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Note that any eigenvalue $\lambda$ of $A$ satisfies $Av = \lambda v \tag{1}$ for some vector $v \ne 0$, so that $A^2 v = A(Av) = A(\lambda v) = \lambda(Av) = \lambda(\lambda v) = \lambda^2 v, \tag{2}$ whence $0 = (A^2 + I)v = A^2v + v = (\lambda ^2 + 1)v, \tag{3}$ and since $v \ne 0$ this implies $\lambda^2 + 1 = 0; \tag{4}$ but equation (4) has no ...

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 $$\left(2-a-b\right)^2\geq ... 2 Well, the proof doesn't claim that v is an eigenvector, so there's no problem if it isn't. The vectors (v,Tv,\dots,T^nv) are linearly dependent regardless of the values of v and T. It doesn't matter whether v is nonzero. It doesn't matter whether T^i is nonzero. It doesn't even matter that T is a linear operator. Any set of n+1 vectors is ... 2 The equations can be written as \dot{p} = Ap, with p \in \mathbb{R}^3 and A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}. Note that A= v v^T -I, where v=(1,1,1)^T, so it has one eigenvalue at 2 corresponding to the eigenvector v, and two at -1 corresponding to the eigenspace \{v\}^\bot. (Note that ... 2 Your differential equation is of the form \vec u'=A\vec u, where \vec u=\begin{bmatrix} x\\ y\\ z\end{bmatrix} and A=\begin{bmatrix} 0 & 1 & 1\\ 1& 0 & 1\\ 1 & 1 & 0\end{bmatrix}. If there is a choice of eigenvectors of A that form a basis of \mathbb R^{3\color{grey}{\times 1}}, then, assuming the eigenvectors are v_1, ... 2 The reason why eigenvalues are so important in mathematics are too many. Here is a very short and extremely incomplete list of the main applications I encountered in my path and that are coming now in mind to me: Theoretical applications: The eigenvalues of the Jacobian of a vector field at a given point determines the local geometry of the flow and the ... 2 Every linear combination of EV_{1}=\pmatrix{1\\0\\0} and EV_3=\pmatrix{0\\1\\0} is a eigenvector with eigenvalue 1. EV_{1,3} = span\{\left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 1 \\ 1 \\ 0 \end{array} \right)\} is the same as EV_{1,3} = span\{\left( ... 2 Counterexample:$$ 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 } $$As for your newest question, note that Laplacian matrices are always positive semidefinite (i.e. have only non-negative eigenvalues). Since the sum of two positive semidefinite matrices is itself ... 1 Pretty sure you are wrong. My counterexample:$$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}$$1 Update: I have undeleted my answer because I think it is fixed now. You got$$ 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 ...

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} + ...

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