New answers tagged

0

Since you have multiple groups you're trying to match against, you need to optimize $$ min((x - a)^2, (x - b)^2) $$ for the values of a and b.


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Let us assume that $Av = 0$ and $A y = b$, then the vector $u = \lambda v + y$, where $\lambda$ can be any scalar from the field beyond the vector space, due to the linearity of $A$ fulfills $$ A u = A(\lambda v + y) = \lambda (Av) + (Ay) = \lambda 0 + b = b $$ Those $u$ are all solutions, if the solution space of the homogeneous system $A v = 0$ is not more ...


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Apply Cayley Hamilton's theorem. You have that the characteristic polynomial is : $ P(λ) = λ(λ-1)(λ-2) $ Then, $ λ^{10} = P(λ)π(λ) + υ(λ)$ which is the euclidean diaereses. $υ(λ)$ will be a second-degree polynomial, written as : $ α_2λ^2 + α_1λ + α_ο$ . Putting the eigenvalues into $λ^{10} = P(λ)π(λ) + υ(λ)$, will make $P(eigenvalue)π(eigenvalue) = 0$ ...


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Notice the following: $B = P^{-1}AP \iff A = PBP^{-1}$. Now consider $A^2$. We see that $A^2 = (PBP^{-1})(PBP^{-1})$. But since $PP^{-1} = I$, this simplifies to $A^2 = PB^2P^{-1}$. Inductively it should be clear that $A^n = PB^nP^{-1}$.


1

Yes that is true. The coordinates are just the coefficients of the base vectors in the linear combination $2e_1+70e_2$ that results in your given vector $A=2e_1+70e_2$.


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The line is independent of the length of the direction vector, so you can assume without loss of generality that $||(a,b)^t|| = 1$ Because if $||(a,b)^t|| \neq 1$, consider the vector $(a',b')^t := \frac{1}{||(a,b)^t||}(a,b)^t $ that obviously still points in the same direction. Then set $a=\cos(\alpha/2)$ and $b=\sin(\alpha/2)$ and solve for $\alpha$.


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Hint : Draw the cartesian coordinate system ($x'x$ and $y'y$). Then draw a random line and a random vector. Then define geometrically or through analytic geometry the reflection of your given vector and form the matrix of the operator that reflects the vector over a random line.


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The vectorial space of matrices $n\times m$ has a canonical basis given by all matrices $n\times m$ of the type $$(e_{ij}) = 1 \;\mbox{at (ij) and 0 the rest.} $$ Any nxm matrix $M$ can then be written as a linear combination of these basis matrices $$M = \sum_{ij}\,M_{ij}\,e_{ij}.$$ On the other hand it is $e_{ij}={\mathbf e}_i\otimes {\mathbf e}_j \equiv ...


1

for your first question, I would advise the different books named "Linear Algebra" with different subtitles of Gilbert Strang (MIT Press). for your second question; yes, there is such a theorem that you will find in A property of positive definite matrices


1

The parametrization that you found requires $u\ge0$, $v\ge0$, and $u+v\le1$. To handle the case $u+v\ge1$, you can apply the same parameterization but replacing coefficient $u$ with $1-u$ and coefficient $v$ with $ 1-v$ (since in that case $1-u+1-v\le1$, while $1-u$ and $1-v$ continue to be nonnegative). Then the coefficient $1-u-v$ becomes $u+v-1$. This ...


1

"Basis expansions" for matrices - that is, writing a matrix as a linear combination of other matrices like $A = \sum c_jB_j$ is typically less useful than looking for a way to write a matrix as a product of matrices like $A = BC$. The latter is done all the time in numerical linear algebra - see for instance the classic book Matrix Computations. The reason ...


1

The map $(u,v)\mapsto A+u(B-A)+uv(C-B)$ should work, but it is not bijective (all points of the form $(0,v)$ get mapped to the point $A$.


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Hint: The matrix $$ \pmatrix{\alpha & 1\\1&1 } $$ will have linearly independent eigenvectors $u$ and $v$. $f$ will have the eigen"vectors" $uu^T,uv^T,vu^T,vv^T$.


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Let us demostrate something else: $$\frac{ \|u\|_1}{\sqrt{2}} \leq \|u\|_2 \leq \|u\|_1$$ for every $u\in\mathbb{R}^2$. Note that $\|u\|_1=|u_1|+|u_2|\ge\sqrt{|u_1|^2+|u_2|^2}=\|u\|_2$ (for the inequality, is enough with power both sides). On the other hand, it is easy to show that for all $x,y\in\mathbb{R}$ we have $xy\le\frac{x^2+y^2}{2}$, which ...


1

It suffices to show that for $a = (a_1, a_2) \in \mathbb{R}^2$, $b = (b_1, b_2) \in \mathbb{R}^2$, it holds that $$\frac{1}{\sqrt{2}}\|a - b\|_1 \leq \|a - b\|_2 \leq \|a - b\|_1.$$ This is quite straightforward to check by definition: \begin{align*} & \|a - b\|_2 \\ = & \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2} \\ \leq & \sqrt{(a_1 - b_1)^2 + (a_2 - ...


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$\mathcal{B} = \{B_1, B_2, B_3,B_4 \}$ is a basis for $M_{2\times 2}(\mathbb{F})$ (where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$). Where $B_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, B_2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, B_3 = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, B_4 = \begin{pmatrix} 0 ...


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Indeed, the given matrix is irreducible. Your claim is correct that if a matrix is reducible, then it can be written in the form (after some reordering of the states): $$A' = \begin{pmatrix} B & C\\ \color{blue}{\mathbf{0}} & D \end{pmatrix}.$$ But notice that $B$ and $D$ must be square blocks! Apparently, you think that the blue block represents ...


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Notation : We note $K_i=\ker (T^i)$. As you said we have $\{0\}=K_0 \subset K_1 \subset K_2 \subset \dots \subset K_n=\mathbb{C}^n$. But we have a stronger result. Proposition : $\forall 1 \le i \le n-1, \dim(K_i) - \dim(K_{i-1}) \geq \dim(K_{i+1}) - \dim(K_i)$. Proof : Consider the morphisms : $\begin{array}{ccccc} & K_{i+1} & ...


2

Since $x\not=y$ and $x+y=0$, there are only two possible options for the intercepts. Let's assume that $x=a$ is positive and $y=-a$ is negative. The two intercepts will be $(a,0)$ and $(0,-a)$, and the slope of a line between those points is $\frac{a-0}{0-(-a)}=\frac{a}{a}=1$ and therefore has positive slope. Likewise, assume that $x=-a$ is negative and ...


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The fact that $x+y=0$ tells you that $x$ and $y$ have opposite signs. This tells you that the slope cannot be negative. Since $x \neq y$ the line can't go through the origin, so no monkey business there. The slope can't be zero, or infinite, because then one of $x$ or $y$ would be undefined. So you can say that the slope is defined, and strictly ...


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You really should specify what you mean by "canonical form". I will assume that you meant rational canonical form since you did not specify that you were dealing with a complex vector space. In this particular problem, because of the nilpotency, these are also the Jordan forms, but that is just luck, the distinction of which canonical form you are talking ...


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The minimal polynomial of $J_n(0)^t$ is $x^n$. This is its characteristic polynomial as well. This tells you that the Jordan canonical of $J_n(0)^t$ is $J_n(0)$. But then they are similar.


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Let's take a simple example and build on it. Consider the basis of $\mathbb R^6$ : $span[(1,0,0,0,0,0),(0,1,0,0,0,0),(0,0,1,0,0,0),(0,0,0,1,0,0),(0,0,0,0,1,0),(0,0,0,0,0,1)] $ It's clear that if you remove the vector $(0,0,0,0,0,1)$, then none of the other vectors generate a $6th$ dimension, so yes, indeed, removing that vector, you get the standard basis ...


2

Removing a vector from a basis of $\mathbb{R}^n$ you always have a basis of some subspace $S$ of dimension $n-1$. This is true because you have $n-1$ linearly independent vectors that spans a subspace. But If you want a particular subspace $S$ then the statement is not true in general and you have to find $n-1$ linearly independent vectors that span this ...


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To show a matrix $M$ is symmetric, you just need to show that $M=M^T$. So we want to show that $ABA^T$ is symmetric by showing that $ABA^T=(ABA^T)^T$. Observe: $$ (ABA^T)^T = (A^T)^T(B)^T(A)^T=AB^TA^T. $$ Since $B$ is symmetric, then $B^T=B$. Which means the equation continues as $ABA^T$. QED


2

Your reasoning is correct; yet, you can do the job faster as follows: let $u\in\ker(A)$ be a (non-zero) eigenvector. Put $X=uu^T$ (a symmetric matrix). Thus $AX+XA=(Au)u^T+u(Au)^T=0$ and we are done. EDIT. To @ Asaf Shachar . Let $M=P_{\geq 0}$. Then $M$ is a closed cone; in particular, it is a convex set. Yet, the edge $E$ can be complicated; if $n=2$, ...


3

Here is a surprising application. First of all, I claim that $\forall n\in \mathbb{N}:\exists x,y\in \mathbb{Z}$ such that $5^n=x^2+y^2$. We can easily prove this by induction, however the induction proof I will show uses a nice trick in the indcution step. Obviously, $5^0=1^2+0^2$, now suppose that $5^n=x_{n}^2+y_n^2$ for some $n>0$. Then ...


1

Hint: $$ (ABC)^T = (A(BC))^T = (BC)^TA^T = C^TB^TA^T $$


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@Coolwater was correct. Multiplying the quaternion by the other's inverse, then converting to Euler angles, gives the correct answer. I will be trying out different configurations in the lab tomorrow. I made a Quaternion class which incorporates the definitions in the link @Coolwater gave. >>> np.rad2deg((Quaternion(np.deg2rad([0, 90, 0])) ...


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I am assuming that you mean $x,y \in \mathbb{R}^d$. Then $xy$ is not a defined object. There is however the inner product of $x$ and $y$, which is equal to: $x^T y$ and is in fact a scalar, i.e. an element in $\mathbb{R}$. Since the transpose of a scalar is just the scalar itself, we have: $(x^Ty)^T=y^Tx$ i.e. the scalar product is commutative. Note ...


1

This statement is false. Consider some $x \in V$ such that $x \neq 0$, and $H = \{kx| k \in \mathbb{Z}\}$. $H$ veryfies the condition on the right, but is not a subspace. Maybe the context assumes that $H$ veryfies some other property (like closed scalar multiplication), but this sentence alone is false.


1

If we have two vectors $\vec a = (x_0, y_0, z_0)$ and $\vec b = (x_1, y_1, z_1)$, we can consider them to be column matrices $$ A = \begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix} \ \text{and} \ B = \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix}$$ Now, the dot product $\vec a \cdot \vec b$ becomes $$ \begin{bmatrix} x_0 & y_0 & z_0 \end{bmatrix} ...


0

Lets try to develop the formula: $$p' = q p \hat{q} = \left(\begin{array}{cc} \vec{p'} \\ 0\end{array} \right) = \left(\begin{array}{cc} \vec{u}\sin{\phi\over2} \\ \cos{\phi\over2}\end{array} \right) \otimes \left(\begin{array}{cc} \vec{p} \\ 0\end{array} \right) \otimes \left(\begin{array}{cc}- \vec{u}\sin{\phi\over2} \\ \cos{\phi\over2}\end{array} ...


0

Orthogonals pairs are {v1,v4}, {v3,v4} only


0

The answer is no. For example, take $X=I$, $A=0$, and $Q=(1/100)I$.


2

An isomorphism between two objects should be a one-to-one, onto (hence invertible), and should preserve the structure of the objects in question. When we consider vector spaces, the structure we care about is the vector space structure. You know the vector space structure is preserved if there is a mapping $$f : V \to W$$ such that $$f(a v_1 + b v_2) = a ...


1

The characteristic polynomial is $$ \det(\lambda I-A)= \det\begin{bmatrix} \lambda-\cos\theta & \sin\theta & 0 \\ -\sin\theta & \lambda-\cos\theta & 0 \\ 0 & 0 & \lambda-1 \end{bmatrix} = (\lambda-1) \det\begin{bmatrix} \lambda-\cos\theta & \sin\theta\\ -\sin\theta & \lambda-\cos\theta \end{bmatrix} $$ and, finally, $$ ...


1

first of all, I believe your rotation matrix is incorrect: your rotation matrix is simply collapsing the $z$-axis, while it should leave the axis unchanged. The correct matrix would be $$ R(\theta) = \begin{bmatrix} cos(\theta) & \sin(\theta)& 0 \\ -sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ Notice that the last ...


0

First, your eigenvalues are wrong $det(xI - A) = - x( (\cos(\theta) - x)^2 + \sin(\theta)^2) = - x(1 + x^2 -2x\cos(\theta)) = -x(x-e^{i\theta})(x-e^{-i\theta}) $ Which gives as eigenvalues $0,e^{i\theta},e^{-i\theta} $. For the eigenvectors, if there are no obvious solutions, calculate a basis of $\ker(A-\lambda I)$ PS : Please, in the future, avoid ...


3

Let's expand on question one a little bit. Here is something that you can say for a matrix over a general (commutative) ring. For any matrix $M$, there is always a matrix $M^{adj}$, the adjugate of $M$, which satisfies the following equations. $$ MM^{adj} = \det(M) \cdot I \qquad\qquad M^{adj}M = \det(M) \cdot I $$ So this is a matrix that can be ...


2

$T^3 = T$ implies that T is diagonalizable. Hence T is diagonalizable and 0 is its only eigenvalue. What can you conclude ?


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So I followed the hints and this is what I got: $ D_{11} = f''(x+g(y))$ $D_{12}=f''(x+g(y))g'(y)$ so then if we combine everything $D_2FD_{11}= f'(x+g(y))g'(y)f''(x+g(y))$ $D_1FD_{12} = f''(x+g(y))g'(y)f'(x+g(y))$ and we see they are indeed equal.


2

Let $A\in\mathbb{C}^{n\times n}$ be Hermitian such that $A$ has positive eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$, then we may let $\{v_1,v_2,\ldots,v_n\}$ be an orthonormal basis for $\mathbb{C}^n$ consisting of eigenvectors of $A$ corresponding to $\lambda_1,\lambda_2,\ldots,\lambda_n$, respectively. Given a non-zero vector $x\in\mathbb{C}^n$, ...


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Hint:Let $A=U\Lambda U^H$ be its Eigen decomposition where $U=[u_1,\dots,u_n]$ is a orthonormal matrix with columns as eigenvectors of $A$ and $\Lambda$ is a diagonal matrix with eigenvalues of $A$. Then $$x^HAx=x^HU\Lambda U^Hx=\sum_{i=1}^{N}\theta_i\lambda_i(A)$$ where $\theta_i = |x^Hu_i|^2\geq 0$ and $\sum_i \theta_i=1$. Now try to use ...


0

We have diagonally dominant matrices if absolute values of diagonal elements are larger than sum of absolute values of the others in that row. Such matrices are invertible, ie. nonzero determinant.


0

$\Vert f\Vert_{\infty}$ is just the largest value that $|f|$ takes on the interval $[0,1]$, which has to be finite by continuity. There is no inner product that induces this norm because it fails to satisfy the parallelogram law: $$\Vert f+g\Vert^2 +\Vert f-g\Vert^2 =2\Vert f\Vert^2 +2\Vert g\Vert^2$$ for all continuous $f$ and $g$ on $[0,1]$. Every inner ...


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If there are $L,U$ such that $A=LU$, then you find them without pivoting. However, not all (invertible) matrices admit an LU decomposition. For this case there os a related method with pivoting, sometimes called LUP. See the Wikipedia page of LU decomposition for details on.


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Put $\mathbf{X}_i$'s next to each other to form a matrix. Oh, in fact you did. Yes, it is $X$. Notice that $$\mathbf{X}_i^{\top}\mathbf{\Omega}^{-1}\mathbf{X}_i$$ is the $i$'th element on the diagonal of $$\mathbf{X}^{\top}\mathbf{\Omega}^{-1}\mathbf{X}$$ So, since you're summing these up, $$\begin{equation} \sum_{i=1}^N ...


2

The characteristic polynomial has no real roots, so it must be of even degree. Hence $n$ is even. To finish, construct a square matrix of any even size that squares to $-I$.


0

WLOG out generality assume that $m < n$. Then you only have to prove it for $m=1$ and $n=2$. The rest is covered by Fermat's Last Theorem for the case $n \ge 3$ which guarantees there is no such solution of this form. So, for the sake of contradiction, assume that $$a + b = c$$ and $$a^2 + b^2 = c^2$$ Now, $$a^2 + b^2 = c^2$$ $$\implies a^2 + b^2 = ...



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