead of the squared norm are Bregman divergences, or Bregman distances. I believe the key properties that ... 0 This is Hadamard's maximal determinant problem. There is a vast literature on it. The Wikipedia essay will get you started. EDIT: Will Orrick has a useful website on the problem. 1 The notation A^H means the hermitian (or conjugate) transpose of A. You want to show that, for any vector v\in N(A) and any vector w\in C(A^H), the (standard) inner product$$ v^Hw=0 $$The definition of C(A^H) says that w=A^Hu for some u; then$$ v^Hw=v^HA^Hu=(Av)^Hu=0 $$because, by assumption, v\in N(A). 0 If it is the matrix algebra \mathfrak{sl}_n over a field then yes, because \mathfrak{sl}_n is a simple Lie algebra and [\mathfrak{sl}_n,\mathfrak{sl}_n] is a nontrivial ideal of \mathfrak{sl}_n thus equal to it. No, look at the matrix with diagonal (1, -1). See 2. See 2. 1$$\det(A-\lambda I) = (1-\lambda)(-2-\lambda)(1-\lambda) + 3(2+\lambda)3=\\ =(1-2\lambda + \lambda^2)(-2-\lambda) + 9(2+\lambda) = \\ =(-2-\lambda + 4\lambda + 2\lambda^2 - 2\lambda^2 - \lambda^3) + 18 + 9\lambda = \\ = -\lambda^3 + 3\lambda - 2 + 18 + 9\lambda = -\lambda^3 + 12\lambda + 16,$$not what you got... 1 you should treat it as a normal polynominal when you are trying to find the roots (which are eigenvalues).$$\det(A-\lambda \cdot I) = (\lambda-4)(\lambda+2)^2$$3 Hint Consider the canonical matrices E^{ij},E^{ii}-E^{jj} , i\neq j. The first group will help you prove A is zero off the diagonal, the second will help you prove that A must be a diagonal matrix of the form {\rm diag}(a,\ldots,a), i.e. a\cdot 1 for some scalar a. 0 As an example, let's pick your matrix:$$A=\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$$To find the eigenvectors for \lambda=1, we need to solve A-\lambda I=0, which is:$$\begin{bmatrix}0&0&0\\0&-1&1\\0&1&-1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$$... 4 Let A = \begin{bmatrix}0&6\pi&0\\-6\pi&0&0\\0&0&0\end{bmatrix} and B = \begin{bmatrix}0&0&0\\0&0&8\pi\\0&-8\pi&0\end{bmatrix}. Using the formula I derived here, e^A = e^B = e^{A+B} = I. Hence, e^{A+B} = e^Ae^B. However, AB-BA = ... 0 The definition: Av=\lambda v this means that (I is the identity matrix)$$Av-\lambda v =0(A-\lambda I)v=0$$So solve the system of equations$$(A-\lambda I)v = 0$$and find the solutions - these are the eigenvectors, v. Show that \lambda=5 is an eigenvalue of A=\begin{bmatrix} 1 & 2 \\4&3 \end{bmatrix} and determine the eigenspace of ... 2 The Woodbury Matrix Identity states that:$$(W+XYZ)^{-1} = W^{-1}-W^{-1}X(Y^{-1}+ZW^{-1}X)^{-1}ZW^{-1}$$Setting W = A, Y = D, and X = Z = I, we get:$$(A+D)^{-1} = A^{-1}-A^{-1}(D^{-1}+A^{-1})^{-1}A^{-1}$$EDIT: If the entries of D are very small, we get the approximation:$$(A+D)^{-1} \approx A^{-1}-A^{-1}DA^{-1}$$1 Suppose you know that \lambda is an eigenvalue of the matrix A. By definition, the corresponding eigenspace is U_\lambda=\{v\in V|Av=\lambda v\}. Writing v=(v_1,\ldots,v_n), the equation Av=\lambda v becomes a system of n linear equations in n variables. The fact that \lambda is an eigenvalue of A implies that the space of solutions of this ... 5 If A, B are square matrices with same inverse C, then AC=CA=I and BC=CB=I. Therefore,$$ A =AI= A(CB)= (AC)B = IB = B. $$The odd thing about matrices is this: If A, B are n\times n matrices over a field, then AB=I iff BA=I. This is a direct consequence of the fact that the N\times N matrices form a finite-dimensional linear ... 2 Specific counterexample to the non-square case: Let A=\left(\begin{smallmatrix}1&0&0\\0&1&0\end{smallmatrix}\right). Then there is a matrix B such that AB=I, namely B=\left(\begin{smallmatrix}1&0\\0&1\\x&y\end{smallmatrix}\right). Note that x,y can each be anything, so this right inverse is not unique. Also note ... 16 More generally, in any situation where the associative law holds, if some x has both a left-inverse l and a right inverse r, then l=r. The reason is that l=l(xr)=(lx)r=r. In particular, if x has a 2-sided inverse, then that's unique. On the other hand, it is entirely possible for some x to have many different left-inverses if it has no ... 3 Yes, it is unique. To show this, assume a matrix A has two inverses B and C, so that AB=I and AC=I. Therefore AB=AC \implies BAB=BAC \implies B=C. So the inverse is indeed unique. For the second question, note that (A^{-1})^{-1}=A so that if A and B both have inverse A^{-1}, then A^{-1} has a unique inverse as well. Since A and B ... 3 Note that GL(n, \mathbb{F}), the set of invertible n\times n matrices over the field \mathbb{F}, is a group. In any group, inverses are unique, so if a^{-1} = b^{-1}, by taking inverses it follows that a = b. In particular, this applies to the group GL(n, \mathbb{F}). 0 Some thoughts on your question : denote by C_1,C_2,C_3,C_4 the columns of A. Let V be the all-ones column (1,1,1,1), and let U be the matrix of all whose columns are equal to V (so that Z=sU). Then \det (A + sU) =$$\begin{array}{lcl} &=& [C_1+sV,C_2+sV,C_3+sV,c_4+sV] \\ &=& [C_1,C_2,C_3,C_4] ...

0

Assuming that the $a_i$ are real then $E=iC$ is Hermitean so $E=VF{V ^{-1}}$ with $V$ unitary and $F$ diagonal with real entries (the eigenvalues of $E$). Generically $V$ and $F$ are unique (up to a phase factor in $V$). Now $C=V(-iF)V^{-1}$ so we can take $U=V$ and $D=-iF$. But then $U{e ^D}U ^{-1}$ is not a real matrix. Why do you think it is? And ...

4

Let $x = \sqrt{a_1^2+a_2^2+a_3^2}$. You can verify that $C^3 = -(a_1^2+a_2^2+a_3^2)C = -x^2C$. Hence, $C^{2m+1} = (-1)^mx^{2m}C$ and $C^{2m} = (-1)^{m-1}x^{2m-2}C^2$. Therefore, $e^C = \displaystyle\sum_{n=0}^{\infty}\dfrac{1}{n!}C^n = I + \sum_{m=0}^{\infty}\dfrac{1}{(2m+1)!}C^{2m+1} + \sum_{m=1}^{\infty}\dfrac{1}{(2m)!}C^{2m}$ $= \displaystyle I + ... 1 Any multiple of an eigenvector is still an eigenvector for the same eigenvalue, even if this multiple is negative. So if$(2,5)^T$is an eigenvector then so are$(-2,-5)^T$,$(10,25)^T$,$(1,5/2)^T$,$(-6,-15)^T$, ... If the machine marking the answers is clever enough (for example it's using MapleTA) it should accept any of these. As to why it has chosen ... 1 Remember that an eigenvector defines a subspace of the domain of the linear transformation. That subspace has many bases --- in fact, each nonzero multiple of the eigenvector is also a basis for the eigenspace. So both$[2,5]$and$[-2,-5]$are bases, and either can be thought of as "representing" the one-dimensional eigenspace. The choice of$[-2,-5]$is ... 3 You can use a continuity argument. The function $$A\longmapsto \det(A+Z)-\det(A)-Sm$$ is a polynomial function from the vector space$M_n(\mathbb C)$of all$n\times n$complex matrices to the complex numbers, so in particular it is continuous. You claim that it is zero on the set of matrices with non-zero determinant. Since this set is dense in$M_n(\mathbb ...

0

What you are asking for, is essentially a standard exercise in numerical analysis, which is called Cholesky decomposition, see http://en.wikipedia.org/wiki/Cholesky_decomposition. But you should have a positive definite symmetric matrix. If not, you should try a $LDL^T$ decomposition, often also called Cholesky decomposition.

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If you are over the field $\mathbb{C}$ and your matrix is self-adjoint (which is already imlplied by symmetric), using the spectral theorem you can diagonalize your matrix, i.e. $D:=UAU^{\dagger}$ is diagonal and $U$ a unitary matrix. To calculate $B$ now, simply take the square root of all diagonal entries in the diagonal form of $A$ and simply transform ...

0

Are you sure it's not span{(6,-1,0,-1),(12,-3,-2,0)} ? Well anyway, homogeneous solution means taking that matrix, multiplying it by an appropriate column vector and setting it equal to the zero vector of appropriate length. Just take s=-1 and t = -2 below to get span{(6,-1,0,-1),(12,-3,-2,0)}. If you want, you can say span{(-6,1,0,1),(-6,1.5,1,0)}. ...

2

The elements of the open Bruhat cell are precisely the matrices where the determinant of the $k\times k$ minor of the bottom left hand corner of the matrix is non-zero for all $k\leq n$. You noted one of these conditions for $k=1$. This condition is true for $\omega_nA$, and won't be changed by multiplication by upper triangular matrices. You can check ...

0

If you just want to define a heading direction based on absolute pitch and yaw angles, then the code they gave you is correct, but they swapped Y and Z from your definition. This is what you'd want: double sRadians = Math.toRadians( yaw ); double tRadians = Math.toRadians( pitch ); double x = radius * Math.cos( sRadians ) * Math.sin( tRadians ); ...

2

Your answer is correct, but not in the form they expected. It looks like you have $33\sqrt{1/11}$, but \begin{align*}&=3*11*\sqrt{1/11} \\ &= 3 \sqrt{11^2/11} \\ &= 3 \sqrt{11} \end{align*}

1

Thanks to WimC's comment on my question, I have gotten my code working. While all implementations of the Hungarian Method I have seen only assign zeroes if they are unique in their row and column, assigning random zeroes from those remaining seems to help with the line-drawing step. Doing so before drawing the lines has allowed my program to find an answer ...

1

If the geometric multiplicity of $0$ is two which means that the dimension of the eigenspace of $0$ is $2$ then there's two linearly independent eigenvectors associated to $0$ and then the given matrix would be similar to $$\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}$$ and this is a contradiction. Think to the rank of the two ...

2

This is a a more complete write up using the ideas of @QuangHoang. It can also be viewed as an analytic (instead of algebraic) version of the ideas of @AndreasCaranti As before, define $f(t)=(A+tB)^n$. This is a matrix valued function, but by examining the individual entries we have $n^2$ real valued functions. We have that $f(t)_{ij}$ is a polynomial of ...

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