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If $n=1$, there's nothing to prove, as all matrices are scalar multiples of the identity. If $n\ge 2$, take two linearly independent vectors $v_1,v_2$ with corresponding eigenvalues $\lambda_1$, $\lambda_2$. Since every vector is an eigenvector, so is $v_1-v_2$ with corresponding eigenvalue $\lambda_3$. So then $A(v_1-v_2) = \lambda_3 (v_1 -v_2) = ... 3 Let$\rho=\max_{\|x\|=1}\|Ax\|$and$u=\arg\max_{\|x\|=1}\|Ax\|$. The case$\rho=0$(i.e.$A=0$) is trivial. Suppose$\rho\ne0$. Then$Au=\rho v$for some unit vector$v$and$\|Av\|\le\rho$. Since$\langle u,Av\rangle=\rho$, we must have$Av=\rho u$. Therefore,$(-\rho,u)$(when$v=-u$) or$(\rho,u+v)$(when$u+v\ne0$) is an eigenpair of$A$. Normalize the ... 3 Note that $$AW = \lambda BW \implies\\ (A - \lambda B)W = 0$$ Thus, to find$W$, we should simply ensure that each column$x$of$W$is a solution to the homogeneous system of equations $$(A - \lambda B)x = 0$$ In Matlab, use null(A - lambda * B) to find a basis to this solutions space. 2 call the eigenvectors$u_1, u_2$and$u_3$the eigenvectors corresponding to the eigenvalues$1, -2, $and$2.$then $$A = 1\dfrac{u_1u_1^T}{u_1^Tu_1} - 2\dfrac{u_2u_2^T}{u_2^Tu_2} + 2\dfrac{u_3u_3^T}{u_3^Tu_3}$$ you can verify this by computing$Au_1, \cdots$. this expression for$Ais called the spectral decomposition of a symmetric matrix. 2 Writing the matrix down in the basis defined by the eigenvectors is trivial. It's just $$M=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{array} \right).$$ Now, all we need is the change of basis matrix to change to the standard coordinate basis, namely: S = \left( \begin{array}{ccc} 1 & 1 & ... 2 It seems to me that this is pretty much just bash-it-out linear algebra using definitions. I'll sketch the reasoning and let you fill in the deets: First, let C=B^{-1}A for ease of notation. Now let's let \mathbf v be an eigenvector of C with eigenvalue \eta. First, (I+k\theta C)\mathbf v = (1+k\theta\eta)\mathbf v; multiply by the inverse of the ... 2 Hint: let \mathbf {S := (B)^{-1}A}, then the eigenvalues of \mathbf{I}+k\theta \mathbf S are the roots of the following polynomial: \begin{align} \lambda(\mathbf I + k\theta \mathbf S) &= \mathrm{det} (\lambda \mathbf I - (\mathbf I + k\theta \mathbf S)),\\ &= \mathrm{det} ([\lambda-1] \mathbf I - k\theta \mathbf S). \end{align} Let \eta := ... 2 They are always non-negative. Suppose \lambda is an eigenvalue of B^TB corresponding to an unit eigenvector v, then \langle v, B^T B v \rangle = \lambda = \langle Bv, B v \rangle = \|Bv\|^2. Hence \lambda \ge 0. 2 Let x an eigenvector of B^TB associated to the eigenvalue \lambda, which is real since B^TB is symmetric, then\lambda ||x||^2=\langle B^TB x,x\rangle=\langle Bx,Bx\rangle=||Bx||^2\implies \lambda\ge0$$2 You're computing the determinant of A, which is not the characteristic polynomial. The characteristic polynomial is rather$$ \det(A-XI)=\det\begin{bmatrix}a-X&-1\\0&a-X\end{bmatrix}=(a-X)^2 $$which has one root (equal to a) with multiplicity 2. The matrix is not diagonalizable, because$$ A-aI=\begin{bmatrix}0&-1\\0&0\end{bmatrix} $$... 2 I'm not sure this helps. Note that A is symmetric positive semidefinite. The biggest eigenvalue will be \max_{\|v\|=1} \langle v , Av\rangle . Since \langle v , Av\rangle = \sum_k (x_k^T v)^2, we can let X=\begin{bmatrix} x_1^T \\ \vdots \\ x_n^T \end{bmatrix} and then we have \langle v , Av\rangle = \| X v \|^2 and so \|A\| = \|X\|^2. 2 AFTERTHOUGHT: it occurs to me that there is something that does not need determinant, although the concept is implicit. As you can easily confirm, we have a matrix A such that \color{red}{A^2 = I}. Now, if you are willing to accept the proposition that every square matrix has an eigenvalue (possibly complex) then we can write$$ Av = \lambda v $$for ... 2 The problem is that you forgot to normalize your vector. The answer should indeed be$$ \pmatrix{1/3\\1/3\\1/3} $$1 Just to clarify for those who may see this question in the future, the actual matrix was$$A = \begin{pmatrix} -a & -b & -b\\ c & -d & 0\\ 0 & d & 0 \end{pmatrix}$$so that the polynomial to solve was$$\lambda^3 + (a+d)\lambda^2 + (ad + bc)\lambda + bcd,$$which has an obvious root \lambda = -d, and hence factors out. 1 For n=4, the eigenvalues are 4, -4, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, -2, -2, -2, -2, -2, -2, -2, -2. For n=5, the eigenvalues are the roots of (x - 5) \cdot (x + 5) \cdot (x - 1)^5 \cdot (x + 1)^5 \cdot x^{24} \cdot (x^2 - 5)^6 \cdot (x^2 - 5\cdot x + 5)^8 \cdot (x^2 + 5\cdot x + 5)^8 \cdot (x^2 - 2\cdot x - 4)^{10} \cdot (x^2 + 2\cdot x - ... 1 Note that A^*A is necessarily positive semidefinite. To show this, note that for any x \in \Bbb C^n,$$ x^*(A^*A)x = (x^*A^*)Ax = \|Ax\|^2 \geq 0 $$The result you showed regarding (x,Mx) is called Rayleigh's theorem. See also the min-max theorem (AKA the Courant-Fischer principle). 1 If the distinct eigenvalues of the adjacency matrix of a k-regular graph are k=\theta_1\ge\cdots\ge\theta_m, the eigenvalues of its Laplacian in non-decreasing order are $0, k-\theta_2,\ldots,k-\theta_m.$ So in this case the algebraic connectivity and the spectral gap coincide. If the graph is not regular then, in general, there is no simple or ... 1 Use Jordan canonical form. For a k \times k Jordan block B = \lambda I + N, where N^k = 0, \exp(tB) = \exp(t\lambda) \exp(tN) where \exp(tN) = I + tN + \ldots + t^{k-1} N^{k-1}/(k-1)! is polynomial in t. Thus if \text{Re}\; \lambda < 0, \exp(tB) \to 0 as t \to \infty. 1 You know that the matrix of the endomorphism is$$ A=\begin{bmatrix} 1 & 2 & a \\ 0 & 3 & b \\ -1 & 1 & c \end{bmatrix} $$for some a,b,c. Elimination gives$$ \begin{bmatrix} 1 & 2 & a \\ 0 & 3 & b \\ 0 & 3 & c+a \end{bmatrix} $$and, since we know that the rank must be 2, we can conclude that b=c+a. We ... 1 Let \Lambda_A and \Lambda_B be the spectra of A and B, respectively. If \mathrm{tr}(AB) was dependent only on the spectra of A and B, that is,$$ \mathrm{tr}(AB)=f(\Lambda_A,\Lambda_B), $$then for any orthogonal matrices U and V, \tilde{A}:=U^TAU and \tilde{B}:=V^TBV would be still SPD with the spectra \Lambda_A and \Lambda_B and ... 1 Suppose Av=\lambda v. We can scale v to have unit norm, so that v=(\cos\alpha,\sin\alpha). Writing out Av=\lambda v gives$$ (\cos\theta\cos\alpha+\sin\theta\sin\alpha,\sin\theta\cos\alpha-\cos\theta\sin\alpha) = \lambda(\cos\alpha,\sin\alpha). $$We can simplify the LHS:$$ (\cos(\theta-\alpha),\sin(\theta-\alpha)) = \lambda(\cos\alpha,\sin\alpha). ... 1 That transformationA$is a reflection$P_x$along the$x$-axis followed by a rotation$R_\theta$by an angle$\theta$: $$A = \left( \begin{matrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{matrix} \right) = \left( \begin{matrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{matrix} \right) \left( \begin{matrix} 1 ... 1 the matrix \pmatrix{\cos \theta & \sin \theta\\ \sin \theta & -\cos \theta} represents the reflection on a mirror along the line y = \tan (\frac{\theta}{2})\ x. therefore \pmatrix{\cos (\theta/2) \\\sin(\theta/2)} is an eigenvector corresponding to the eigenvalue 1 and \pmatrix{\sin (\theta/2) \\-\cos(\theta/2)} is an eigenvector ... 1 For any n \times n matrix A, the characteristic polynomial is$$c_A(\lambda) := \det(\lambda I - A) = \lambda^n + \text{(lower order terms in$\lambda)} ,which in particular has degree n. In general, the degree of the minimal polynomial c_A has degree \leq n, and is at least the number of distinct eigenvalues of A. 1 A minimal polynomial of a matrix and its characteristic polynomial have the same irreducible factors So the minimal polynomial of the matrix will be (x-a)^k(x-b)^l where 1\leq k\leq p;1\leq l\leq q also note that for two distinct eigen values their eigen vectors are linearly independent So the dimension will be k+l where k,l are in the given ... 1 Try this : Define T:\mathbb R^3\rightarrow \mathbb R^3 by T(1,0,0)=0;T(0,1,0)=(0,1,0);T(0,0,1)=(0,0,-1) NOTE:So the transformation becomes T(c_1,c_2,c_3)=(0,c_2,-c_3) 1 Suppose that B^{-1}Av = \eta v. Then, first \begin{align} (I-k(1-\theta)B^{-1}A)v & = v - k(1-\theta)B^{-1}Av\\ & = v-k(1-\theta)\eta v \\ & = (1-k(1-\theta))\eta v. \end{align} On the other hand, similarly, you can prove that(I+k \theta B^{-1}A) v = (1+k\theta\eta)v.$$This, in turn, implies$$(1+k\theta\eta)^{-1}v = (I+k \theta ... 1 Letf(x)=\langle Ax,x\rangle$. The gradient is better computed as a directional derivative: $$\langle\nabla f(x),u\rangle=\lim_{t\to 0}\frac{f(x+tu)-f(x)}{t}=\cdots=\langle Ax,u\rangle+\langle Au,x\rangle=\langle 2Ax,u\rangle,$$ where the dots are straightforward and the last equality follows because$A$is symmetric. Thus,$\nabla f(x)=2Ax\$. Now the ...