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1

Your differential equation is of the form $\vec u'=A\vec u$, where $\vec u=\begin{bmatrix} x\\ y\\ z\end{bmatrix}$ and $A=\begin{bmatrix} 0 & 1 & 1\\ 1& 0 & 1\\ 1 & 1 & 0\end{bmatrix}$. If there is a choice of eigenvectors of $A$ that form a basis of $\mathbb R^{3\color{grey}{\times 1}}$, then, assuming the eigenvectors are $v_1, ... 1 The equations can be written as$\dot{p} = Ap$, with$p \in \mathbb{R}^3$and$A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$. Note that$A= v v^T -I$, where$v=(1,1,1)^T$, so it has one eigenvalue at 2 corresponding to the eigenvector$v$, and two at -1 corresponding to the eigenspace$\{v\}^\bot$. Hence if ... 0 Different solution The first equation gives$$\ddot{x}=\dot{y}+\dot{x}$$ using the second and third one we get $$\ddot{x}=x+z+x+y=2x+y+z$$ Using the first $$\ddot{x}=2x+\dot{x}\\ \ddot{x}-\dot{x}-2x=0$$ This equation assumes$x=e^{mt}$and hence we get $$m^2-m-2=0\implies m=2,-1$$ and so $$x=c_1e^{2t}+c_2e^{-t}$$ Similarly $$y=c_3e^{2t}+c_4e^{-t}\\ ... 1 Update: I have undeleted my answer because I think it is fixed now. You got$$ V_{\lambda_2} = \left(\begin{array}{ccc} 0 \\ 1 \\ 1 \end{array} \right) $$correct but then copied it down wrongly.(I think..) Then you correctly wrote down the case \lambda_1. From$$ \left(\begin{array}{ccc } 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 ... 2 Every linear combination of$EV_{1}=\pmatrix{1\\0\\0}$and$EV_3=\pmatrix{0\\1\\0}$is a eigenvector with eigenvalue$1$.$EV_{1,3} = span\{\left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 1 \\ 1 \\ 0 \end{array} \right)\}$is the same as$EV_{1,3} = span\{\left( ...

0

It seems very doubtful that there could be a simple closed form for the eigenvalues in general (i.e. simpler than explicitly taking the characteristic polynomial and solving this quartic polynomial in radicals). Case in point: take $$a_{{1,1}}=-3,a_{{1,2}}=3,a_{{2,1}}=0,a_{{2,2}}=1,b_{{1,1}}=-3,b_{{1,2 ... 1 Well, multiply them on the left by the matrix, divide elementwise, and see if every resulting quotient is the same (within some numerical tolerance). In other words, be sure that Ax-\lambda x = 0. 2 I'm going to assume that D is symmetric. Let x be an eigenvector of A corresponding to eigenvalue \lambda. Let y=Dx. Then$$A'y = (A'D)(D^{-1} y) = DAx =\lambda Dx = \lambda y.So then y is an eigenvector of A'. 0 I'm not sure I understand your question but I am assuming you want to solve the trust region problem? \begin{align} \min_x\qquad & \frac{1}{2}x^TQx-b^Tx\\s.t.\qquad &x^Tx\leq \Delta\\ \mathcal{L}(x,\lambda)&=\frac{1}{2}x^TQx-b^Tx- \lambda(\Delta-x^Tx)\\ \nabla_x\mathcal{L}(x,\lambda)&=Qx-b+ 2\lambda x=0\\ b&=(Q+2\lambda I)x\\ ... 2 Let a = x^\dagger Ax, b = y^\dagger Ay, and c = x^\dagger Ay. The eigenvalues of B are determined by\lambda_\pm =\frac{a+b\pm\sqrt{(a-b)^2+4\vert c\vert ^2}}{2}.$$Since 0\leq a,b\leq 1 by assumption, then having 0\leq \lambda_{\pm}\leq1 is equivalent to$$(a+b)^2\geq (a-b)^2+4\vert c\vert^2\qquad (1)$$and$$ \left(2-a-b\right)^2\geq ...

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It's not really clear what you want. If it's just a list of graphs then: (a) Paths. (b) Cayley graphs for abelian groups. (c) Strongly regular graphs. There are other classes, but these would be the simplest.

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The statement is equivalent to asking whether $$A - I = -\left(\pmatrix{0&0\\0&I} + L\right)$$ Is necessarily invertible. My intuition is that this is not the case, but I'm still hunting for a counterexample.

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Let $J$ be the matrix $\begin{bmatrix}I&0\\0&0\end{bmatrix}$. Let's assume by absurd that the maximum eigenvalues of $A$ is $1$, then $$\label{eq} \left ( J-L\right )x_{max,A}=x_{max,A}$$ where $x_{max,A}$ is the maximum eigenvector of $A$ related to $\lambda_{max,A}=1$, that is the maximum eigenvalue of $A$. Right ...

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I'm not sure what you're looking for and this might just be a reiteration of that you did, but here it goes anyway. Let $\mu _1$ and $\mu_2$ be such that $\mu _1^2=\lambda _1$ and $\mu _2^2=\lambda _2$. Assuming $A=K\text{diag}(\lambda _1, \lambda _2)K^{-1} = KDK^{-1}$, it follows that \begin{align} A&=K\text{diag}(\mu_1, \mu_2)\text{diag}(\mu_1, ... 1 Pretty sure you are wrong. My counterexample:B=\begin{bmatrix}0 & 0 & 0 & 0\\ 0 & -1 & 1 & 0\\ 0&0&-1&1\\ 0 & 0 & 0 & 0\end{bmatrix} \\A=\begin{bmatrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$$2 Counterexample:$$ A = \pmatrix{ 1&0&0\\ 0&0&0\\ 0&0&0 },\; B= \pmatrix{ 0&0&0\\ 0&-1/2&1/2\\ 0&1/2&-1/2 } $$As for your newest question, note that Laplacian matrices are always positive semidefinite (i.e. have only non-negative eigenvalues). Since the sum of two positive semidefinite matrices is itself ... 1 Let A=\begin{bmatrix} 1 & 0\\ 0 & 2\end{bmatrix} and B=\begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}. Clearly the first equality always holds, however \begin{bmatrix} 1 \\ 1\end{bmatrix} is an eigenvector of B, but not of A. 0 As far as I'm concerned, the biggest deal about eigen-anything is that there are many canonical cases where the eigen-vectors/functions/states form a complete basis for the space that you're working in. For instance, in the case of Quantum Mechanics, the "eigenvectors" of operators which corresponds to observable quantities, like energy, position, etc., form ... 1 I think we can use the following inequality due to Weyl. Let A and B be Hermitian matrices with eigenvalues \lambda_i and \mu_i, respectively (indexed in descending order so that \lambda_1 and \mu_1 are the top eigenvalues). Let \nu_i denote the eigenvalues of A+B, indexed in the same way. Then:$$\nu_{i+j+1} \leq \lambda_{i+1} + ...

0

More generally, $AB$ and $BA$ will have the same non-zero eigenvalues. If $AB v = \lambda v$, and $\lambda \neq 0$, then $BABv = \lambda Bv$, and so $\lambda$ is an eigenvalue of $BA$ (since $Bv \neq 0$). Hence $AA^*$ and $A^*A$ have the same non-zero eigenvalues. Hence all other eigenvalues must be zero.

2

The reason why eigenvalues are so important in mathematics are too many. Here is a very short and extremely incomplete list of the main applications I encountered in my path and that are coming now in mind to me: Theoretical applications: The eigenvalues of the Jacobian of a vector field at a given point determines the local geometry of the flow and the ...

0

You know that $v_{1,1} = -v_{1,2}$. This is always the case, as your matrix is not inversible. If the matrix were inversible (ie, rows not multiple of each other) then you would find $(0,0)$ as the only solution (ie, no eigenvector !). This is when there is a mistake in the computation of eigenvalues.

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To find the geometric multiplicity of an eigenvalue $\lambda$, you want to find $nullity(A-\lambda I)=n-rank(A-\lambda I)$, where n is the number of columns of A. (Notice that you want to use n instead of rank(A).) Any basis of $\ker(A-\lambda I)$ will give you a set of linearly independent eigenvectors for the eigenvalue $\lambda$. In your example, you ...

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Regarding counting eigenvectors: Algebraic multiplicity of an eigenvalue = number of associated (linearly independent) generalized eigenvectors. That is, the characteristic polynomial of $A$ will be of the form $$p(x) = (x - \lambda_1)^{j_1} \cdots (x - \lambda_n)^{j_n}$$ and $j_i$ is the number of generalized eigenvectors associated with $\lambda_i$. ...

2

Well, the proof doesn't claim that $v$ is an eigenvector, so there's no problem if it isn't. The vectors $(v,Tv,\dots,T^nv)$ are linearly dependent regardless of the values of $v$ and $T$. It doesn't matter whether $v$ is nonzero. It doesn't matter whether $T^i$ is nonzero. It doesn't even matter that $T$ is a linear operator. Any set of $n+1$ vectors is ...

2

What we really want is to find a basis of the kernel of the transformation, that is, find a basis for the subspace of $\vec x = \pmatrix{x_1 & x_2 & x_3}^T$ such that $$A\;\vec x = \pmatrix{x_1 + 2x_2 + 3x_3\\0\\0} = \pmatrix{0\\0\\0}$$ Here's one way to do this: set $x_2 = 0$, this reduces to $$x_1 +2(0) + 3 x_3 = 0 \implies \vec x = ... 1 If you're curious how to get the row reduction method to work: \begin{bmatrix}3-\lambda & -2 \\1 & -1-\lambda\end{bmatrix} \sim \begin{bmatrix}1 & -1-\lambda\\3-\lambda & -2 \end{bmatrix} \sim \begin{bmatrix}1 & -1-\lambda\\0 & -2+(-1-\lambda)(-3+\lambda) \end{bmatrix}. This looks messy, but, as alex says, this matrix must be rank ... 1 You've gotten the eigenvalues correctly. Now, given a matrix A with an eigenvalue \lambda, an eigenvector for \lambda is just a non-zero element of the null space of A-\lambda I. So, let's find the null space of$$\begin{bmatrix}3 \strut& -2\\1 & -1\end{bmatrix}-\begin{bmatrix}1+\sqrt{2} & 0\\0 & ...

1

The SVD can be obtained by computing the eigenvalue decomposition of the symmetric matrix \begin{align} \begin{bmatrix} 0&X\\X^T&0 \end{bmatrix} &= \begin{bmatrix} U&0\\0&V \end{bmatrix}⋅ \begin{bmatrix} 0&Σ\\Σ^T&0 \end{bmatrix}\cdot \begin{bmatrix} U&0\\0&V \end{bmatrix}^T \\&= \frac1{\sqrt2} \begin{bmatrix} ... 1 Quoting the wikipedia article on SVD: The left-singular vectors of M (i.e. the columns of U) are eigenvectors of MM^T. The right-singular vectors of M (i.e. the columns of V) are eigenvectors of M^TM. The non-zero singular values of M (i.e. the non-zero diagonal elements of \Sigma) are the square roots of the non-zero eigenvalues of both ... 1 It's clear that 1 - r is an eigenvalue whose corresponding eigenspace is n - 1 dimensional (it's the null space of a matrix consisting only of 1s). On the other hand by inspection the column vector consisting only of 1s is an eigenvector with eigenvalue 1 + (n-1) r and the multiplicity must be 1 since we already have n - 1. 1 The matrix A that you have is symmetric. So it has an orthonormal basis of eigenvectors. The eigenvectors you have found are mutually orthogonal (which they must be because they correspond to different eigenvalues.) So, if you normalize your eigenvectors and make those normalized vectors the columms of a matrix U, then U^{T}U=I is automatic, and ... 0 To show a map is the zero map, show that it takes an entire basis to zero. Then it must take every vector, a linear combination of basis vectors, to zero by linearity. To show that your expression takes a basis to zero, compute:f(T)=(T-a_1)(T-a_2)(T-a_3)\dots(T-a_n)$$which is some operator and let it act on the only basis you've got: ... 0 Just apply f(T) to a generic linear combination v=\sum \lambda^i v_i. The last factor$$ (T-a_nI)\left(\sum \lambda^i v_i\right)=\sum \lambda^i (Tv_i - a_nv_i)=\sum \lambda^i (Tv_i - a_nv_i)=\sum \lambda^i (a_iv_i-a_nv_i)=\sum \lambda^i (a_i-a_n) v_i $$so now you have only n-1 coefficients left... 0 We think of f(T) as a map from V to V given by f(T)(v)=(T-a_1I)(T-a_2I)\dots(T-a_nI)(v). It is not hard to see that f(T) is a linear transformation. A nice fact about linear transformations is that their image is completely determined by where they send the vectors in a given basis. So, what is f(T)(v_i) for any i = 1,2,\dots,n? 1 Suppose V_x is the x-eigenspace, and V_y is the y-eigenspace, and 0 \ne w \in V_x \cap V_y. Then since w \in V_x, Aw = xw. Likewise w \in V_y, so Aw = yw. Then xw = yw so (x - y)w = 0. But x \ne y, so x - y \ne 0. Then multiplying through by (x - y)^{-1} yields w = 0, contradicting the assumption that w \ne 0. Thus 0 ... 0 Consider the quadratic form u^\dagger Au. Partition u into two subvectors x and y of equal lengths, i.e. u^T=(x^T,y^T). Then u^\dagger Au = \|M^\dagger x\|^2 + \|My\|^2 + 2\operatorname{Re}(x^\dagger Fy). Therefore A is positive seimidefinite only if \ker(M)\subseteq\ker(F) and \ker(M^\dagger)\subseteq\ker(F^\dagger). In particular, A ... 0 I think I can answer this myself. Since the matrix A is hermitian, the smallest eigenvalue is bounded above the smallest diagonal element. Since M M^\dagger and M M^\dagger are hermitian, I can diagonalize them with unitary matrices, U and V. Then I can transform A by  A^\prime = \begin{pmatrix} U^\dagger & 0 \\ 0 & V^\dagger \end{pmatrix} ... 0 The evaluation of the maximal eigenvalue can be reduced to a system of equations like the following:$$ \pmatrix{ 0&1&0& &0&0&0\\ 1&0&1&\cdots&0&0&0\\ 0&1&0& &0&0&0\\ &\vdots&&&&\vdots\\ 0&0&0&&0&1&0\\ ...

1

As I just realized, the problem has nothing special to do with primes. The question boils down to a subset of equation of the overall eigenvalue equation: $$P\cdot \vec x - \lambda \vec x =0$$ You can restrict this problem by picking the kind of connected (via reflection on the diagonal) vector components: $$\begin{eqnarray} x_2-\lambda x_1&=0\\ x_3+ ... 2 You can find the Eigenspace (the space generated by the eigenvector(s)) corresponding to each Eigenvalue by finding the kernel of the matrix A-\lambda I. This is equivalent to solving (A-\lambda I)x=0 for x. In your case: For \lambda =1 the eigenvectors are (1,0,2) and (0,1,-3) and the eigenspace is gen\{(1,0,2);(0,1,-3)\} For \lambda =2 ... 0 Denote A your matrix. To find the eigenspace of \lambda solve for X=(x,y,z)^T the equation$$AX=\lambda X$$1 The reason why it shows up in control theory is because the matrix A, while constant, will contain unknown parameter variables K_1,...,K_n. In this case a closed-form solution to the resulting differential equation, while theoretically available, is not so easily analyzed. What is most important in control theory is not finding the exact values of the ... 2 Without appealing to other theorems or arguments, your analysis is fine until you say that \lambda^{m-1} = 1 therefore \lambda = \exp 2\pi i \frac{k}{m-1}. The error here is that it excludes the possibility of repeated eigenvalues. That is, the set of all \lambda \neq 0 is not equal to the set (of unique values) generated by \exp 2\pi i ... 6 Since we have P^2=P then the polynomial x^2-x=x(x-1) annihilates P so the set of eigenvalues of P is a subset of the set of the roots of this polynomial:$$\operatorname{sp}(P)\subset\{0,1\}$$0 Let us denote by \sigma_p(T) the set of all eigenvalues of an operator T. Then, for any operator K acting on a complex space, we have$$\sigma_p(K^2)=\{ \lambda^2;\; \lambda\in \sigma_p(K)\}.$$As observed by Yiorgos, one inclusion is obvious: if \lambda\in\sigma_p(K) and if u is an associated eigenvector, then K^2(u)=K(Ku)=K(\lambda ... 0 No. Repeated Eigen values don't necessarily have repeated Eigen vectors. Counter Example:$$L =\begin{bmatrix}2&0\\0&2\end{bmatrix}\lambda_{1,2}=2v_1 =\begin{bmatrix}1\\0\end{bmatrix}v_2 =\begin{bmatrix}0\\1\end{bmatrix}$$0 Since the matrix is symmetric, it is diagonalizable, which means that the eigenspace relative to any eigenvalue has the same dimension as the multiplicity of the eigenvector. It doesn't make sense to speak about a “repeated eigenvector”; you can find a basis of the eigenspace, which is the null space of the matrix L-\lambda I (where \lambda is the ... 0 In your second matrix(e=3), suppose v=(v_1,v_2,v_3) then v_1+v_2+v_3=0 Let v_2=t,v_3=s\implies v_1=-t-s and the solution set is$$r=(-t-s,t,s)=t(-1,1,0)+s(-1,0,1) i.e. the eigenvectors are $x_2=(-1,1,0),x_3(-1,0,1)$ Note: any multiples of these vectors are also dependent eigenvectors

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Actually, you have $2$ different eigenvectors for which $(L-3I) v = 0$. One is $[1,0,-1]$ and one is $[1,-1,0]$, for example.

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