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$A^2=-I$ does not imply that $A=iI$. For example, one such $A$ is $$\left(\begin{array}{rr} 0 & -1 \\ 1 & 0\end{array}\right).$$ Now, for its exponential, note first that $$A^{2n}=(A^2)^n=(-I)^n=(-1)^nI$$ and $$A^{2n+1}=(A^2)^nA=(-I)^nA=(-1)^nA.$$ Then $$\exp(t A)=\sum_{n=0}^\infty\frac{t^n}{n!}A^n= ... 4 Your approach is only correct if n=m. If n\not=m, then A^{-1} does not exist. P is symmetric if P^T=P, so that's what you have to show.$$P^T=\left(A\left(A^TA\right)^{-1}A^T\right)^T=\left(A^T\right)^T \left(\left(A^TA\right)^{-1}\right)^TA^T$$... 3 If the square matrix A is positive definite then the hypersurface defined by$$Y=X^T.A.X$$is convex. Symmetric real-valued matrices are diagonizable, i.e., even when viewed as complex matrices their eigenvalues are all real. For general complex square matrices replace 'symmetric' with 'hermitean'. 3 No, there isn't (at least assuming f(A) is supposed to always be nonzero and n>1). For instance, let n=2 and consider the matrices A_t=\begin{pmatrix} 1+t & 0 \\ 0 & 1\end{pmatrix}. For t>0, f(A_t) must be a multiple of (1,0), and for t<0, f(A_t) must be a multiple of (0,1). By continuity of f, we get that f(A_0) ... 3 Q is non-singular, such that Q^{-1} exists and$$QRx=0 \iff Rx=Q^{-1}0 \iff Rx=0$$3 Hint. If Q = (q_{ij}) is orthogonal, then due to (Q^t Q)_{ii} = 1, we have$$ \sum_{i=1}^n q_{ij}^2 = 1 $$for all j. Hence if all q_{ij} are integers, at most one in each row/column is non-zero, namely \pm 1. So for each i, there is exactly one j = \pi(i) such that q_{i,\pi(i)} \in \{\pm 1\}. Now, the permutation \pi and the choice of the ... 2 If Ax=\lambda x, then (A^3-I)x=(\lambda^3-1)x. Hence, if \lambda is an eigenvalue, then \,\lambda^3-1=0, and thus \lambda=1 or$$ \lambda=-\frac{1\pm i\sqrt{3}}{2}. $$2 You are being asked to show that the non-zero rows of the reduced matrix form a basis of the row-space of A. They already span the row space. You are just left with linear independence. If the different non-zero rows are identified as A_{R\,i}, you must show that the only way to combine them to get 0 is with the trivial combination. Imagine you have a ... 2 I feel compelled to write this answer, based on the fact that the answer which was accepted makes no sense. What the OP mentions seems to be an embedding of \mathbb{R}^5 inside the space M of hermitian 2\times 2 matrices over the quaternions, which is Sp(2)-equivariant: where Sp(2) acts on \mathbb{R}^5 via the canonical double cover Sp(2) \to ... 2 An easy induction shows that$$ A^n=\begin{cases}\hphantom{-}I&\text{if}\enspace n\equiv 0\mod4\\ \hphantom{-}A&\text{if}\enspace n\equiv 1\mod4\\ -I&\text{if}\enspace n\equiv 2\mod4\\ -A&\text{if}\enspace n\equiv 3\mod4 \end{cases} There results that \begin{align*} \exp (\varphi A)&=\sum_{k=0}^\infty \frac{\varphi^{2k}}{2k!} ... 2 The positive definiteness (as you already pointed out) is a property of quadratic forms. However, there is a "natural" one-to-one correspondence between symmetric matrices and quadratic forms, so I really cannot see any reason why not to "decorate" symmetric matrices with positive definiteness (and other similar adjectives) just because it is in "reality" ... 2 Question: Let A=(a_{i,j})\in\{0,1\}^{n\times n}, find I\subset [0,\infty) such that\alpha \in I\qquad \iff \qquad \begin{cases}\alpha\ \rho(A)<1 \\ \alpha\ a_{i,j}<1 & \forall i,j=1,\ldots,n\end{cases}$$Answer: - If \rho(A)>0, then I=\big[0,\min\{1,\rho(A)^{-1}\}\big). - If \rho(A)=0 and A\neq 0, then I= [0,1) ... 2 Hint: if X^3 can be diagonalized as S D S^{-1}, then S D^{1/3} S^{-1} works, where D^{1/3} is the diagonal matrix whose diagonal elements are cube roots of the diagonal elements of D. 1 Hint: 1 Both sides satisfy the same differential equation and have the same initial conditions, so they agree. More explicitly, let f(x)=e^{iAx} and let g(x)=I\cos x + i A \sin x. These are matrix-valued functions of x. Compute the following:$$f'=iAe^{iAx},\quad f''=-e^{iAx},\quad g'=-I\sin x+i A\cos x,\quad g''=-I\cos x -iA\sin x.$$Therefore f''=-f and ... 1 HINT: You need solve this problem$$ \min_{x\in D}\frac{1}{2}\|x\|^2,\qquad D=\{x\in\mathbb{R}^n~:~x^{T}c-\beta=0\}. $$For this, define the Lagrangian by$$ L(x,\lambda)=\frac{1}{2}\|x\|^2-\lambda(x^{T}c-\beta). $$Imposing the condition L_{x}(x,\lambda)=0, we have x=\lambda c. As x must belong to D we have that \lambda=-\beta/\|c\|^2. So ... 1 Note: This answer is essentially based on this one by Jason DeVito. I have merely added some details. I assume A is real matrix. Note that it's rank as a real matrix equals its rank when considered as a complex matrix. So from now on we consider A as a complex matrix. It is proved here that all the eigenvalues of A are purely imaginary. Also, we ... 1 Hint: If you can find κ, λ such that$$κ\begin{pmatrix}1 & 1\\-1&1 \end{pmatrix}+λ\begin{pmatrix}1 & -1\\1&0 \end{pmatrix}=\begin{pmatrix}1 & 2\\3&4 \end{pmatrix} then, the answer is yes. You have $4$ equations with $2$ unknowns, the first of the equations being $κ\cdot1+λ\cdot1=1$.