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we can rewrite the matrix equation as s system of two linear equations. they are \begin{align} (a-k)x + by &= 0\\cx + (d-k)y &=0\end{align} multiply the first equation by $(d-k)$, the second one by $b$, and subtracting gives $$\left((a-k)(d-k)-bc\right)x = 0.$$ we have two choices: (a) $x = 0$ or (b) $$(a-k)(d-k)-bc = 0.\tag 1$$ the first ...

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As an aside, it's not too hard to show that the space $O(n, \Bbb F)$ of $n \times n$ orthgononal matrices over the field $\Bbb F$ (which we take to be $\Bbb R$ or $\Bbb C$) is an $\frac{1}{2}n (n - 1)$-dimensional (resp., real or complex) manifold, which in particular means that for any orthogonal matrix $A \in O(n)$ there is a local homeomorphism between ...

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Yes, what you want is possible. Here is a conceptual explanation. Let $S_n$ be the symmetric group acting on vectors by permutations. The matrix $J$ commute with all the matrix of $S_n$. It follows that the eigenspaces of $J$ are invariant subspaces of the $S_n$ action. It is well-known that the action of $S_n$ has just two invariant subspaces, namely the ...

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Yes. Let $S$ be the subset of $\{-1, 1\}^{2n}$ where each vector $v$ in $S$ has $n$ $1$s and $n$ $-1$s. It suffices to show that $\mathrm{span}(S)$ is equal to the $2n-1$ dimensional subspace perpendicular to the vector $(1,1,\dots, 1)$. (As you've noted, the all one's vector $(1,1,\dots, 1)$ already spans the other one-dimensional eigenspace of $J$). To ...

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let us count the number of constraints: (a) to make the first column orthogonal to the remaining $n-1$ columns, you need $n-1$ constraints. al together one needs $(n-1)+(n-2) + \cdots + 2 + 1=\frac12 n(n-1).$ (b) to make all columns of length $1,$ one needs $n$ constraints. therefore, to make an $n \times n$ orthonormal matrix, you will have $$n^2 ... 0 I came across a solution to this question a while ago, I just did a little Google search but unfortunately I couldn't find the original post to give credits to the person who solved this problem. I try to restate his/her elegant answer to the best of my recollections. Let's assume the inequality is correct for real positive numbers a, b, c, i.e. ... 0 The matrix \begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix} is a counterexample. It is positive-definite since for (x,y) \neq (0,0) we have$$ \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 2x^2 - 2xy + y^2 = x^2 + (x-y)^2 > 0. $$On the other hand, its ... 1 HINT: What would happen if you multiplied this matrix by \pmatrix{1 \\ 1 \\ \vdots \\ 1}? What does this tell you? 1 A corollary to Gershgorin's theorem states: More generally: let M\in M_{n\times n}(\mathbb{R}) be such that: $$(\forall i\in {1,..,n}) (\sum_{j\neq i} a_{i,j}) < a_{i,i}.$$ However, Gershgorin's theorem implies that each eigenvalue \lambda of M satisfies: \lambda \in \cup_{i=1}^n Ball_{(\sum_{j\neq i} ... 3 Yes it is. Apply Gershgorin's Theorem. 0 If the matrix A has an inverse A^{-1} than AB=AC \Rightarrow B=C. But isf A is not invertible B and C can be different. As noteed in other answers you can have as example a matrix :$$ A=\begin{bmatrix}a&0\\0&0 \end{bmatrix} $$and you see that for$$ B=\begin{bmatrix}0&0\\b&0 \end{bmatrix} $$You have AB=0, so for different ... 1 If AB=AC then A(B-C)=0, and we can use the fact that the product of two non-zero matrices can be zero, so that both A is non-zero and B-C is non-zero, so that B\neq C. 0 You just have to find a counter-example, like$$A=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix},\quad B=\begin{bmatrix} 1\\0 \end{bmatrix},\quad C=\begin{bmatrix} 2\\0 \end{bmatrix}$$3 Hint: You can find a counterexample with$$A=\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}$$0 The implications give a directed graph with X \Rightarrow Y yielding an arrow from X to Y: From D you can hop to A, then to F and finally each E. In fact for this graph you can reach any point Y from any point X, and vice versa, thus X \Rightarrow Y \wedge Y \Rightarrow X thus X \iff Y for any pair X, Y. 0 yes. the determinant have the properties that$$det(AB) = det(A)det(B), det(I) = 1.$$therefore using the two properties we have,$$det(A^2) = \left(det(A)\right)^2 = 1 \implies det(A) = \pm 1.$$2 You have 1 = \text{det}I = \text{det}(A^2) = \text{det}A\text{det}A  It follows that \text{det}A = \pm 1. 5$$\det{A^2}=\det{AA}=\det{A}\det{A}=(\det{A})^2=\det{I}=1\therefore\det{A}=\pm 1$$0 Your first chain shows that a,b,c,d are all equivalent, because it forms a cycle of implications (because it begins and ends with a), hence any one implies any other. So now all you need to show is that at least one of them (doesn't matter which) is equivalent to e and f. This is because equivalency is transitive, i.e. if x is equivalent to z and ... 0 let rank(C) = c, rank(A) = a. let us look at the null space of M. we have$$(x, y)^\top \in \ker(M) \iff Ax + By = 0, Cy = 0 $$pick a particular y which has \text{number of columns of } C - c free variables in it. we can solve Ax = -By which will have \text{number of columns of } A - a free variables in it. therefore the total number of ... 0 Suppose, m=rank(A), n=rank(C), p=rank(M) The partial matrix A has m linear independent rows and the partial matrix C has n linear independent rows. If you fill up the rows of A with zeros, and take the zeros of the zero-block to C, you can easily see that you get m+n linear independent rows of M. So, p\ge m+n. 8 Yes. Equality holds iff AB = BA. Hint: Note that AB - BA is skew-Hermitian, and that$$ 2\operatorname{trace}[(AB)^2] - 2\operatorname{trace}(A^2B^2) =\\ \operatorname{trace}(ABAB + BABA -ABBA - BAAB)=\\ \operatorname{trace}[(AB - BA)^2] $$Note: The inequality assumes that both \operatorname{trace}[(AB)^2] and \operatorname{trace}(A^2B^2) are ... 1 Hints. The answer should be a closed disc of radius \frac12 centred at the origin of the Argand plane. Let x=\pmatrix{u\\ v}\in S^1 and y=\pmatrix{e^{i\theta}u\\ v}. Note that y lies inside S^1 too. What is the relationship between x^\ast Ax and y^\ast Ay? The numerical range of every matrix is compact and convex. So, in view of item 1, what ... 0 \left({\begin{array} \ x_1 \\ x_2 \end{array}}\right)^T \left( {\begin{array} \ 0 & 1\\ 0 & 0 \end{array}} \right) \left({\begin{array} \ x_1 \\ x_2 \end{array}}\right), so we have to maximize x_1x_2 given that x^Tx = 1. Here you can use Lagrange multiplier. 2 if the numerical range is defined as$$\{x^\top Ax : x^\top x = 1\},$$then with$$z^\top = (z_1, z_2), z^\top Az = (z_1, z_2)(z_2,0)^\top=z_1z_2 \text{ subject to } |z_1|^2 + |z_2|^2 = 1.$$let$$z_1 = \cos t \, e^{is_1},z_2 = \sin t \, e^{is_2}, s_1, s_2, t \text{ are real.}$$with that the numerical range of A is$$\left\{\frac12 ...

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\begin{gather*} \begin{bmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{bmatrix} \\ = \begin{bmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3 - a^3\end{bmatrix} \\ = \begin{bmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3 - a^3\end{bmatrix} \\ = (b^2-a^2)(c^3 - a^3)-(b^3-a^3)(c^2 - a^2) = ...

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$\left( \begin{array}{ccc} 1-\lambda & -1 & 2 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right)$ = $\left( \begin{array}{ccc} 2-\lambda & -2+\lambda & 0 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right) =(-2+\lambda) \left( \begin{array}{ccc} -1 & 1 & 0 \\ -1 ... 1 For any two ONBs there is a unitary mapping one to the other. So, if one knows $$\|T\|_2^2 = \sum|\langle e_i, Te_i\rangle |^2$$ and wishes to have an arbitrary pair of UNBs here, one can $$\|T\|_2^2 = \|UT\|_2^2= \sum|\langle e_i, UTe_i\rangle |^2= \sum|\langle U^*e_i, Te_i\rangle |^2$$ Choose$U^*$so that$U^*e_j=f_j$, and you have the claimed ... 0 Since$A-2I=\begin{pmatrix}0&0&0\\a&0&0\\a+3&a&1\end{pmatrix}$,$\;\;\text{nullity}(A-2I)=2\iff \text{rank}(A-2I)=1 \iff a=0$, so A is diagonalizable$\iff a=0$. 1 You could definitely get a valid answer using the diagonalization formula. However, I find that for questions like these, it's easier to find the eigenvectors with an "educated guess". In particular, try the basis $$\left\{ \pmatrix{1&0\\0&0}, \pmatrix{0&0\\0&1}, \pmatrix{0&1\\1&0}, \pmatrix{0&-1\\1&0} \right\}$$ 0 You'll get for$a \neq 0$, that the nullspace of$A-2I$is less than$2. To see why, you have that \begin{align*} A - 2I = \left(\begin{matrix} 0 & 0 & 0 \\ a & 0 & 0 \\ a+3 & a & - 3 \end{matrix}\right) \end{align*} is row equivalent to \begin{align*}\left(\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ a+3 & a ... 1 Assuming you got the correct set of equations, from the first equation we getx$can be anything, the second equation gives,$y$can be anything but$ax=0$. From here either we have$a=0$or$x=0$. If$a=0$, then the third equation gives$z=0$. But since$x$and$y$have no restrictions, therefore you can generate two independent eigen vectors by choosing ... 0 If you write$x=(x_1, x_2)^t$, with$x_1=\alpha+i\beta$,$x_2=a+ib$, then you're looking for $$1+2*max\{a\alpha+b\beta\}$$ given $$(a^2+b^2)(\alpha^2+\beta^2)=1$$ And you can conclude with Cauchy-Schwartz 0 Here is a much simpler way to do the calculation, if you are familiar with quadratic forms: Consider the quadratic form$Q[x]=x^TAx$given by this symmetric matrix. Then $$Q[x]=2x_1x_j+2x_2x_j+...+2x_{j-1}x_j+2x_{j+1}x_j+..+2x_{2n+1}x_j=2(x_1+..+x_{j-1}+x_{j+1}+..+x_{2n+1})x_j \\$$ Now, make the orthogonal change of variable $$y_1=\frac{1}{\sqrt{2n}} ... 0 I have not seen a solution that also give the eigenvectors, so here it goes: We use the standard notation of e_k being a vector in \mathbf{R}^{2n+1} with one on position k and zeros elsewhere. It is straight forward to show that the vectors$$e_1-e_k$$where k\not\in\{1,n+1\} give 2n-1 eigenvectors corresponding to eigenvalue zero. Let us give ... 0 You can fully characterize the eigenvalues as follows. There are 2n-1 zero eigenvalues; this follows because there are only two linearly independent columns. There is at least one eigenvalue of either +\sqrt{2n} or -\sqrt{2n}. You can see this through the characterization of the SVD: the first singular value is the largest possible norm of the image of ... 0 Write down det(A-xI) and you do basic determinant operation then you will find characteristics~polynomial from there you can find eigen values. Hint: The characteristics polynomial will be det(A-xI)=x^{2n+1}-2nx^{2n-1} From here you will find that eigen values of this matrix are 0~and~\pm \sqrt{2n} 0 The rank of your matrix is 2, which implies that \lambda=0 is an eigenvalue with multiplicity at least 2n+1-rank=2n-1. Now, if \lambda_1, \lambda_2 are the remaining eigenvalues, since tr(A) is the sum of the eigenvalues you get \lambda_1+\lambda_2=0. [I assume the row and column have the same index]. This Yields \lambda_2=-\lambda_1. ... 1 Pick rank-many independent rows from \begin{pmatrix}-AB&0\\B&1\end{pmatrix}. Some of these rows live in \begin{pmatrix}-AB&0\end{pmatrix} and some in \begin{pmatrix}B&1\end{pmatrix}, where they are still linearly independent hence certainly not more than \operatorname{rank}\begin{pmatrix}-AB&0\end{pmatrix} and ... 0 I think the answer is yes. Every matrix A \in \Re^{m\times n} should have a singular value decomposition, i.e.,$$A=U\Sigma V^T \tag{1}$$where \Sigma \in \Re^{d\times d} is a diagonal matrix with all non-zero singular values on the dianonal and U \in \Re^{m\times d}, V \in \Re^{n \times d}. d here denotes the # of non-zero singular values of ... 0 Sketch of the proof: The classical proof of the existence of the SVD is based on the fact that for any matrix A, there is a unit vector v such that$$\tag{1} \sigma=\|Av\|_2=\max_{w:\|w\|_2=1}\|Aw\|_2\geq 0 $$and using the induction. Given such a v, take a unit vector u such that Av=\sigma u. Then you construct two square unitary matrices:$$ ... 0 Your decomposition is even more than an SVD since it gives you not only singular values but eigen values of a square matrix. SVD is a bit more general and applies to rectangular matrices (not only square) and allow to give a decomposition with unitary matrices$M=U\Sigma V^*$(! not the same on left and right sides of the diagonal matrix, wich is not ... 0 I'll start with an explanation of "$*$". In general$A^*$is called the adjoint of$A$. This is defined relative to some inner product, often the Euclidean one. It is characterized by the equation$(x,Ay)=(A^*x,y)$. For the real Euclidean inner product, the adjoint is the transpose. For the complex Euclidean inner product, the adjoint is the conjugate ... 1 To answer your question about minimal polynomial and similar matrices, if you take the two following matrices: $$\begin{pmatrix} \lambda & 1 &0 &0\\ 0 & \lambda &0 &0\\ 0 & 0 &\lambda &1\\ 0 & 0 &0 &\lambda\\ \end{pmatrix}$$ $$\begin{pmatrix} \lambda & 1 &0 &0\\ 0 & \lambda &0 &0\\ 0 ... 1 This question is the complex counterpart of if matrix such AA^T=A^2 then A is symmetric? With the inner product \langle X,Y\rangle=\operatorname{Re}\operatorname{tr}(XY^\ast) defined on the real linear space M_n(\mathbb C), Hermitian matrices are orthogonal to skew-Hermitian matrices. Now, if we denote the Hermitian and skew-Hermitian parts of ... 0 Set A=(a_{ij}), a_{ij}\in K. Then A=\sum_{i,j}a_{ij}E_{ij} where E_{ij} is the n\times n matrix which have 1 on the (i,j)th entry and 0 otherwise. Now consider the F-subspace of K generated by all a_{ij} and choose c_1,\dots,c_m an F-basis for this. Then a_{ij}=\sum_{k}f_{ij,k}c_k with f_{ij,k}\in F. Thus ... 7 Let us first do a calculation (where we use the assumption)$$ (A-A^*)^*(A-A^*)=(A^*-A)(A-A^*)=A^*A-(A^*)^2-A^2+AA^*=AA^*-(A^*)^2. $$Now$$ ((A^*)^2)^*=A^2=A^*A, $$so$$ (A^*)^2=(A^*A)^*=A^*A. $$Inserting this into the calculation above,$$ (A-A^*)^*(A-A^*)=AA^*-A^*A $$But then the trace of that matrix on the left-hand side is zero, since (here we use ... 3 Hints: Note that the matrix A has rank 2. So, for n>2, A will have eigenvalue 0 with multiplicity n-2. Note that A is symmetric. It follows that the kernel of A is the orthogonal complement of its image. That is, the eigenspace associated with \lambda = 0 is the orthogonal complement of the space spanned by the vectors$$ ... 1 If$A$has at least one positive eigenvalue, then the maximal eigenvalue$\lambda_\max(A)$is positive and can be characterized as $$\tag{1} \lambda_\max(A)=\max\left\{\frac{x^*Ax}{x^*x}:x\neq 0\right\}$$ (I suppose you know that this holds; see the Courant-Fischer theorem.) Note that$x^*Ax$is real for any complex vector$x$. Since$\lambda_\max(A)>0$, ... 0 First, it is better to use normalized color components in range$[0,1]$. Second, we need to handle black color$(0,0,0)$as well. Here is one possible way to go: add a fourth component,$1$for all colors. Given four pairs of RGB-colors, let$T$be a transformation matrix,$U$is a non-singular matrix, which columns are the original colors, and$V\$ is a ...

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