New answers tagged

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 $A=2e_1+70e_2$.


1

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 $\alpha$.


0

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.


1

You did the same calculation for $\Delta_3$ as for $\Delta_0$. Instead, you need to replace the $z$ column by the right-hand column, like you did for $x$ to get $\Delta_1$.


0

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.


0

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


1

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


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

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


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 ...


0

From your question link, I can see that you’re interested in seeing how solving the eigenvalue problem $S_w^{ - 1} S_b w - Jw = 0$ shows that the vector $w = S_w^{ - 1} \left( {\mu _1 - \mu _2 } \right)$ maximizes $$J = \left( {\frac{{\left( {\tilde \mu _1 - \tilde \mu _2 } \right)^2 }}{{\tilde s_1^2 + \tilde s_2^2 }}} \right) = \left( {\frac{{w^T S_B ...


1

There is no hope for an upper bound. To spell out @Keith's comment, assume that $\|L(\sigma)^{-1}\|\le g(\sigma)$ for all such families. Then $g(\sigma)\ge\frac{e^{b/\sigma}}B$. Now consider the family \begin{equation*}L(\sigma)=\begin{pmatrix}Be^{-b/\sigma}&0\\0&g(\sigma)^{-2}\end{pmatrix}\end{equation*}to arrive at a contradiction.


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

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

Comment 2: Even more precisely, the derivative at $x \in R^2$ acts on the vector space of all directions emanating from x, but by representing them with vectors in $R^n$ we are implicitly using the fact that this vector space is canonically the isomorphic to vector space of directions emanating from $0 \in R^n$. If we are taking derivative of a function ...


0

Watch! I'm gonna nail this question. Remember this: Derivative of a multi-variable function, is not a number, but rather a linear transformation, a matrix. If $u:\mathbf{R}^2\to\mathbf{R}$ then its derivative at $x \in \mathbf{R}^2 $ is a matrix, here from $\mathbf{R}^2$ to $\mathbf{R}$. So, what does it do? It takes a vector $e$ in $\mathbf{R}^2$, and ...


0

I don't know how this helps, but thought of writing it up since some one here might be able to extend it. We can write $$h^n(X)=A^nX(A^n)^T+\sum_{i=0}^{n-1}A^iQ(A^i)^T$$ where ($A^{0}=I$).


1

There are two kinds of matrices most often used to represent simple undirected graphs, adjacency matrices and incidence matrices. [Their variations for graphs that allow loops/self-edges, multi-edges, and/or directed edges are common, but we mostly avoid discussing them here.] The adjacency matrix for a bipartite graph is mentioned already in the Question. ...


1

May I take it that setting density=foo in random_element isn't working for you? See the reference manual. From my experimentation MS.random_element(density=1) definitely gives me the fully non-zero matrix over $\mathbb{F}_2$ every time. That said, I agree the documentation is not very clear on exactly how to get what density, though if set to None, ...


0

Well, $0.5$ is the expected density you would get, so you ought to get that as an average. If you need a different expected density and sage doesn't provide a builtin method to do it, you can just sample each bit with different probabilities of being $1$ and being $0$. Of course, the probability of being $1$ is exactly the expected bit density of the matrix. ...


1

We have $L:\mathbb{R}^{2\times 2}\rightarrow\mathbb{R}^{2\times 2}:\begin{pmatrix} a&b\\c&d\end{pmatrix}\mapsto \begin{pmatrix}a&c\\b&d \end{pmatrix}$. Or after the identication $\mathbb{R}^{2\times 2}\cong \mathbb{R}^4$, we get the same linear map $L:\mathbb{R}^4\rightarrow \mathbb{R}^4:(a,b,c,d)\mapsto (a,c,b,d)$. Notice that ...


0

Let $D= diag(v^t)$. Then, you have $(X D)^T (X D) = D^T X^T X D$. Now, if the SVD of $X$ is $U S V^T$, $X^T X = V S^T U^T U S V^T = V S^2 V^T$ where $S$ is the square matrix of the appropriate dimensions containing the squared singular values of $X$. Since $D$ is diagonal, $D^T = D$, and diagonal matrices commute with other matrices, so $D^T X^T X D = D X^T ...


1

Hint: The matrices $$\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}0&-1\\1&0\end{bmatrix}\right\}$$ form a basis of $\mathbb{R}^{2 \times 2}$ and each of them is an eigenmatrix of $L$, i.e. $L(A) = \lambda A$ where $\lambda$ is a scalar. ...


1

Note that if $A$ is similar to $B$, i.e. $A = S B S^{-1}$ with $S$ invertible, then $F_A(M) = M S B S^{-1} - S B S^{-1} M = S (S^{-1} M S B - B S^{-1} M S) S^{-1} = S F_B (S^{-1} M S) S^{-1}$, so $F_A$ and $F_B$ are conjugate and thus will have the same characteristic polynomial. I'll use the basis $e_{11},e_{12},e_{21},e_{22}$ of $M_2(\mathbb C)$, where ...


0

The characteristic polynomial is $$ \det \begin{pmatrix} 1-X &0 &0 &0 \\ 2&1-X & -3 & -2\\ 3& 0 & 0-X &-9 \\ -1& 0& -1& 0-X \end{pmatrix} =(1-X)^2(3-X)(-3-X) $$ You want to determine the geometric multiplicity of the eigenvalue $1$, that is the dimension of the eigenspace, which is the null space of ...


2

In other words you want $\det(B)$ to be odd and not divisible by $13$. The "not divisible by $13$" isn't a problem, because $\det(B) \equiv \det\pmatrix{2 & 4\cr 3 & 2} \equiv -8 \mod 13$. If $A_1 = \pmatrix{a & b\cr c & d\cr}$, then $B \equiv \pmatrix{a & b\cr 1+c & d\cr} \mod 2$, so $\det(B) \equiv ad + b(1+c)\mod 2$. You want ...


0

The number of leading 1's in RREF tells us the rank of the matrix which is equal to the dimension of column space of $A$. The column space is defined as the span of the columns of $A$, we should be able to find 3 columns of $A$ that span it if the dimension of column space is 3.


0

We have a minimal $n\geq 1 $ such that $X^{n-1}v\neq0,X^{n}v=0$ as $V$ finite dimensional. Now $Yv = 0$ implies $0=X^iYv=(\lambda +(\lambda+2)+...(\lambda+2(i-1)))X^{i-1}v+YX^iv=0$. Putting $i=n$ gives a contradiction.


1

Consider the factorization of the matrix $A$ which is always possible since $A$ is a real symmetric matrix, and the : $A=P^{-1}\Lambda P =P^T\Lambda P$, where the matrix $P$ is orthogonal because $A$ is symmetric, and the diagonal matrix $\Lambda$ contains the eigenvalues of $A$. Then the ellipsoid can be rewritten as: $$x^TP^T\Lambda Px = (Px)^T\Lambda(Px) ...


1

You should recall the formula $$ \chi_A(X) \overset{def}= \det(X\, \mathrm{id}_n - A) = \sum_{i=0}^n [(-1)^k \mathrm{tr}(\Lambda^k A)] X^{n-k}, $$ which, if you have never seen it proved, I have written down here (Your question holds for a general commutative unital ring $R$, but if you only care about a vector space over a field or the real numbers, feel ...


1

We consider the equations (1) $A^2+B^2=AB$, (2) $A^2+B^2=2AB$. A couple $(A,B)$, solution of (2), is simultaneously triangularizable, and, moreover, when $n=2$, $AB=BA$. Equ (1) has not the second property: indeed $A=\begin{pmatrix}0&-1\\1&-1\end{pmatrix},B=\begin{pmatrix}\dfrac{1-i\sqrt{3}}{2}&0\\\dfrac{3+i\sqrt{3}}{2}&-1\end{pmatrix}$ is a ...


3

Let $$X=\left[\begin{matrix} a &b\\ c &d\end{matrix}\right].$$ Expanding the matrix product, you need to solve $$\begin{cases}2a+3c=3\\4a+5c=5\\2b+3d=4\\4b+5d=6\end{cases}$$ Hint: $$\begin{cases}2a+3c=3\\4a+5c=5\end{cases}$$ $$\begin{cases}2b+3d=4\\4b+5d=6\end{cases}$$ The real, efficient method is by solving the system (by Gaussian ...


3

It's not possible to multiply $2\times 2$ matrix with $2\times 1$ matrix and give a $2\times 2$ matrix. But if in your question $B$ is a $2\times 1$ matrix, you can solve this system of equation by elimination: $$\begin{cases}2x+3y=a\\4x+5y=b\end{cases}$$


4

Hint: Think about inverting Matrix A and rewriting it as $X=A^{-1}B$. The inverse of a 2 by 2 matrix is simple to generate by hand. Your approach is the right one but you will have to introduce a 2 by 2 matrix for $X$ instead of a vector.


0

Note that the eigendecomposition of $A$ can be written as $ U S D U^H$ where $U$ is a unitary matrix, $D$ is a diagonal matrix containing the absolute value of the eigenvalues of $A$ and $S$ is a diagonal matrix with $+1$ on the diagonal if the corresponding eigenvalue in $D$ is positive and $-1$ if negative. Then, note that $(US) D U^H$ is an SVD of $A$, ...


1

From Hadamard's determinant bound one can deduce that the absolute value of the determinant is at most $54$. This is done as follows. Let $M$ be an $n\times n$ matrix whose elements lie in the interval $[0,a]$, for some positive number $a$. Then $$ \det M=\left(\frac{a}{2}\right)^n\det\left[\begin{array}{c|ccc}1 & 1 & 1 & 1\\ \hline 0 & ...


0

Since the matrix is upper triangular, the determinant is just the product of the diagonal elements. Hence $$D = \left[x + i\left(\frac{3}{4} + y \right)\right] \cdot \left [\left(x- \frac{5}{4} \right) + iy \right ] \cdot \left[ \left(x + \frac{3}{4} \right) + iy \right]$$ Setting $D$ to zero and computing the values of $x$ and $y$ will give you the ...


1

Your answer should be given by scaling * rotation * translation. But your rotation matrix is wrong. From the figure, you need to rotate the object by -$\pi/3$, so your rotation matrix is $$\begin{pmatrix}0.5 && \sqrt 3 /2&&0\\-\sqrt 3 /2&& 1/2 &&0\\0&&0&&1\end{pmatrix}$$


0

An orthogonal matrix is one which satisfies $A A^T = A^T A = I$. That is, $A^{-1} = A^T$. Equivliently, $A$ has orthonormal columns (and orthonormal rows). You've shown the columns are orthogonal for $A$, but each column of $A$ has norm 9. So, if you divide $A$ by 9, each column will still remain orthogonal to each other and then the columns will be ...


0

If the problem is as simple as I think it is : Let $ B = \frac{1}{9} A $ . Do the multiplication by each unit, calculate B and check if the columns and rows are orthogonal unit vectors. All this by supposing what you meant in the start was $ \frac{1}{9} A $/


1

There is no solution over $\mathbb R$ if $n \ge 3$ is odd and $\det(A) < 0$. $\det(\text{adj}(B))= \det(\det(B) B^{-1}) = \det(B)^{n-1}$, which can't be negative in this case.


2

The case $n=1$ is immediate. For the case $n\ge 2$, by user1551's comment, $${\rm adj}({\rm adj}{(A)})=(\det A)^{n-2}A.$$ Also, there is a fact that ${\rm adj}(kA)=k^{n-1}{\rm adj}(A)$ for any scalar $k\in\mathbb{C}$. Thus we have \begin{align} {\rm adj}((\det A)^{-\frac{n-2}{n-1}}{\rm adj}{(A)}) &=(\det A)^{-(n-2)}{\rm adj}({\rm adj}{(A)})\\ &=(\det ...


0

One way to proceed is as following: Start by constructing one orthonormal basis for each of the vectors $\vec{n_1}$ and $\vec{n_2}$. This can be done by the trick given in an answer to another question [1]. This will result in two transformation matrices $$R_1=\begin{bmatrix}\vec{u_1} & \vec{v_1} & \vec{w_1}\end{bmatrix}$$ and ...


0

As stated, this is false, since even $A$ itself could be singular under the given assumptions. To conclude that $A-B$ is nonsingular when $B$ is "small", one needs: $A$ to be nonsingular $\|B\| < \|A^{-1}\|^{-1}$ The reason this works is that $\|A^{-1}\|^{-1} = \min_{\|x\|=1}\|Ax\|$. Hence, assumption 2 implies $\|Bx\|<\|Ax\|$ for all $x$, making ...


1

In general, to find the derivative $Df(A)(H)$ you need to write $f(A+H)-f(A)=L(H)+O(\Vert H\Vert^\alpha)$, where $L(H)$ is linear on $H$, and $\alpha>1$. Let's do this with $g(A)=A^{-1}$: For all $H$ sufficiently small, $A-H$ in invertible, and ...


0

The map: $\Phi(A) = A^{-1}$ is differentiable over $GL_n$ and $D\Phi(A) H = - A^{-1} H A^{-1}$ for all $A \in GL_n$ and all $H \in M_{n \times n}$. Letting $B(M,N) = MN$, $B$ is bilinear map on a finite dimensional normed vector space, hence continuous and so differentiable with $DB(M,N)(H,K) = HN + MK$. Now $f = B \circ (\Phi, \Phi)$, so: $$Df(A) H = ...


3

If you have $AB=I$, then it follows $ABA=A$ , and therefore $BA=I$. So, a matrix always commutes with its inverse. This is exactly what you need to prove your claim.



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