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Set $X:=\begin{bmatrix} A & B\\ -B & A\end{bmatrix}$ and $Y:=\begin{bmatrix} BA & AB\\ -AB & BA\end{bmatrix}$. Note that $XY=\begin{bmatrix} ABA-BAB & A^2B+B^2A\\ -B^2A-A^2B & -BAB+ABA\end{bmatrix}=\begin{bmatrix} I & \bf 0\\ \bf 0 & I \end{bmatrix}$. So $XY=YX=\begin{bmatrix} BA^2-AB^2 & BAB+ABA\\ -ABA-BAB & ... 5 As$A$is normal, it is orthogonally diagonalizable. So, there is an orthogonal matrix$P$such that: $$A=P^*DP,$$ where$D$is a diagonal matrix, and$A^*=P^*D^*P$. Then there is a polynomial$p=p(x)=\sum_{k=0}^mc_kx^k$, such that:$p(D)=D^*$. If$D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$, then his polynomial takes$\lambda_j$to ... 5 The answer is YES. More precisely, we have: Proposition: Let$A$,$B$and$C$be positive semidefinite Hermitian matrices of the same size. If$D:=ABC$is Hermitian, then$D$is also positive semidefinite. Proof: Since$A$,$B$,$C$and$D$are Herimitian, $$D=ABC=CBA.$$ Firstly, suppose that$C$is invertible, so there exists a unique positive ... 4 This is another way, even though Mark Bennet is more elementary and perfectly clear. I'm using the properties of the function "$det$" defined over the set of square matrix with value in the base field. By Cauchy - Binet formula $$det(AB)=det(A)det(B)$$ So in this problem, with an iterated reasoning, $$det(A^{k}) = det(A)^{k}$$ Now, you need to prove ... 3 Let us write $$G_{m} = G = mA (B + mA^{*}A)^{-1}$$ in order to clarify that$G_{m}$depends on$m$. Then we claim that$G_{m} \to H$in operator norm. To this end, we first check the following two simple lemmas: Lemma 1. Let$J = (AB^{-1}A^{*})^{-1}$. Then we have $$G_{m} = J \left( AB^{-1} - \frac{1}{m}G_{m} \right) = H - \frac{1}{m} JG_{m}. ... 3 There's no equality that looks like this, no. I'll give a reason below. There's an inequality:$$ \| M \cdot x \| \le \|M \| ~ \| x \|. $$But when you see the definition of \|M\|, it's pretty disappointing:$$ \|M \| = \max_{\|x\| = 1} \| M\cdot x \|. $$What about the first part? Well, look at$$M = \begin{bmatrix}1 & 0 \\ 0 & 0 ... 3 Assume that$(I+A-P)x=0$for some vector$x$then$x+Ax=Px$. Note that$PA=A$because$P$is a transition matrix hence$P^kx+Ax=P^{k+1}x$for every$k\geqslant0$, that is,$P^{k+1}x=x+(k+1)Ax$. Each coordinate of$P^{k+1}x$stays bounded when$k\to\infty$since every$P^{k+1}$is a transition matrix, hence$Ax=0$and$Px=x$. Since$P$is irreducible, its ... 3 Usually when we talk about the determinant without any other information attached, the only really relevant information is whether or not it is zero. However, if we are interested in geometry, there is some significance to matrices with determinant$1$. Namely, an important subset of them form the so-called special orthogonal group, which is just a fancy way ... 3 NOTE: My original answer was about a different question. (I have considered only adding integer multiples of some row.) I have completely changed it, hopefully the new answer is correct. Other answers have already gave explanations about geometric meaning of$\det(A)=1$, so let me address the questions about row operations. Let us start by checking what ... 3 Because$\varphi$is is analytic, it is enough to show that it agrees with the exponential in some small neighbourhood of zero. Let$\varepsilon>0$such that$\varphi$maps any trace-zero matrix with all entries less that$\varepsilon$is absolute value to$SL(n)$. In particular, for any$s$with$|s|<\varepsilon$, $$... 3 If you can find a pair of matrices that don't commute for N, then for all n \geq N you can take those two matrices as upper left blocks in a matrix where the rest of the columns are fixed ( geometrically, if you've found two linear transformations for dimension N, then for any n > N, perform those linear transformations on a subspace of dimension ... 2 Yes, if A is triangular matrix and invertible, then the inverse A^{-1} of A is also triangular. To show this, note that (see here)$$adj(A)\cdot A=\det(A)\cdot I$$where adj(A) is the adjugate of A and I is the identity matrix with same size as A. Therefore, we have$$A^{-1}=\frac{1}{\det(A)}adj(A).$$Since A is triangular, we can show that ... 2 It's only true if A is symmetric. And as for intuition, consider the one-dimensional case: the derivative of ax^2 is 2ax. I always recommend to write out the quadratic form and calculate the derivative by hand. Once you've done that, you'll understand and you'll never forget it anymore. 2 Let L_A be left multiplication by A. Then you can split F^{n\times n} into the direct sum of L_A-invariant n copies of F^n (the columns of M), on each of which L_A acts by the usual multiplication of a matrix by a column vector. Hence \det(L_A)=[\det(A)]^n. Similarly, \det(R_B)=[\det(B)]^n. Hence ... 2 The derivation becomes a lot simpler if we take the derivative with respect to the entire x in one go:$$\frac{\delta}{\delta x}(Ax-b)^T(Ax-b) \ \ = \ \ 2(Ax-b)^T\frac{\delta}{\delta x}(Ax-b) \ \ = \ \ 2(Ax-b)^TA$$This follows from the chain rule:$$\frac{\delta}{\delta x}uv \ \ = \ \ \frac{\delta u}{\delta x}v+u\frac{\delta v}{\delta x}$$And ... 2 Actually we do not need quaternions, because we are working only with one rotation so we can assume that R is rotation around z-axis. We can restrict ourselfs only to xy-plane. Rotations in 2d can be expressed by unit complex numbers and skew-symmetric matrices correspond to pure imaginary numbers. Cayley transformation for skew-symmetric matrices:$$ ... 2 but I found it to be wrong Then you must have made a mistake somewhere. If you post your counterexample, we could find out which mistake. Let$\beta$be the symmetric bilinear form. Then there are two possibilities,$\beta(v,v) = 0$for all$v \in K^2$, or$\bigl(\exists v_1\in K^2\bigr)\bigl(\beta(v_1,v_1)\neq 0\bigr)$. In the first case, pick any ... 2 They are correct. What the characteristic polynomial $$x^2(x-1)(x+1)$$tells you is that there are at least$3$cages, of which the cages for eigenvalues$1$and$-1$have size$1$. This only leaves the eigenvalue$0$, for which you have$2$options: There are$2$linearly independent eigenvectors for the eigenvalue$0$. In that case, your matrix has$4$... 1 First note that for any matrix$A$, the matrix$A^*A$is self-adjoint and hence it is diagonalizable. Let$diag(\lambda_1, …., \lambda_n)$be the diagonal matrix where$\lambda_j$is the$j$-th eigenvalue of$A^*A$. Then we can let$\sqrt{A^*A} = diag(\sqrt{\lambda_1}, …, \sqrt{\lambda_n})$. Now these values,$\sqrt{\lambda_j}$are real and are known as ... 1 Since$\|Ax\|^2 = \langle x , A^* A x \rangle $, and$A^* A \ge 0$and is Hermitian, we have$\|Ax\|^2 =\langle x , A^* A x \rangle \le \lambda_\max \|x\|^2$and the maximum is attained for an eigenvector corresponding to the maximum eigenvalue. That is,$\|Ax\| \le \sqrt{\lambda_\max} \|x\|$, or${ \|Ax\| \over \|x\| } \le \sqrt{\lambda_\max}$if$x \neq ...

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using linearity with respect to the first column, those matrix have the same determinant: \begin{bmatrix} b+c & a^2 & a \\ c+a & b^2 & b \\ a+b & c^2 & c \\ \end{bmatrix} \begin{bmatrix} b+c + a& a^2 & a \\ c+a + b& b^2 & b \\ a+b +c & c^2 & c \\ \end{bmatrix} whose determinant is ...

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Normally, say if $M$ is diagonalizable, one can think of $M = V^{-1}DV$ as $\vec{x}_1 = V \vec{x}$ is rotating the space. $\vec{x}_2 = D \vec{x}_1$ is rescaling in the newly rotated space. $\vec{x}_3 = V^{-1} \vec{x}_2$ is rotating the space back to the original coordinates.

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And yet another way of looking at this: Saying that an element $a$ is invertible in a ring $R$ is equivalent to saying that multiplication by that element $R → R, x ↦ ax$ is invertible/bijective. If $A^{2014}$ is invertible, then multiplying by $A^{2014}$ is bijective which is the $2014$-fold self-composition of multiplying by $A$. So multiplying by $A$ ...

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Induction is easily avoided. Suppose that vector $u = \sum_i^n c_i v_i$ has that unique representation in terms of ordered basis $\mathscr{B}=\{v_1,\ldots,v_n\}$. Now $id(u)=u$ has that representation and $I_n (c_1,\ldots,c_n)^T = (c_1,\ldots,c_n)^T$. Thus $I_n$, the identity matrix, represents $id$, the identity map, with respect to any ordered basis.

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The displayed inequality is a direct consequence of Youla decomposition of complex skew-symmetric matrices. More specifically, we can always decompose a complex skew-symmetric matrix $A$ as $U(D_1\oplus\cdots\oplus D_k)U^\ast$, where $U$ is a unitary matrix and every $D_j$ is either a $2\times2$ complex skew-symmetric matrix or a zero block. As  ...

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Just because you have a certain multiplicity of an eigenvalue does not mean you only have one Jordan block containing that eigenvalue! For example if you had an eigenvalue of multiplicity $4$ that could be $1$ block of $4$, it could be $1$ block of $3$ and $1$ of $1$, it could be blocks of size $2$ and $2$, it could be $2$, $1$, and $1$, or finally it could ...

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The most straightforward way to see that your answer is incorrect is to calculate the rank of $A$, which is $1$, since the columns (equivalently, the rows) span a space of dimension $1$. Your answer has rank $2$. Your argument is that a $2\times 2$ Jordan matrix with two linearly independent eigenvectors having eigenvalue zero must be ...

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Ok so I figured it out. Notice $w^\top X y = y^\top X^\top w$. So continuing the above expression: $p(w|X,y) \propto exp(-{1 \over 2}(2\sigma_n^{-2}y^\top X w -w^\top(\sigma_n^{-2}XX^\top -\Sigma_p^{-1} )w))= exp(-{1 \over 2}(w^\top A^{-1}w-2 \bar w^\top A^{-1}w))$ where $A=(\sigma_n^{-2}XX^\top -\Sigma_p^{-1})^{-1}$ and the answer follows.

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Thomas, here there are no traces. I assume that $h:M_n\rightarrow M_n$ and $||.||$ is the euclidean norm on $\mathbb{R}^n$. The derivative of $h$, in the point $XW$, is denoted by $D_{XW}h\in L(M_n,M_n)$ and is a jacobian. If $f:W\rightarrow ||h(XW)\alpha-y||^2$, then $D_{W}f:H\in M_n\rightarrow \mathbb{R}$ is a linear application. If $g:W\rightarrow ... 1 When working with the Frobenius norm, we can stack the columns of$X$into a single vector$\widehat X\in \mathbb R^{n^2}$and take the norm of that. This simplifies the computation: the gradient is$2\widehat X$and the Hessian is$2I$where$I$is the identity matrix acting on$\mathbb R^{n^2}$; i.e., the identity matrix of size$n^2\times n^2\$. If you ...

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