New answers tagged

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An easy way? You need to check this and only this condition. First of all, the transformation law must be linear (as @Michael Joyce just remembered). So $C$ is not an isomorphism. Secondly, domain and codomain must have the same dimension. So how D could be an isomorphism? $\mathrm{dim}(\mathbb{R^3}) = 3$ while $\mathrm{dim}(P^3) = 4$. For $A$ and $B$ ...


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I don't think there is relevance between the number of solutions of the linear system $Ax = b$ and that of the system of ODE $x' = Ax$. For the system of ODE $x' = Ax$, there must be $n$ linearly independent solutions arouse from the eigenvalues of $A$, irrespective of the rank of $A$. For completeness, I would like to add that the only relation between ...


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Quadratic form $$ P(x) =\left( \begin{array}[t]{cc} x & 1 \\[0.5 ex] \end{array} \right) \left( \begin{array}{cc} a_{2} & \frac{a_{1}}{2} \\ \frac{a_{1}}{2} & a_{0} \\ \end{array} \right) \left( \begin{array}{c} x \\ 1 \\ \end{array} \right) $$


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$T(S(u))\ne S(u)$ in general. $T$, being non singular, is an isomorphism of linear space, so it preserves the dimension of subspaces it transforms. So the $\mathrm{Rank}(S)$ is the dimension of the image of $S$. Such an image is transformed by $T$ in a subspace of the same dimension. Such transformed subspace is the image of $TS$, whose dimension is the ...


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Partial answer.If $\rho (I-\omega A)<1$ then $\omega \ne 0.$ And $\rho (I-\omega A)<1$ implies that $(I-\omega A-x I)$ is invertible whenever $|x|\geq 1,$ which implies that $I(1-x)/\omega -A $ is invertible whenever $|x|\geq 1,$ whence $|(1-x)/|\omega|>\rho (A).$ Putting $x=-1$ we have $2/|\omega|>\rho.$


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You're absolutely correct (as per linked question) that, when reducing a matrix, you can pivot on any suitable column at each step and still get a reduced matrix with the same kinds of useful properties (e.g. column vectors corresponding to pivoted columns form the basis of a column space, proof of consistency/inconsistency etc). But pivoting on these other ...


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You could apply a change of basis transformation:$$\pmatrix{1&1&1\\0&1&1\\0&0&1}^{-1}\pmatrix{a_0\\a_1\\a_2}=\pmatrix{a_0-a_1\\a_1-a_2\\a_2}.$$ The matrix in this case is pretty easy to invert by inspection.


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(a) is a subspace since the span of any subset of a vectors space is a subspace (b) is not a subspace of $\mathbb{R}^4$ since it consists of vectors from $\mathbb{R}^5$. (c) is a subspace since it's the null space of a matrix (d) is not a subspace since the zero vector is not an element (e) is not a subspace since the zero vector is not an element


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[Please don't up vote: just summarising comments so question not in unanswered state forever] A linearly independent set means no vector in the set can be expressed as a linear combination of the others. A linearly dependent set means at least one of the vectors can be expressed as a linear combination of the others. Please forget the faulty definition ...


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Given $p(x) = a_0 + a_1 x + a_2 x^2$, need to find $\alpha_1, \alpha_2, \alpha_3$, such that $$p(x) = \alpha_1 * (1) + \alpha_2 (1 + x) + \alpha_3 (1 + x +x^2) = a_0 + a_1 x + a_2 x^2$$ Expand LHS, $$(\alpha_1 + \alpha_2 + \alpha_3) + (\alpha+2 + \alpha_3) x + \alpha_3 x^2 = a_0 + a_1 x + a_2 x^2 $$ We have $$\alpha_3 = a_2$$, $$\alpha_2 + \alpha_3 = ...


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The RHS can be written as a general quadratic by simplifying,$$p(x) =a_0+a_1x+a_2x^2$$ Since we already know that $\{1,1+x,1+x+x^2\}$ for a basis we know that we can write any polynomial of degree $2$ in terms of the elements of the above set.


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Suppose we could diagonalize the matrix $A$. It isn't necessarily true that we can, but if we could we would have all of the eigenvalues along the diagonal. If we subtract the identity matrix, what happens to each eigenvalue? Now in this case, it isn't necessarily true that we have a diagonal representation, so consider an eigenvector $v$ corresponding to ...


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Your statement "a set of functions are only linearly dependent if you can show that one function is simply a scaled version of another in the same set" is not quite right. A set of functions $\{f_1, \dotsc, f_n\}$ is linearly dependent if you can write one as a linear combination of the others, i.e. if there are numbers $a_2, \dotsc, a_n$ such that $$ f_1 = ...


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Perhaps you're looking for something like this: The algebraic and geometric multiplicities of $\lambda$ are different iff there is a vector $v$ such that $(A - \lambda I) v \ne 0 $ but $(A - \lambda I)^2 v = 0$. You could write this condition (in the case $\lambda \ne 0$) as: there is a vector $v$ such that $ A^2 v = 2 \lambda A v - \lambda^2 v \ne ...


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$B$ could be $A^T$ or $A$ with either one row or column switched.


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Assume for the sake of contradiction that one of them is generatable by a linear combination of the other two. Without loss of generality, say $a = l_1 b + l_2 c$ for some $l_1,l_2 \in \mathbb{R}.$ We see that $a$ is of odd magnitude, so $l_1 b + l_2 c$ should be of odd magnitude. We also know that $a - c = l_1 b + (l_2 - 1) c$ is of odd magnitude. Can it be ...


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You've sort of answered your own question. Suppose $A$ has an eigenvalue $\lambda$ with algebraic multiplicity $m_a$. If $$ \DeclareMathOperator{null}{null} \dim\null(A-\lambda I)\neq m_a $$ then $A$ is not diagonalizable. So, to check if $A$ is not diagonalizable, we need only find such an eigenvalue.


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Yes, you're right. Since the range of the linear map is $\mathbb{R}$, because, for instance, the vector $(1,0,0,0)$ maps to $2\ne0$, the rank-nullity theorem tells you that the kernel has dimension $4-1=3$. You find a basis by solving with respect to one of the unknowns and give the other values that ensure linear independence. So, since we can do $$ ...


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In the second component you have the term $(\frac23x_1+x_2)e^t$, and so you can take $x_2=-\frac23 x_1$ (the numbers $x_2$ and $x_3$ need not be zero). The third line gives $x_3=0$. So, you get all scalar multiples of the basis for the stable space (as it should).


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Here's a partial answer: Suppose $n$ is a solution. Then $n(n+1)$ is a solution if $n+1$ is prime, since $n+1\mid n(n+1)$ and $n\mid n(n+1)$, and the condition is clearly satisfied for all primes less than $n$. So in addition to $2$, $3$, $6$, and $42$, we also get \begin{align*} 42(43) = 1806 = 2\cdot 3\cdot 7\cdot 43 . \end{align*} This fails at the next ...


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Let $$n=p_1p_2\cdots p_k$$ where all $p_i$ are distinct and $p_i<p_j$ iff $i<j$ (this number is square-free, since $p|n$ implied $p^2\not|n$ for a prime $p$). Now, since $p_i-1|n$, we know that either $k=1$ and $n=2$, or $k>1$ and $2|n$, so $$n=2p_2p_3\cdots p_k$$ now $p_2-1|n$ and $p_2-1$ can only have a factor $2$, since $p_2$ is the smallest ...


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Hint: the subspace is the set of orthogonal vectors to $(2,-1,2,4)$ and so one should find a basis for the orthogonal complement of the subspace spanned by that vector. Added: and yes, this orthogonal complement is the kernel of the linear map $T\colon\mathbb R^4\to\mathbb R$ defined by $$T(x_1,x_2,x_3,x_4)=2x_1-x_2+2x_3+4x_4.$$ Since the range of $T$ has ...


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It needs to be complete, in the metric generated by the inner product. This answer assumes your working in an infinite dimensional Inner product space. The problem is that an infinite linear combination may no longer be in the space.


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Solution 2 (without assuming $D$ to be linear) This is continuation of your solution: you got to the point where $\forall y\in Y\,\exists x\in X: Tx=y$ and therefore $T$ is an open mapping. This means that $T$ maps the open unit ball $B_X(0,1)\subset X$ in an open set in $Y$ which contains $0_Y\Rightarrow\exists r>0:\,\overline{B_Y(0,r)}\subset ...


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The formula is as follows $\det(A) = \prod_{i=1}^N \lambda_i$, where $\lambda_i$ is the $i^{th}$ eigenvalue of the matrix. The formula follows from the definition of the matrix characteristic equation $\det(A-\lambda I) = 0$.


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$\mathbb{R}^2$ Fix $z$, let $x$ run around the unit circle, and forget $y$ for now. The locus of projections of $z$ on $x$ is the circle with diameter 1 containing the origin and $z$. The same locus for $y$. Since the diameter is $||z||=1$, any two points are no farther away from each other than 1. QED. $\mathbb{R}^n$ The projection $z_{xy}$ of $z$ ...


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In fact, this is a classical issue: these matrices are called (symmetric) "Pascal matrices", and have many properties. The fact that their determinant is an easy consequence from their Cholesky factorization (because these matrices are positive definite ). see web.mit.edu/18.06/www/Essays/pascal-work.pdf http://mathworld.wolfram.com/PascalMatrix.html ...


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$||A||$ in a sense of $\det(\det( A))$ makes absolutely no sense. Even if you define $\det(A) := (\det (A))$, i.e. as a $1\times 1$-matrix, $\det (\det A)$ would be just $\det (A)$... The determinant is a scalar.


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Since $\det (A)=4(4-i)\neq 0$, all three row vectors are linearly independent over $\mathbb{C}$, hence also over $\mathbb{R}$ by the definition of linear independence. Edit: The new matrix $A$ now is indeed singular and we have $\dim_{\mathbb{C}} (S)=2$. Now solve $\lambda_1v_1+\lambda_1v_1+\lambda_1v_1=0$ for real $\lambda$, to see that $\dim_{\mathbb{R}} ...


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Usually, in linear algebra (and functional analysis, and many other fields) double vertical lines $\|\cdot\|$ denote a norm. If $A:V\to W$ is a linear operator between normed vector spaces (or a matrix in the finite-dimensional case with a fixed choice of basis), then often the following norm is used $$\|A\| = \sup_{x\in V}\frac{\|Av\|_W}{\|v\|_V},$$ where ...


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Take a basis $\{u_i\}$ of $U$, and a basis $\{v_i\}$ of $V$ (not necessarily of the same size of course). Then note you have the canonical basis of $U \times V$ consisting of $(u_i,0)$ and $(0,v_j)$. If you assume $f$ is linear, then you can show that $\Gamma_f$ is the linear span of the vectors $(u_i, f(u_i)) \in U \times V$. Conversely, assume that ...


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Are you sure? Define the matrix $A$ as $$A=\begin{pmatrix}1&-2&1\\-2&4&-2\\1&-2&1\end{pmatrix}.$$ Clearly, the matrix $A$ is symmetric and the sum of the coefficients of each of its rows is nil. Moreover, the eigenvalues of $A$ are $0$ (double eigenvalue), and $6$, hence $A$ is positive semi-definite. Yet, $a_{13}=1>0$.


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The right space is the second! For completeness, let $\mathbb{V}$ be a complex vector space of dimension $n$; I recall that a flag of $\mathbb{V}$ is a strictly increasing sequence of vector subspaces of $\mathbb{V}$ $$ \{\underline{0}\}=\mathbb{V}_0<\mathbb{V}_1<\dots<\mathbb{V}_r\leq\mathbb{V}\,\text{where:}\,\forall ...


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Procede with the usual row reduction, trying to avoid pivoting on the element with $a$ as long as possible: $$\left(\begin{array}{ccc|c} 3 &-6 &6 &15\\ -2 &7 &a &-25\\ 2 &-6 &6 & 20 \end{array}\right) \xrightarrow{\begin{matrix}R1~/~3\\R3~/~2\end{matrix}} \left(\begin{array}{ccc|c} 1 &-2 &2 ...


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First, look at this problem: given two values $a, b$ with $a \ge b$, what are the maximum and minimum values of $z = ax + by$, for various $x, y \ge 0$, with $x + y = 1$? To solve it, let $y = 1 - x$, then $z = ax +b(1 -x) = b + (a - b)x$. Now $x \ge 0$ and $1 - x \ge 0$, so $x \le 1$. The maximum occurs at the maximum value of $x$, which is $1$. The ...


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We are assumed to know that the inverse of a matrix $A$ exists iff $\det A\ne 0$ (assumed knowledge which I infer from StillLearning's proposed solution). We look at the simultaneous equations $ax+by=e$, $cx+dy=f$, and want to know when they have a solution $x, y$ for every $e$ and $f$. We show that this happens iff the matrix ...


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I will write $f_x$ instead of $\frac{\partial f}{\partial x}$ because this requires fewer keystrokes. By the chain rule you have \begin{equation} g_r(r,\theta) = f_x(x,y) \cos(\theta) + f_y(x,y) \sin(\theta) \end{equation} and \begin{equation} g_\theta(r, \theta) = f_x(x,y) r (-\sin(\theta)) + f_y(x,y) r \cos(\theta) \end{equation} In both equations $x = ...


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Bilinear is nonlinear. It's linear in both main variables, but not in any superposition. Naively speaking, it's linear if you cut along $x$ or $y$ axis, but you're not allowed to rotate the frame (which is what a proper linear function allows, even requires, as linearity is independent of choice of coordinates). A typical example is bilinear interpolation ...


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A (non-trivial) linear function from $\mathbb{R}^n$ for $n > 1$ will never be a multilinear function (bilinear when $n=2$) if it is a linear function. Instead, a multilinear function is a linear function from a tensor product. So a bilinear function on $\mathbb{R}$ is a linear function $\mathbb{R}\otimes\mathbb{R} \to X$, but $\mathbb{R}$ is the unit ...


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a function cannot be linear and bilinear at the same time. Bilinearity is a kind of non linearity.


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My interpretation is that diagonalization, when possible, simplifies the interpretation (and subsequent computations) of a matrix. In the eigenbasis, operations can be performed "coordinate-wise", or basis vector by basis vector, almost independently. Suppose you have a vector $x^a = (x_1,x_2,\dots,x_n)$, and you already have computed $y^a = Ax^a$. If you ...


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A linear operator is not hermitian "with respect to a basis". When there is a basis of eigenvectors as you mention, $Av_k=\lambda_kv_k$, then $A$ is already diagonal. The point of diagonalization is to find such basis.


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Adding to achille hui's answer, there is a variety of nice formulations. Let the vertex coordinates be $v_i = (x_i,y_i,z_i)$ for $i\in\{0,\ldots,7\}$ such that $v_i$ and $v_j$ are connected by an edge if and only if the binary representations of $i$ and $j$ differ in exactly one bit. Furthermore, let us gather all control point coordinates in the matrix ...


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You know by linear algebra that, if $\mathbb K$ is a field, a matrix $A\in \mathcal{M}_{n\times n}(\Bbb K)$ is invertible if and only if its columns are linearly independent. Now: if $\#\mathbb K=q$, how many $n$-tuples $(v_1,\cdots,v_n)$ of linearly independent vectors are there? $v_1$ can be anything except $0$. So it can be chosen in $q^n-1$ possible ...


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You can check https://en.wikipedia.org/wiki/General_linear_group where there is a formula and a little explanation for the number of matrices in GL(n,q), which is $$ (q^n-1)(q^n-q)(q^n-q^2)\cdots(q^n-q^{n-1}) $$


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We assume $D$ to be a subspace. If $T$ is injective, then let $y_n\to y\in Y$ with $\{y_n\}\subset D$. Then $\exists \{x_n\}\subset X: Tx_n=y_n$ and $\|x_n\|\leq C\|y_n\|\Rightarrow \|x_n-x_m\|\leq C\|y_n-y_m\|\,(\text{because $T$ is injective})\,\Rightarrow$ the sequence $\{x_n\}$ is Cauchy in $X$ and therefore $\exists x\in X: x_n\to x$. But then ...


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I think you need the assumption that $T$ is surjective. For otherwise, let $D$ be some dense subspace of $Y$, and let $T$ be the inclusion mapping $D \rightarrow Y$, then your question would not make any sense. If $T$ is surjective, $T$ is an open map. For any $y \in Y$, there is a sequence $\{y_n\}$in $D$ such that $y_n \rightarrow y$. Let $x$ be such that ...


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since U not equal to W, the basis for W must have at least one vector linearly independent from U, so $\dim(\operatorname{span}(U + W))$ is at least 4. but they are subspaces of $\mathbb R^4$, so $\dim (U \cup W)=4$. Use the fact that $\dim U +\dim W=\dim(U + W)+ \dim(U\cap W)$, i.e. $6=4+\dim(U\cap W)$


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Clearly we can assume that $y$ spans $Y$ (otherwise replace $Y$ with $Z = \operatorname{span} \{ y_i : 0 \leqslant i < n\}$). Then some subsequence of $y$ is a basis of $Y$. Without loss of generality, suppose that $\{ y_i : 0 \leqslant i < m\}$ is a basis of $Y$. Let $\{\psi_i : 0 \leqslant i < m\}$ be the dual basis of $\{ y_i : 0 \leqslant i < ...



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