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4

(I assume $M$ is supposed to be real; if it is allowed to be complex, then the exercise is trivial.) Hint What are the possible eigenvalues of $M$? What can one say about the eigenvalues of real $2 \times 2$ matrices...?


3

One way to prove this is through eigenvalues. Since $A$ is positive definite, it is a symmetric matrix. $$A \text{ is positive definite } \iff \text{ all eigenvalues are positive. }$$ It is known that for any matrix $M$: $$\lambda \text{ eigenvalue of matrix } M \implies \lambda^k \text{ eigenvalue of matrix } M^k,\quad k=1,2,\ldots$$ Also, you can use ...


3

I assume you are working with $n \times n$ matrices over the complex numbers. Let $A^\ast$ denote the Hermitian adjoint of $A$, i.e. the complex conjugate of the transpose. Then, by definition, $A$ is Hermitian if and only if $A = A^\ast$. Now suppose that $A$ is an arbitrary complex $n \times n$ matrix. Set $B = \frac{1}{2}(A + A^\ast)$ and $C = ...


2

$$A-I=\begin{bmatrix} 3 & -2 & 3\\ 0 & -2 & 3\\ -1 & 2 & -3 \end{bmatrix}$$ And you can see the two last columns are proportionate so the matrix is not invertible and $\lambda=1$ is an eigenvalue My computation of $\det(A-\lambda I)=-\lambda^3+\lambda^2+13\lambda-13$


2

Let $M=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$. Then we want $M^2=\left(\begin{matrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{matrix}\right)=\left(\begin{matrix}-1&0\\0&-1-\epsilon\end{matrix}\right)$. Now we can't have $b=0$ or $c=0$ (why?). Therefore $a=-d$ and ...


2

no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$ for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$ pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2 = ...


1

If $x_2-x_1 = \frac12$ then $x_2^2-x_1^2 = (x_2-x_1)(x_2+x_1) = \frac12$ is equivalent to $x_2+x_1 = 1$ The book may have wanted you to use Euler Lagrange equations?


1

Assume $v$ is an eigenvector of $M$ with eigenvalue $\lambda$. Then $v$ is eigenvector of $M^2$ wih eigenvalue $\lambda^2\ge0$. Since $M^2$ has only negative eigenvalues $-1$ and $-1-\epsilon$, $M$ has no eigenvectors. Thus the matrix $A$ that maps $e_1\mapsto e_1$ and $e_2\mapsto Me_1$ is invertible and since $M$ maps $Me_1$ to $-e_1$ we find that ...


1

Here is an analogy that might help: I suppose that the "vector" that you normally think of is a list of coordinates. For example, $(-1,-2,-3,-4)$ is a vector in $\Bbb R^4$. Note, however, that we could also represent this list of numbers as a function. In particular, if we define $f:\{1,2,3,4\} \to \Bbb R$ by $$ f(1) = -1\\ f(2) = -2\\ f(3) = -3\\ f(4) = ...


1

I don't understand the question entirely (in particular "If x is a set of values rather than a symbol then how can y remain a vector if for each element of x, y is scalar?"), but perhaps this will help: The ODE $$y''(x) + y(x) = 0$$ is homogeneous and linear, and so its space of solutions is a vector space $\Bbb V$, and its constituent vectors are functions ...


1

It depends on the conditions you have available to you, that you have information about. (Sufficent)You can compute the characteristic polynomial $Det ( A- \lambda I) $ and check there are different eigenvalues, i.e., no n-ple roots for $n \times n$ matrix. You can also just compute the eigenspaces if you have a repeated root; the eigenspaces associated to ...


1

The general matrix is given by the sum between the identity matrix and a circulant matrix, hence its characteristic polynomial over $\mathbb{Q}$ is given by: $$ p(\lambda)=(\lambda-1)^n-1.$$ Over $\mathbb{F}_2$ such a matrix cannot be invertible since the sum of the elements in every row/column is zero, hence $(1,1,\ldots,1)$ is an eigenvector associated ...


1

Its simply a matter of adding the areas of three parallelograms. To see this draw the vector $w$ from the origin, and connect its end to $u+w$ and $v+w$. So then its just the sum of the three determinants... Edit: see image:


1

The key is, as the book says, to normalize each of the vectors. That is, we want to replace each vector with a multiple of that vector which has length one. In other words, we want the unit vector in the same "direction". The length of a vector $\vec x = (x_1,\dots,x_n)$ is given by $$ \|\vec x\| = \|(x_1,\dots,x_n)\| = \sqrt{x_1^2 + \cdots + x_n^2} $$ ...


1

This is an awfully complicated way of deriving the derivative or gradient. It is simpler to show that $\|J(\theta+h)-J(\theta) - (X \theta-y)^T X h\|$ is bounded above by $\|X^TX\| \|h\|^2$ from which we see that $DJ(\theta)(h) = (X \theta-y)^T X h = \langle X^T (X\theta -y), h \rangle$, where $\langle \cdot,\cdot \rangle $ is the usual inner product in ...


1

If $x^2=y^2$, then $y=\pm x$, which is the same as $x=\pm y$. But $\sqrt{x^2}=|x|$.


1

Note that $x = y$ if and only if $-x = -y$. Also, $x = -y$ if and only if $-x = y$. So there are really only two possibilities: $$x = y \qquad \mbox{or} \qquad x = -y.$$ In other words, once you know that $x^2 = y^2$, then you know that $x$ and $y$ have the same magnitude (the same absolute value); you also know that either $x$ and $y$ are exactly the ...


1

Your matrix is the sum between an identity matrix and a circulant matrix, so the characteristic polynomial is given by: $$ p(\lambda)=(1-\lambda)^n-(-1)^n \tag{1}$$ and the determinant is given by $(-1)^n p(0)$, so it is $2$ if $n$ is odd and zero otherwise.


1

If $b$ is anything other than 1 then the matrix is diagonalizable. If $b=1$ then the matrix is diagonalizable only if $a=0$ (its already in diagonal form when this is true, in fact), since if $a \neq 0$ then \begin{equation} A-I=\begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix},\end{equation} the space of solutions of the homogenous system associated with ...


1

First understand the following $\mathbf{Thereom:}$ Let $\{v_1,…,v_n\}$ be any basis of an inner product space V. Then there exists an orthonormal basis $\{u_1,…,u_n\}$ of V such that the change of basis matrix from $\{v_i\} to \{u_i\}$ is triangular i.e. for $k=1,2.., n$, $u_k= a_{k1}v_1+a_{k2}v_2+..+a_{kk}v_{k}$ The proof comes from applying the Gram ...


1

(1) Consider the map $$ \rho : S_n \rightarrow M_n(\mathbb{R}^n) $$ where $S_n$ is symmetric group. So if $A$ is a permutation, then it is an image of $\rho$. Let $\rho(t)=A$. And note that if $t=t_1\cdots t_m$ where $t_i$ is a single permutation, then $$ A_i:=\rho(t_i),\ A=A_1\cdots A_m$$ To show that $ADA$ where $D$ is diagonal, we suffice to show that ...



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