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(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...?

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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 ...

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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 $$A-I=\begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix},$$ 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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