# Tag Info

4

Yes, it is true. You can consider the transpose map $\phi^*\colon W^{**}\to V^{**}$ and the canonical isomorphisms $\omega_W\colon W\to W^{**}$, $\omega_V\colon V\to V^{**}$. Set $\psi=\omega_V^{-1}\circ\phi^*\circ\omega_W$ and finish up.

3

HINT You know a polynomial satisfied by $A$, namely $(A^2-1)^2$. So the minimal polynomial has to be a factor of $(x-1)^2(x+1)^2$. Which of the possible factors could give the correct minimal polynomial for $A^2$? If neither power of $(x-1)^i(x+1)^j$ is $2$ then we would have one of $A-I=0,A+I=0,(A+I)(A-I)=0$. In all these cases $A^2=I$, a contradiction. ...

3

Assuming that you meant “linear map”, $T$ is a linear map if you see $\Bbb C$ as a real vector space, since, if $z,w\in\Bbb C$ and $\alpha,beta\in\Bbb R$, we have\begin{align}T(\alpha z+\beta w)&=\overline{\alpha z+\beta w}\\&=\alpha\overline z+\beta\overline w\text{ (because $\alpha,\beta\in\Bbb R$)}\\&=\alpha T(z)+\beta T(w).\end{align} But $T$ ...

3

Assuming we're diagonalizing over the complex numbers, yes. In fact, we can perturb $A$ such that all its eigenvalues are distinct, which will be more than good enough. Some notation: let $A^{\mathsf L}$ be the $(n-1)\times(n-1)$ leading principal submatrix of $A$: the matrix with the last row and column of $A$ removed. Lemma. If $\lambda$ is an eigenvalue ...

2

Yes, denote $J$ the matrix with 1 above the diagonal and $0$ otherwise. $$X=I+J+J^2$$ $$X^{-1}=(I+J+J^2)^{-1}=(I-J)(I-J^3)^{-1}=(I-J)(I+J^3+J^6+\cdots)=I-J+J^3-J^4+J^6-J^7+\cdots$$ Note that $J^n=0$. The sum above has only finite terms which is not $0$. Wish it helps.

2

You can use the classical method consisting in writing $A=I+N$ where $I$ stands for the identity and $N$ is nilpotent or order $n$ (check it, $N^{n}=0$). Then, Use Newton's binomial expansion on $A^{n}$ to obtain a formula for the inverse by gathering all powers of $A$ on one side and identity on the other side.

2

$\{1,X,X^{2}\}$ is a basis for your space. So the space is three dimensional. This implies that any three linearly independent vectors automatically span the space.

2

Note that for $|A|=n$ you can form at most $\binom n 2 + n = \frac{n(n+1)}{2}$ different sums in $A+A$. In fact, if $A=\{1,10,100,\dots,10^{n-1}\}$ you do get $\binom n 2 + n$ different sums, since those will be all numbers with at most $n$ digits and either exactly two of them $1$ and all others $0$ or exactly one of them $2$. Hence, $$|A+A| \le \frac{|A|(|... 2 Hint: Note that the dimension of the kernel (AKA nullspace) of A is given by 7 - \rho(A) = 4. However, the nullspace of A is also an eigenspace of A. 2 p(x)=x^3(x^4-x^2+1) which has a triple root at 0 and four simple roots at$$\pm\sqrt{\frac{1+\pm i\sqrt3}2}$$(The second factor is a quadratic in x^2.) Since eigenvectors corresponding to distinct eigenvalues are linearly independent, A, considered as a complex matrix, has rank 4. But the rank of a matrix doesn't change if we make the ground ... 1 Short answer is no - you need to go about this in two steps The stretch - This can be defined the matrix$$M_S = \begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}$$The rotation - This is a standard rotation matrix$$M_R = \begin{bmatrix}\cos (- \frac{\pi}{3}) & -\sin(-\frac{\pi}{3}) \\ \sin(-\frac{\pi}{3}) & \cos(-\frac{\pi}{3})\end{bmatrix}$$... 1 This is a linear transformation. The areas are multiplied by the Jacobian (the determinant of the transformation) which is 1. So the area remains the same, equal to \pi. 1 Any transformation is represented by a matrix. Now, for a \mathbb{R}^2 \to \mathbb{R}^2 transformation, this would be a 2 \times 2 matrix$$T = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$$Now, we are given that$$T\vec{x} = \vec{x}\implies \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}...

1

You can scale the eigenvectors, i.e., if $v$ is an eigenvector then $\lambda v$ is as well for all real $\lambda \ne 0$. Thus, you can scale $\|B\|$ as you wish. PS: $B$ is similar to the diagonalization of $A$, i.e. $B^{-1} A B=D$, a diagonal matrix.

1

Seems you want a function of the form $f(x) = k \cdot x^\alpha$ where $0 \lt \alpha \lt 1$ Just pick whatever $k$ and $\alpha$ constants suit you best. The requirement that $f(0) = 0$ is satisfied. So try for example $\alpha=1/2$, $\alpha=2/3$, $\alpha=3/4$, etc. and then pick the appropriate values for $k$ to satisfy $f(100)=1$. Plot these functions and you ...

1

Hint. $T$ is a $\mathbb{R}$-linear map. $T$ is not a $\mathbb{C}$-linear map. Notes. $\mathbb{C}$ is one-dimensional vector space over the filed $\mathbb{C}$ where $\alpha=(1)$ is a basis. $\mathbb{C}$ is two-dimensional vector space over the filed $\mathbb{R}$ where $\beta=(1,i)$ is an (ordered) basis. When you talk about "linear maps", you ...

1

Each successive application of the $\nabla$ operator increases the order of the derivative of $g(\lambda)$ by one. It also increases the tensorial order of the result by one by liberating another $a$-vector from the function argument $\,\lambda=a^Tx,\;$ i.e. \eqalign{ \nabla f(x) &= g^{\prime}(\lambda)\;&a \\ \nabla^2f(x) &= g^{\prime\prime}(\... 1 Well, since s is a map from a 4-dimensional vector space to a 3-dimensional vector space, it cannot be invertible (or for that purpose the inverse of T). However, you can demand that the restriction of S to \mathrm{Im}(T) be the inverse. There's of course many ways to achieve that but here is one that undoes the particular T that you picked: ...

1

The constant $u_0$ is the input signal value around which you perform linearization. Suppose that the nonlinear system is given by $$\dot{x}(t) = f(x(t),u(t)).$$ For the linearization around the equilibrium $(x_0, u_0)$ it holds $f(x_0,u_0)=0$. In other words, $u_0$ is such an input signal that makes the state $x_0$ an equilibrium of the autonomous system $\... 1 The first three input vectors are linearly independent, so we can evaluate$\phi$on the standard basis vectors by linearity:$\phi(1,0,0)=\phi(\frac12((4,2,1)-(2,2,1))=\frac12((4,1,0)-(2,3,3))=(1,-1,-\frac32);\phi(0,1,0)=\phi((2,2,0)+(2,2,1)-(4,2,1))=(3,2,1)+(2,3,3)-(4,1,0)=(1,4,4)$; and$\phi(0,0,1)=\phi((2,2,2)-(2,2,1))=(0,0,1)-(2,3,3)=(-2,-3,-2).$So ... 1 Another approach you may like is as follows: Let the input vectors are$u_1,u_2,u_3,u_4$and the corresponding output vectors are$v_1,v_2,v_3,v_4$so that for a$3\times 3$matrix$A=\left (x_{ij}\right )$, the system$Au_i=v_i, (i=1,2,3)$gives you a set of$9$equations involving$9$variables$x_{ij}, (1\le i\le 3, 1\le j\le 3)$. Solving the above you ... 1 Well, if$v=\lambda_1 a_1 + \lambda_2 a_2 + \lambda_3 a_3$for some$\lambda_1, \lambda_2, \lambda_3 \in \mathbb R$, then$v=A\left(w\right)$where$w=(\lambda_1, \lambda_2, \lambda_3)\in\mathbb R^3$and$A$is linear transformation with matrix $$\left[\begin{matrix} -1& 4 & 1 \\ h & -2 & 6 \\ 7 & 5 & 2\end{matrix}\right].$$ So, every ... 1 Essentially, you're considering$A'=\begin{pmatrix} a_0 & b^\top\\ c & A \end{pmatrix}$for scalar$a_0$, column vectors$b,c$and a nonsingular matrix$A$. A tool which is applicable for matrices of this form is the Schur complement. This is based on performing block-Gaussian elimination. In particular, we note the following block-diagonalization of ... 1 You don't specify what you mean by a "minimum" change, so for my own convenience I'll assume that you want to minimize$\|A - f(A)\|_F$, where$\|\cdot\|_F$denotes the Frobenius norm. Note that a matrix$M$will satisfy$MB = 0$if and only if each row$r$of$M$satisfies$rB = 0$. In fact, for any row-vector$r$, the nearest vector for which$rB ...

1

When talking about the Euclidean space $\mathbb{R}^n$, the default basis is the standard basis. Let $A$ be a matrix where the column vectors are $v_1,\cdots,v_n$. Then if you consider the row vector $$w=(\underbrace{1,\cdots,1}_{\textrm{n terms}})$$ it follows from the matrix multiplication that $wA=0$, which implies that $A$ is not of full rank and ...

1

Hint. In general, for a map $L:V\to W$, in order to know $L$, you need to know $L(v)$ for every single element $v$ in $V$. But if $L$ is a linear map where $V$ is a vector space, then it suffices to know $L(v)$ for some special elements, i.e., on a basis of $V$. Once $L$ is known on a basis, it is uniquely determined. Once $L(v)$ for every element $v$ in a ...

1

Since your matrices are symmetric and real, your trace is equal to $\operatorname{Tr} A^\top X$, which is the Frobenius inner product, and therefore you want to maximize $\langle A,X\rangle_F$ subject to $\|X\|=1.$ By Cauchy-Schwarz, $\langle A,X\rangle_F \leq \|A\|_F\cdot \|X\|_F$. This is maximized for $X=\lambda A$ (as a consequence of Cauchy-Schwarz), ...

1

Your $T$'s are in the wrong places when you write the relations among the orthonormal vectors; they should be on the first element to give a scalar, so $$\begin{bmatrix}u_1\\u_2\end{bmatrix}^T\begin{bmatrix}u_1\\u_2\end{bmatrix}=1$$ and so on.

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