# Tag Info

6

Let $I = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$. Since $A = -I+B$ and the matrices $I$ and $B$ commute, we have $e^{tA} = e^{t(-I+B)} = e^{-tI}e^{tB}$. Trivially, $e^{-tI} = \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{-t}\end{bmatrix}$. Also, since $B^2 = 0$, we have $e^{tB} = ... 4 Let$T(v_i) = \sum_j a_{ji} \cdot v_j$, so that$(a_{ji})$is the matrix of$T$w.r.t. to$(v_1,\dotsc,v_n)$. The wedges$v_{i_1} \wedge \dotsc \wedge v_{i_k}$with$i_1<\dotsc<i_k$form a basis of$\Lambda^k(V)$. We have: $$\Lambda^k(T)(v_{i_1} \wedge \dotsc \wedge v_{i_k}) = T(v_{i_1}) \wedge \dotsc \wedge T(v_{i_k})=\sum_{j_1,\dotsc,j_k} ... 3 Let p be the total amount of hours necessary for the project. Let j and v be the amount of constant hours per day that John and Vince can contribute to the project respectively. Furthermore, we have also to assume, that both persons are additively productive, which might not be the case in the real world :). Both sentences translate to the equations ... 3 Let me do this with real methods only. Let \displaystyle P(X)=\sum_{i=1}^k \frac{1}{\alpha_i}\prod_{j\neq i}^k \frac{\alpha_j-X}{\alpha_j-\alpha_i} be the Lagrange interpolating polynomial for x\to\frac{1}{x} at points \alpha_i. Then \displaystyle P(0)=\sum_{i=1}^k(\frac{1}{\alpha_i}\prod_{j\neq i}^k\frac{\alpha_j}{\alpha_j-\alpha_i}) Let us ... 2 Vince needs x days for the task, John x+3. This means that Vince completes {1\over x} projects per day, and John {1\over x+3}. Together they complete {1\over2} of a project per day, which leads to the equation$${1\over x}+{1\over x+3}={1\over2}\ .$$2 The matrix$$ A(x):=\sum_{n=0}^\infty A_n x^n $$is invertible at x=1. The set of invertible matrices is an open set in the space of all matrices. Thus, an idea to solve the problem is to show that x\mapsto A(x) is continuous. In order to prove that you need additional assumptions. One would be to assume that the power series$$ \sum_{n=0}^\infty ... 2 Note: The question changed dramatically after this answer was written. Since$\sum_n A_n$exists, we see that$A_n \to 0$and hence the radius of convergence of$\|A_n\|$is at least one, so the function$f(x) = \sum_n A_i x^n$is real analytic on$(-1,1)$. Furthermore, Abel's theorem (applied to each component) shows that$\lim_{x \uparrow 1} f(x) = ...

2

Here's how I solve this problem: Notice I am writing vectors in columnar form; thus, the OP's $(a, b)$ is my $\begin{pmatrix} a \\ b \end{pmatrix} \tag{0}$ etc.; these things being said , we have: The line $y = mx$ is in fact the set of points (vectors) $L = \{(x, mx)^T = x(1, m)^T \mid x \in \Bbb R \}$; this is the one-dimensional subspace of $\Bbb R^2$ ...

1

See the general formula for functions of Jordan blocks, which this is. You'd have to figure out how matrix exponents were defined in your course. There is no general formula for triangular matrices. The hard way is to notice that $A=-I+J$, where $J^2=0$, and then to write $e^{tA}$ as a power series, powers of $-I+J$ can be computed explicitly by induction. ...

1

Logarithmic spiral Approximate solution The logarithmic spiral is self-similar. As a consequence of that, the tangents make the same angle with the position vector in each point of the spiral. You can use that to draw a reasonable approximation of the spiral simply by taking unit steps in the direction of the current tangent. Perhaps it's easiest to look ...

1

Let $X=L^2[0,1]$ and define $T:X \to X$ by $(Tf)(x) = x f(x)$. Then $T$ is bounded and $T^* =T$. Let $c\neq 0$ be a constant, and let $\gamma(x) = c$ for all $x$. Then we see that $\|Tf-\gamma\|>0$ for all $f \in X$ (since $x \mapsto { c \over x}$ is not in $X$). Hence $\gamma$ is not in the range of $T$. If we choose $g \in X$, we can define $f_n(x) = ... 1 I thought it might be worthwhile to add an explicit example. Consider$T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$with $$[T] = \left[ \begin{array}{ccc} a & d & g \\ b & e & h \\ c & f & i \end{array}\right]$$ In particular, using the usual$e_1 = [ 1,0,0 ]^T=(1,0,0)$,$e_2 = (0,1,0)$,$e_3=(0,0,1)$we have $$T(e_1) = (a,b,c), \ \ \ ... 1 Compare my answer to this question. The equality is in fact true for all distinct non-zero complex numbers \alpha_1,\ldots,\alpha_k, and can be proved as follows. Since I will be using complex numbers I don't want to use i as a variable, so I am slightly changing your notation. Write$$f(z)=\frac{1}{z^2}\prod_{j=1}^k \frac{\alpha_j}{\alpha_j-z}$\$ and ...

Only top voted, non community-wiki answers of a minimum length are eligible