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.


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