New answers tagged

0

Using $$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+xh(x+h)-f(x)}{h}$$ So $$f'(x) = \lim_{h\rightarrow 0}\frac{f(h)}{h}+\lim_{h\rightarrow 0}\frac{xh(x+h)}{h} = \lim_{h\rightarrow 0}\frac{f'(h)}{1}+\lim_{h\rightarrow 0}x(x+h)$$ So $$f'(x) = f'(0)+x^2 = -1+x^2$$ So $$f(x) = -x+\frac{x^3}{3}+\mathcal{C}$$ Now ...


0

$$y+6=(2x-4)x$$ $$x^2-4x+3=2x^2-4x-6$$ $$9=x^2, x=\pm3$$ Therefore, the answer is both lines tangent to the point when $x=\pm3$. $$y=2(x-3)\text{ and }y-24=-10(x+3)$$


5

Since $xy=-1-xz-yz=-1-z(x+y)$, we obtain $$\frac{(x+1)(y+1)}{x+y}=\frac{xy+x+y+1}{x+y}=\frac{(-1-z(x+y))+x+y+1}{x+y}=-z+1$$


0

Hint: Let $u=\langle 2,2,5\rangle$ and $v=\langle 6,5,4\rangle$, and find $d=\frac{|u\times v|}{|v|}$.


3

Note that the function $f(A) = det(I - A)$ is continuous. So $U = f^{-1}(\mathbb{R} - \{0\})$ is open as a preimage of an open set under a continous map.


0

The distance from $P$ to the point on $l$ at time $t$ will be given by: $$d(t) = \displaystyle \sqrt{(2+6t)^2 + (2+5t)^2 + (5 + 4t)^2}$$ So our problem can be solved like a standard minimization problem from calculus: the value of $t$ that minimizes $d(t)$ corresponds to the point on the line closest to $P$. However, taking the derivative of $d(t)$ is ...


1

More Generally. First you are going to want to set this matrix up as an Augmented Matrix where $Ax=0$. $1)$ To find the rank, simply put the Matrix in REF or RREF $\left[\begin{array}{ccc|c} 0 & 0 & 0 &0 \\ 0 & 0.5 & -0.5 & 0 \\ 0 & -0.5 & 0.5 & 0 \end{array}\right] \longrightarrow RREF \longrightarrow \left[\begin{...


0

If by "scaling" you mean a multiple of the identity matrix, then this is not possible. Such a matrix commutes with any rotation, so given a combination of scaling and rotation, you can do the rotations first and then the scaling; that is, you will have $\lambda R$ for some real number $\lambda$ and real orthogonal matrix $R$. Then $R$ is diagonalizable over ...


1

I am assuming via the title that $S\subseteq \mathbb{R}^{3}.$ Before attempting this problem, it may be best to see what elements of $S$ are like. Since $a+b=0$, $b=-a$ and similarly, $c=-a$. Therefore, we can think of $S$ as $$S=\{[a,-a,-a]\in\mathbb{R}^{3}\}.$$ ( a ) To show a set is not closed under addition, it suffices to provide a counterexample. ...


1

The matrix has one linearly independent row (take the negative of the second to get the third) implying that the rank is 1 and the nullity is 2.


0

The answer is yes. But you don't need they have the same basis. More precisely: Let $V$ be a subspace of the finite dimensional space $W$; if $\dim V=\dim W$, then $V=W$. Proof. Since $\dim V\le \dim W$ for any subspace $V$ of $W$ we can prove the equivalent statement that if $V\subsetneq W$ (proper subspace), then $\dim V<\dim W$. If $\{v_1,v_2,\...


0

Note that saying $V,W$ as having the same basis and same dimension is somewhat redundant; you should think and phrase that sentence as '$V,W$ are both generated by a basis of the same dimension', having the same basis for instance will directly imply they have the same dimension. If they have the same basis then clearly they are the same, since every ...


3

Let's say we have $n\times n$ matrices $A,B,D,E$ and $n\times 1$ matrices $x,y,C,F$. Set up a system $$Gz=H,$$ using the block matrices $$G=\begin{pmatrix} A & B \\ D & E \end{pmatrix}, \qquad z=\begin{pmatrix} x \\ y \end{pmatrix}, \qquad H=\begin{pmatrix} C \\ E \end{pmatrix}. $$ Solve this system for $z$, then you will have obtained the ...


1

$(1)\implies (2)$: According to Andrea , $\lim_{t\rightarrow 0^+}\dfrac{{e^{tA}}_{ij}}{t}=a_{ij}$ and, consequently, $a_{ij}\geq 0$. $(2)\implies (1)$: There is $s>0$ s.t. $A+sI$ is a non-negative matrix. Then $e^A=e^{-sI}e^{A+sI}=e^{-s}\sum_{n=0}^{\infty}(A+sI)^n/n!$. Thus $e^A$ is a non-negative matrix. In the same way, for every $t\geq 0$, $e^{tA}$ ...


0

By row-reducing the matrix, we get $$ \begin{pmatrix} m + 1 & 1 & m \\ 1 & -m & 1 - m \\ 1 & -1 & 2 \end{pmatrix} \xrightarrow{R_1 \leftrightarrow R_3} \begin{pmatrix} 1 & -1 & 2 \\ 1 & -m & 1 - m \\ m + 1 & 1 & m \end{pmatrix} \xrightarrow{R_2 = R_2 - R_1} \\ \begin{pmatrix} 1 & -1 & 2 \\ 0 & -m +...


2

Hint: Via the invertible matrix theorem Given a set of $n$ vectors from $\Bbb R^n$, forming a square matrix $A$ using those vectors as the columns we have: The columns of $A$ are linearly independent $\iff$ $\det(A)\neq 0$ via contrapositive, we have the columns of $A$ are linearly dependent $\iff \det(A)=0$ Calculating the determinant of the matrix we ...


8

Let $\lambda \in \mathbb C$ be an eigenvalue with eigenvector $0 \neq x \in \mathbb C^n$. Then $$(x,x) = (A^tAx,x) = (Ax,Ax) = \lvert\lambda \rvert^2 (x,x).$$ Thus $\lvert \lambda \rvert = 1$ for all eigenvalues. Since $A$ is real, eigenvalues come in conjugate pairs, so either all three are real, or two are complex and one is real, but the two which are ...


1

Your question is how to compute the matrix exponential $\exp(a)$ where $a$ is an element of a finite-dimensional real associative algebra $A$. (Your phrasing is not more general: you're implicitly working in the algebra $M_n(A)$.) This reduces to the case of real matrices, because the left regular representation $$A \ni a \mapsto \left( L_a : b \mapsto ab \...


1

I'll just clarify the question for you. $\mathcal B$ is a basis (thus why they named it $\mathcal B$ ;) ) and the notation $[v]_{\mathcal B}$ and $[T(v)]_{\mathcal B}$ denote the matrix representations (i.e. column matrix containing the coordinates) of the vectors $v$ and $T(v)$ respectively ($T$ is a linear transformation so $T(v)$ is a vector) wrt $\...


1

This matrix you are trying to find depends upon your choice of basis. If you choose the standard basis for $\mathbb{C}^{n}$ then your matrix is given by the $m\times n$ matrix whose columns are $m$-vectors which are the images of basis elements. Let $T:\mathbb{C}^{n}$$\rightarrow$$\mathbb{C}^{m}$ and $A$ be your matrix that induces your map $T$ then the ...


1

One such map could be $f(x) = Ax, x \in \mathbb{C^n}$, with $A$ is a $m \times n$ matrix.


2

The matrix $BA$ cannot be invertible: there exists a vector $x\in\mathbb{R}^7$ such that $Ax=0$ and $x\ne0$, because the homogeneous linear system $Ax=0$ has infinitely many solutions. Then $BAx=0$, which can't happen for an invertible matrix. However, the matrix $AB$ can be invertible. Just to make a simple example, $$ \begin{bmatrix} 1 & 0 \end{...


3

$A$ is a $5 \times 7$ matrix, so its null space must have dimension at least $2$. If $Ax = 0$ then $BAx = 0$. So the null space of $BA$ has dimension at least $2$, and $BA$ is not invertible.


6

$BA$ is a $7 \times 7$ matrix, so it needs to be rank 7 to be invertible, But, since $\mbox{rank} (BA) \leq \min(\mbox{rank}(A),\mbox{rank}(B)) \leq 5$ (since $A,B$ have one dimension as $5$) it cannot be invertible. $AB$ on the other hand can be invertible (for example, generate two matrices with i.i.d. standard normal entries for $A,B$ and then multiply ...


2

I found a polynomial that works, but I am not sure if it the "best" one. Let the $n$ points be $p_1,p_2\dots p_n$ with $p_j=(a_{j,1},a_{j,2}\dots a_{j,d})$. And suppose we want to find a polynomial $P$ with $P(p_j)=b_j$. Notice that the polynomial $P_j=(x_1-a_{j,1})^2+(x_2-a_{j,2})^2+\dots+(x_d-a_{j,d})^2$ is only zero at the point $p_j$ Therefore the ...


0

Thank you very much, Mr. Bhuyan. I rewrite $\operatorname{rank}(\mathbf{A}\mathbf{B}-\mathbf{I}) =\operatorname{rank}((\mathbf{A}-\mathbf{I})\mathbf{B}+(\mathbf{B}-\mathbf{I})) \leq \operatorname{rank}((\mathbf{A}-\mathbf{I})\mathbf{B})+\operatorname{rank}(\mathbf{B}-\mathbf{I}))$, and from $\operatorname{rank}((\mathbf{A}-\mathbf{I}), \mathbf{B}) \leq ...


0

let us compute three points of this equation $$x-3y+2z=5$$ for example $$P_1(5,0,0),P_2(0,1,4),P_3(1,0,2)$$ then you can compute the equation of the plane as follows: $$[x,y,z]=(5,0,0)+\alpha(-5,1,4)+\beta(-4,0,2)$$ with real numbers $$\alpha,\beta$$


0

Hint: Let $M$ be a point with coordinates $(x,y,z)$, $O$ the origin. Think of dot product.


3

The $x$ in the Lagrangian can be any value. However the "subject to" constraints tell us that the domain over which you are to minimize $L$ are only those that satisfy $c^Tx = 0$ and $x^Tx = 1$, i.e. all $x$ in the intersection of the unit $n$-sphere and the plane $c^Tx = 0$. After you have derived the Lagrangian equations $$\frac{\partial L}{\partial \mu} ...


0

Denote the sizes of the blocks by $x_1 \geq x_2 \geq \dots \geq x_5 \geq 1$ with $x_1 = 5$. We must have $x_2 + \dots + x_5 = 10$. The rank of $J_k(0)^l$ is $\max(k-l,0)$. If $A$ is similar to $\operatorname{diag}(J_5(0), J_{x_2}(0), \dots, J_{x_5}(0))$ then $A^3$ is similar to $\operatorname{diag}(J_5^3(0), J_{x_2}^3(0), \dots, J_{x_5}^3(0))$ whose rank is $...


3

A (nilpotent) Jordan block of size $k$ contributes $k-n$ to the rank of $A^n$ if $k\ge n$, and contributes $0$ if $k\le n$. Let $n_k$ denote that number of Jordan blocks of size $k$. From $A^5=0$, we known that all blocks have size $\le 5$, i.e., $n_k=0$ for $k>5$. From $\operatorname{rank}(A)=10$, we known that there are five Jordan blocks, i.e., $$\...


2

The entanglement of the qubits does not matter. All you need to do is tensor $U$ with $\begin{pmatrix}1&0 \\ 0 & 1\end{pmatrix}$, which is the effect of $U$ on the third qubit. You'll get $U' = \begin{pmatrix}U&0 \\ 0 & U\end{pmatrix}$ or a permutation thereof. As a linear transformation, $U' = U \otimes \text{id}$, so you'd have $$U' \left|...


1

Assuming $A$ is a real matrix, using singular value decomposition we can write $$ A = U S V^T$$ where $S$ is a real valued diagonal matrix (i.e., $S=S^T$); $U$ is the left Eigenvector and $V$ the right Eigenvector. Then, you can write $$A^TA = V S^T U^T U S V^T = VS^2V^T$$. However, if $A$ is positive symmetric, then $U=V$ and you can use eigenvalue ...


2

I remember encountering the same confusion as you when presented with such inequalities in optimization problems. $x = (x_1, \dotsc, x_n)^T$ is a column vector of variables which are usually under our control in the problem, but subject to certain constraints. For instance, you might encounter the constraint $x \geq 0$. In the vector terms you may be used ...


1

The complex $1$ dimensional case is the well known one: if $w=a+ib$ and $z=x+iy$ $$wz=ax-by+i(ay+bx)$$ So via the usual $(x+iy)\in\mathbb{C}\cong \mathbb{R}^2\ni(x,y)$ identification, multiplication becomes $$B\begin{pmatrix}x \\ y\end{pmatrix}:=\begin{pmatrix}a & -b \\ b& a\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}.$$ If one would choose the ...


0

A basis for $\;V\;$ is, for example $$x_2=-3x_1\;,\;\;x_3=2x_1\implies\left\{\;v:=\begin{pmatrix}1\\\!-3\\2\end{pmatrix}\;\right\}$$ and thus an orthonormalized basis here is pretty boring: just multiply $\;v\;$ by $\;\frac1{\left\|v\right\|}\;$ : $$u:=\frac v{\left\|v\right\|}=\frac1{\sqrt{14}}\begin{pmatrix}1\\\!-3\\2\end{pmatrix}=\begin{pmatrix}\frac1{\...


0

For the question of minimal polynomial, all that matters about a projection is that it is diagonalisable with eigenspace for $\lambda-1$ (the set of fixed vectors, the image of the projection) and for $\lambda=0$ (the kernel, the subspace parallel to which the projection is done) only. The whole space being the direct sum of these eigenspace means that the ...


1

Given a desired $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$, we wish to find coefficients $w,x,y,z$ such that: $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=w\begin{bmatrix}1\\0\\1\\0\end{bmatrix}+x\begin{bmatrix}0\\1\\0\\1\end{bmatrix}+y\begin{bmatrix}1\\0\\-1\\0\end{bmatrix}+z\begin{bmatrix}0\\1\\0\\-1\end{bmatrix} = \begin{bmatrix}w+y\\x+z\\w-y\\x-z\end{bmatrix}$ ...


1

Describe the (rows') nullspace explicitly (by now it must be clear that its dimension is two): $$\text{Second row}:\;\;y+w=0\implies y=-w\;,\;\;\text{First row}:\;\;x+z=0\implies x=-z\implies$$ basis for the null space: $$\left\{\;\begin{pmatrix}0\\1\\0\\\!-1\end{pmatrix}\;,\;\;\;\begin{pmatrix}1\\0\\\!-1\\0\end{pmatrix}\;\right\}$$ Now check these two ...


3

It's possible for every $n$. Let $D=A-B$. Then $A^2-B^2=D^2+DB+BD$. It suffices to find two positive definite matrices $B$ and $D$ such that (a) $D^2+DB+BD$ has exactly one positive eigenvalue and $n-1$ negative eigenvalues. It also suffices to find two positive definite matrices $B$ and $D$ such that (b) $C=DB+BD$ has exactly one positive ...


2

$$x\begin{pmatrix}1\\2\end{pmatrix}+y\begin{pmatrix}3\\4\end{pmatrix}=\underbrace{\begin{pmatrix}1&3\\2&4\end{pmatrix}}_A\underbrace{\begin{pmatrix}x\\y\end{pmatrix}}_{\bf x}$$ If we change this slightly to $$x\begin{pmatrix}1\\2\end{pmatrix}+y\begin{pmatrix}2\\4\end{pmatrix}=\underbrace{\begin{pmatrix}1&2\\2&4\end{pmatrix}}_A\underbrace{\...


0

Consider the case $n = 2$. Using the invariant of the $2$-norm under rotations, we have $$ ||Q_1'Q_2' - Q_1Q_2||_2 = ||Q_1'Q_2' - Q_1Q_2' + Q_1Q_2' - Q_1Q_2||_2 = \\ ||(Q_1' - Q_1)Q_2' + Q_1(Q_2' - Q_2)||_2 \leq ||(Q_1' - Q_1)Q_2'|| + ||Q_1(Q_2' - Q_2)|| = \\ ||Q_1' - Q_1|| + ||Q_2' - Q_2||.$$ The general case can be handled by induction by writing "$Q_2' =...


0

It's very basic but just too long for a comment. If you take n vectors from the p vectors, there are 2 cases, the n are dependent then by definition, having a dependent subset of vectors, adding more vectors gives always a dependent set the n aren't dependent. Since n is also the dimension of the space, they form a regular basis of the vectorial space. ...


2

They're not asking for all the coefficients, just which ones are zero. Your friend seems to be making arguments to linearity. In the first two (top) graphs, they split into the "1" part and the "sin" part, each 0 on the half of the period. Then he writes the first part as a sum again: a constant 1/2, and an alternating-sign 1/2. (The first two graphs on the ...


1

The solution is not hard: I think: $$a_{n} = \frac{1}{\pi} (\int\limits_{-\pi}^{0}-1cos(nx)dx + \int\limits_{0}^{\pi}sin(x)cos(nx) dx) = \frac{1}{\pi} \int\limits_{0}^{\pi}sin(x)cos(nx) dx = \frac{1}{\pi} \frac{cos(n\pi)+1}{1-n^2}$$ $$b_{n} = \frac{1}{\pi} (\int\limits_{-\pi}^{0}-1sin(nx)dx + \int\limits_{0}^{\pi}sin(x)sin(nx) dx)=\frac{1}{\pi} (\frac{...


3

You're right to start with the identity $$\chi_A(A)=(-1)^nA^n+a_{n-1}A^{n-1}+\ldots+a_2A^2+a_1A+a_0I=0$$ but instead of factoring out an $A$, you can subtract $a_0I$ from both sides and multiply by $A^{-1}$ to obtain $$ -a_0A^{-1}=(-1)^nA^{n-1}+a_{n-1}A^{n-2}+\ldots+a_2A+a_1I$$ Then multiply both sides by $-\frac{1}{a_0}$ to conclude.


0

For two vectors $v,w \in \mathbb{R}^3$, the area of the parallelogram is the square root of the Gram determinant $\text{Gram}(v,w)$. For two three-dimensional vectors you conveniently have $$\sqrt{\text{Gram}(v,w)} = ||v \times w||$$ where $||.||$ is the euclidian norm. So in your case you get $$Vol_2(P(2a+3b, a-2b)) = ||(2a+3b) \times (a-2b)||$$ with (...


0

Thanks to @user1551, please tell me if you want to post the answer yourself I delete mine. Take $S = k I $ with $k = \lambda_{max}(\frac{A + A^T}{2})$. Because $A = \frac{A + A^T}{2} + \frac{A - A^T}{2}$ But $\left< \frac{A - A^T}{2}x,x \right> = 0$


3

Always. That's because every real/complex square matrix admits a singular value decomposition.


0

For any skew-symmetric matrix $K$, we have $x^TKx=(x^TKx)^T=x^TK^Tx=-x^TKx$ and therefore $x^TKx=0$. It follows that if you split $A$ into the sum of its symmetric part $\frac{A+A^T}2$ and its skew-symmetric part $\frac{A-A^T}2$, then $\langle Ax,x\rangle=\langle \frac{A+A^T}2x,x\rangle$. In other words, it suffices to find a positive definite $S$ such that ...



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