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0

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 ...


0

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 ...


1

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 ...


2

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 ...


2

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 ...


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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. ...


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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?


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A corollary to Gershgorin's theorem states: More generally: let $M\in M_{n\times n}(\mathbb{R})$ be such that: \begin{equation} (\forall i\in {1,..,n}) (\sum_{j\neq i} a_{i,j}) < a_{i,i}. \end{equation} However, Gershgorin's theorem implies that each eigenvalue $\lambda$ of $M$ satisfies: \begin{equation} \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 $B\neq C$.


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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 $$A=\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}$$


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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 $\text{det}A = \pm 1$.


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$$\det{A^2}=\det{AA}=\det{A}\det{A}=(\det{A})^2=\det{I}=1$$ $$\therefore\det{A}=\pm 1$$


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, $p\ge m+n$.


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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 ...


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$\left({\begin{array} \ x_1 \\ x_2 \end{array}}\right)^T \left( {\begin{array} \ 0 & 1\\ 0 & 0 \end{array}} \right) \left({\begin{array} \ x_1 \\ x_2 \end{array}}\right),$ so we have to maximize $x_1x_2$ given that $x^Tx = 1.$ Here you can use Lagrange multiplier.


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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 ...


2

\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) = ...


1

$\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 get $x$ 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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