# Tag Info

3

Well, if $A$ is an $n \times n$ matrix, the rank of $A$ plus the nullity of $A$ is equal to $n$; that's the rank-nullity theorem. The nullity is the dimension of the kernel of the matrix, which is all vectors $v$ of the form: $$Av = 0 = 0v.$$ The kernel of $A$ is precisely the eigenspace corresponding to eigenvalue $0$. So, to sum up, the rank is $n$ minus ...

2

Hint (For (b).) Suppose that there are (at least) two nonpositive eigenvalues, and let $\Bbb W$ denote the span of their eigenspaces. What can we say about $v^T A v$ for $v \in \Bbb W$?

0

I believe the lower bound on the eigenvalues is 0. The eigenvalues of $\Gamma=\left[\begin{array}{cc} I & B \\ B^{*} & I \end{array} \right]$ are $1+\beta$ where $\beta$ is an eigenvalue of $B$. Every eigenvalue has a multiplicity of 2 (or, rather, twice the multiplicity of the corresponding eigenvalue of $B$). The eigenvectors are ...

0

An eigenvector $v$ of a matrix $A$ is a directions unchanged by the linear transformation: $Av=\lambda v$. An eigenvalue of a matrix is unchanged by a change of coordinates: $\lambda v =Av \Rightarrow \lambda (Bu) = A(Bu)$. These are important invariants of linear transformations.

0

Unless the smallest eigenvalue of $A$ is $\ge$ largest eigenvalue of $B$ you can do the same trick with diagonal matrices that you did above. Hence the condition is : $\lambda_n(A) \ge \lambda_1(B)$.

1

I wasn't able to figure out an exact formula for eigenvalues and eigenvectors for a pentadiagonal Toeplitz matrix. Knowing them for a tridiagonal Toeplitz, though, is very helpful. (Those can be found, for example, in this paper on the sensitivity of the spectrum of a tridiagonal Toeplitz matrix.) As you've already discovered, the eigenvalues are ...

2

First answer: Your operator can be written in terms of the usual derivative operator. We have $L=(\frac{d}{dx})^2+3\frac{d}{dx}-4$, where $4$ stands for multiplication by the constant. From the point of view of linear algebra, $L$ is a linear operator/mapping/function from a vector space to itself. Second answer: A square matrix as we usually understand it ...

1

(1) No, there is nothing uniquely special about this $L$. (2) The notions of eigenvalues and eigenvectors are defined for any linear map from a vector space $\Bbb V$ to itself: A (nonzero) vector $v$ is an eigenvector, of eigenvalue $\lambda$, for the linear transformation $T: \Bbb V \to \Bbb V$ iff $$T(v) = \lambda v.$$ If $\Bbb V$ is finite-dimensional, ...

1

Hint: $Tr(A) = A_{11} + A_{22} = 4$ (sum of eigenvalues) Another Hint: $Det(A) = A_{11} * A_{22} - A_{12} * A_{21}= 13$ (product of eigenvalues)

1

Hint: What is the trace and determinant of the matrix?

1

Hint: If $a$ and $b$ are real numbers, what are the eigenvalues of $\begin{bmatrix}a & -b \\ b & a\end{bmatrix}$? Alternatively, notice that $\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$ (the matrix which represents a rotation by $180^{\circ}$) has eigenvalues $\pm i$. Can you manipulate this matrix into one which has eigenvalues $2 \pm 3i$? ...

1

Hint: start by finding the characteristic polynomial of $A$, and try working from there.

1

There is no need to take normal forms, you just have to check that we have positive definiteness by its definition. Let $M=A^T A+I$. Obviously $M=M^T$ and we have: $$\forall v\neq 0,\qquad \langle v,Mv\rangle = \|v\|^2 +\|A v\|^2\color{red}{>}0$$ as well as: $$\inf_{v:\|v\|=1}\langle v,Mv\rangle \geq 1.$$

0

Take $p\ge q$ and $rank(A)=p$, i.e., $A$ has full rank. Then $AA^T$ is invertible and positive definit. Let $\lambda_1$ be the smallest eigenvalue of $B$. Let $\sigma$ be the smallest eigenvalue of $AA^T$, i.e., which is the square of the smallest singular value of $A$. Now take $x\in \mathbb R^q$. Then $$x^TXx = x^TABA^Tx \ge \lambda_1 \|A^Tx\|^2 = ... 5 Here's a useful fact: if  x  is an eigenvector of  M  with eigenvalue \lambda, then  x  is an eigenvector of  M + c I  with eigenvalue \lambda + c . Proof:  (M + c I) x = Mx + cx = (\lambda + c) x.  (Similarly, if  x  is an eigenvector of  M + c I  with eigenvalue \lambda , then  x  is an eigenvector of  M  with eigenvalue \lambda ... 5 Well, since A^T A is positive definite, it is diagonalisable with positive real eigenvalues. That is, if A is n \times n, then for some n \times n invertible matrix P, we have:$$PA^TAP^{-1} = \left(\begin{matrix} \lambda_1 & 0 & \ldots & 0 \\ 0 & \lambda_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 ...

1

This seems to fall under eigenvalue perturbation problem. The following assumes that eigenvectors $x_{0j}$ of the unperturbed matrix are chosen such that $||x_{0j}||^2 = x_{0j}^{*} x_{0j} = 1$. An eigenvalue of perturbed matrix is $\lambda_j = \lambda_{0j} + \mathbf{x}_{0j}^{*} \delta A \mathbf{x}_{0j} + o(||\delta A||)$, where $\lambda_{0j} = -2 - 2 \cos ... 1 Hints:$B$is diagonalized by the same$S$. The entries of$D$correspond to the eigenvalues of$A$. 2 Hint:$B$is diagonalized by the same$S$. 0 It depends on$w$:) The eigenvalues of$uv^T$are$v^T u, 0, \dotsc, 0$for any vectors$u,v$. Notice that$\max(B) = ee^T \max(w)^2$, where$e = (1,\dotsc,1)^T$. So we have $$\lambda_\max(F\max(B)F^T) = \max(w)^2 e^T F^T F e$$ and $$\lambda_\max(FBF^T) = w^T F^T F w.$$ In case$w$is a normed eigenvector to the largest eigenvalue of$F^TF$then, we ... 0 I don't know if the following answer addresses your issue, but I'm posting it due to no answers. If$A$is a$n\times n$real symmetric matrix, then all its eigenvalues are real and$A$is diagonalizable, with$ \mathbf\Lambda = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)$and$\lambda_i,\, i =1,\ldots,n$not necessarily distinct. In any case we ... 0 The result as you quoted it is wrong. Perhaps the book has a misprint. The correct statement is that in a$k$-dimensional subspace$K$of the$n$-dimensional space$H$, there are unit vectors$x, y$with$\langle x, Ax \rangle \le \lambda_k$and$\langle y, Ay \rangle \ge \lambda_{n+1-k}$. In this case with$n=5$,$k = 2$,$\langle x, Ax \rangle \le 4$... 2 The answer of @Hippalectryon is morally correct. However, there is this problem of approaching a polynomial with multiple roots, somehow delicate.$\tiny{\text{Can be proved with the argument principle from complex analysis.}}$We'll work directly with the eigenvalues, bypassing the characteristic polynomial. Use this standard observation: if two symmetric ... 8 Let$\chi_n$be the characteristic polyomial of$A_n$and$\chi=a\prod(X-\lambda_i)$be that of$A$, then$\chi_n\mathop{\to}_{n\rightarrow\infty}\chi$. If a sequence of polynomials converges to another, then by continuity the roots of the polynomials in the sequence must converge to the roots of the limit polynomial (with the same multiplicity). Hence the ... 1 The first question seems to be handled quite well by other people but I just want to add one more thing. Positive-definiteness always starts from symmetry. So the converse can never be true. That being said, it's only when we have the guarantee that the matrix is symmetric that we can conclude a matrix is positive-definite. 2 consider on this $$\begin{bmatrix} 1 & 2 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 3 \\ \end{bmatrix}$$ 2 No. The simplest example is any upper triangular matrix whose diagonal entries are real: $$\begin{pmatrix} 1 & 1-i & 2 \\ 0 & 2 & 3i \\ 0 & 0 & 3 \end{pmatrix}$$ has complex entries in the non-diagonal spots but its eigenvalues are$1,2,3$and it is certainly not symmetric! 1 Is the matrix $$\begin{pmatrix}1&1\\0&1\end{pmatrix}$$ symmetric? It has only one positive eigenvalue of multiplicity two. 1 Also see Horn's theorem. It says that two real$n$vectors$\bf{x}$,$\bf{y}$are the eigenvalues and the diagonal of a (real) symmetric matrix if and only if$\bf{x}\succ \bf{y}$in the Schur order, that is \begin{eqnarray}x'_1 &\le &y'_1\\ x'_1 + x'_2 &\le& y'_1 + y'_2 \\ \ldots \\ x'_1 + \ldots + x'_n &=& y'_1 + \ldots + y'_n ... 3 If lowest means the same as minimum to you, then yes. Here is a sloppy outline, I will leave it to you to fill in the gaps. For a real symmetric matrix$n\times n$matrix$A$, consider the Rayleigh quotient $$R_A(u) = \frac{u^\top A\,u}{u^\top u}\quad\text{for}\quad u\in\mathbb{R}^n\setminus\{0\}$$ Every eigenvector$v$is a stationary point of that ... 0 If$m_2<0$, then there is no maximum value of$\lambda$. Indeed, there is$u>0$s.t.$\lambda>u$implies that$m>0$and$Q$is symmetric$<0$. Let$f(x)=x^TQx+x^TB+m$; then$f(0)>0$and, for sufficiently great$||x||$,$signum(f(x))=signum(x^TQx)=-1$. 2 If the given eigenvalues are all distinct, then you can take the companion matrix of the polynomial$(x-\lambda_1)\cdots(x-\lambda_n)$, which has real coefficients because the roots occur in conjugate pairs. 3 In this solution, we only assume that$B$is diagonalizable (i.e., the eigenvalues$\lambda_i$'s need not be distinct). If$v_1,v_2,\ldots,v_n$are the eigenvectors of$B$and$w_1,w_2,\ldots,w_n$are the left eigenvectors of$B$, where$Bv_i=\lambda_iv_i$and$w_i^\top B=\lambda_i w_i^\top$for$i=1,2,\ldots,n. Then, \begin{align} F\left(v_i ... 4 If eigenvalue is a \pm bi, the matrix is \begin{bmatrix} a & -b \\ b & a \end{bmatrix}. $$For more eigenvalue pairs, place 2 \times 2 blocks like this one down the diagonal of your 2k \times 2k matrix. 5 Let e_i be the eigenvectors of B, i.e.$$Be_i=\lambda_ie_i$$and let E_{ij}=e_i e_j^T be the elementary matrices in this basis, i.e.$$E_{ij}e_k=\delta_{jk}e_i.As it turns out, F is already diagonal in the E_{ij}-basis: \begin{align} &BE_{ij}e_k=\delta_{jk}Be_i=\lambda_i\delta_{jk}e_i=\lambda_iE_{ij}e_k\\ ... 1 An example of such matrix is the companion matrix of the polynomial, which in this case is\begin{pmatrix} 0 & 0 & 0 & -D\\ 1 & 0 & 0 & -C\\ 0 & 1 & 0 & -B\\ 0 & 0 & 1 & -A \end{pmatrix}.$$2 The characteristic polynomial of H is$$ p(\lambda)=|\lambda I-H|=(\lambda-1)^n+Tr(vv^T)(\lambda-1)^{n-1} $$for vv^T is a Rank-1 matrix and all principal minors above 2 are 0. Let v=\pmatrix{v_1\\\vdots\\v_n}. Then$$ vv^T=\pmatrix{v_1v_1\cdots v_1v_n\\ \vdots \hspace{15 mm} \vdots \\ v_nv_1 \cdots v_nv_n} $$So ... 1 Householder matrices are orthogonal and symmetric. That is:$$ (I - 2uu^T)^T = I - 2uu^T  (I - 2uu^T)^2 = I - 4uu^T + 4u(u^Tu)u^T = I $$Here u^T u = 1 since u is a unit vector. Eigenvalues of orthogonal matrices have absolute value 1, since multiplication by an orthogonal matrix is an isometry (length preserving). Since the Householder ... 1 Pick any vector w \in V where V is an inner product space of dimension n ; assume the basis has been chosen so that v^{\top} w = \langle w,v \rangle. Then$$ Hw = w - v v^{\top} w = w - \langle w,v \rangle v. $$It follows that if v has norm 1, H is the orthogonal projection operator to the (n-1)-dimensional subspace of V orthogonal to ... 2 The characteristic polynomial of H is$$ p(\lambda)=|\lambda I-H|=(\lambda-1)^n+Tr(2uu^T)(\lambda-1)^{n-1} $$for uu^T is a Rank-1 matrix and all principal minors above 2 are 0. Since Tr(2uu^T)=2Tr(u^Tu)=2$$p(\lambda)=(\lambda+1)(\lambda-1)^{n-1}$$Or eigenvalues of H are ±1. 0 Just compute Hu and Hv where v is orthogonal to u. And remember that eigen spaces of different eigenvalues are orthogonal (for symmetric matrices). 2 As remarked in the comment, the construction is identical to the case where one minimizes R(u). First of all, observe that$$\tilde R(u) = R(u) + \frac{\int_M \alpha |u|^2 }{\|u\|^2_{2}} \ge R(u) - \|\alpha\|_\infty \ge \lambda_1 - \|\alpha\|_\infty.$$Thus \tilde R is bounded below. So we can take u_1, u_2, \cdots \in H, so that \|u_i\|_2 = 1 ... 0 First, there is no advantage to trying to compute only the real eigenvalues versus just computing them all; they could all be real or all be complex. Second, the QR algorithm as you've written it converges extremely slowly compared to state-of-the-art implicitly shifted Hessenberg QR. If you use eig() in Matlab, it will use the faster algorithm and almost ... 0 Let your matrix be A = \lambda u u^T + \mu v v^T, and assume that \lambda \gg \mu. That is, assume your matrix is rank 2, where the only two nonzero eigenvalues are \lambda and mu, and one is much bigger than the other. Assuming you used the power method to compute an eigenpair, you should compute \tilde \lambda and \tilde u (different than the ... 0 Since you want to put bounds on the eigenvalues of such matrices, I think that using Gershgorin's Circle/Disc Theorem is helpful. This theorem says that if A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots ... 1 I took a probabilistic approach to the problem. I wanted to say something like if you choose a "random vector" x then Q(x) will "typically be" c\|x\|^2. I decided on typically be meaning the expected value of Q(x). So I want to show$$E[Q(X)] = c\|X\|^2$$for some c. This also requires a distribution for a random vector X. I would like ... 4 Hint. For a j\times j matrix M we have$$ \det\big(\DeclareMathOperator{adj}{adj}\adj(M)\big)=\det(M)^{j-1} $$Suppose A is n\times n. Then$$ \det\big(\adj\big(\adj(A)\big)\big) = \det\big(\adj(A)\big)^{n-1} = \det(A)^{(n-1)^2} $$Now, since the determinant of A is the product of the eigenvalues of A we know that \det(A) is a positive ... 1 Although @Bey was correct, we can state a more general result in a language natural to matrices. Let us consider the Schatten p-norm with the k largest singular value for p = 1 or p = 2. Then, after changing to an orthonormal basis of A, we may assume A to be diagonal and the problem reduces to \ell_p approximation of k largest Eigen values ... 3 Since$$\mathbf{Q}(\mathbf{x})=\sum_{i=1}^n \sum_{j=1}^n a_{ij}x_ix_j$$We can represent \mathbf{Q}(\mathbf{x}) as a vector \mathbf{q} \in \mathbb{R}^d,d=\frac{n+n^2}{2} with the quadratic basis functions \{x_1^2,x_1x_2,...,x_n^2\}:$$\mathbf{Q}(\mathbf{x})=a_{11}x_1^2+2a_{21}x_1x_2+....+a_{nn}x_{n}^2\mapsto ... 1 We usel_2$norm for matrix as $$\|A\|=\sqrt{\sum\limits_{i,j=1}^{n}|a_{ij}|^2}\tag{1}$$ First we define inequality for matrix as Let$A$be$n\times n$matrix on$\mathbb{C}$and$B$be$n\times n\$ non-negative matrix. $$A\leqslant B \hspace{5 mm} \text{iff } \hspace{5 mm} |a_{ij}|\leqslant b_{ij}, \hspace{2 mm} i,j=1,\cdots,n \tag{2}$$ We ...

Top 50 recent answers are included