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4

Suppose $A$ and $B$ are positive operators on a finite-dimensional inner product space $V$ and $\|A + B\| = \|A\| + \|B\|$. Because $A$ and $B$ are positive, $A+B$ is also a positive operator. Thus there exists $x \in V$ such that$$\|x\| = 1 \quad \text{and} \quad\|A+B\| = \langle (A+B)x, x \rangle. $$ Now \begin{align*} \|A+B\| &= \langle (A+B)x, x ...


3

If it has rank $1$, then all but one eigenvalues are $0$, so the value of the nonzero eigenvalue equals the trace, which is rational.


3

The answer to your question is no. That is, your friends are wrong. In general, applying row-reduction to your matrix will change its eigenvalues. We can see what goes wrong if we take this approach slightly differently. In particular: $$ \pmatrix{ 2 & 1 & 2 \\ 0 & 2 & -1 \\ 0 & 1 & 0 \\ } \to \pmatrix{ 2 & 1 & 2 \\ ...


3

The answer is no. Consider the space of infinite real sequences and let $T(x)=(0,x_1,x_2,\dots)$. Suppose $T(x)=\lambda x$. If $\lambda=0$, then $x$ has to be zero, so $0$ is not an eigenvalue. If $\lambda \neq 0$, then $0=\lambda x_1$, so $x_1=0$, but then $x_1=0=\lambda x_2$. Continuing by induction we again conclude that $x=0$. So $\lambda$ is not an ...


2

No, the shift homomorphism defined on $(e_1,...,e_n,...)$ by $f(e_i)=e_{i+1}$ does not have an eigenvalue.


2

You can find a computation method for the eigenvalues and eigenvectors in the following recent document: "Eigendecomposition of Block Tridiagonal Matrices" by A. Sandryhaila and J.M.F. Moura (arxiv.org/pdf/1306.0217)


2

Here's a counterexample. Let $V$ be the space of functions $\mathbb{Z}\to K$ which have finite support. Define $T:V\to V$ by $(T\varphi)(n)=\varphi(n+1)$ (i.e., thinking of an element of $V$ as a "bi-infinite sequence", shift the terms in the sequence by $1$). Then $T$ is an isomorphism, but if $T\varphi=\lambda\varphi$, then ...


2

The eigenvector you obtained and the one in the book differ only in sign, that is by multiplication by $-1$. Both vectors are eigenvectors for eigenvalue $0$ and equally correct answers. For the other eigenvalue, you just forgot to subtract in the last line. If you correct this you will get two free variables.


1

If $x$ is a unit eigenvector of $A$ associated with $\lambda$, we have $$ (\lambda I-B)^{-1}(A-B)x=(\lambda I-B)^{-1}(\lambda I-B)x=x $$ and hence $$ 1=\|x\|=\|(\lambda I-B)^{-1}(A-B)x\|\leq\|(\lambda I-B)^{-1}(A-B)\|. $$


1

Let $v$ be an eigenvector of $A$ corresponding to the eigenvalue $\lambda$, i.e. $Av=\lambda v$. Then $(\lambda I-B)^{-1}(A-B)v=(\lambda I-B)^{-1}(\lambda I -B)v=v $. Hence by definition of the operator norm, $\left\|(\lambda I-B)^{-1}(A-B)\right\|\geq 1$. Question: Where did we use that $\lambda$ is not an eigenvalue for $B$?


1

In char polynomial, coefficient of $\lambda^{n-i}$ is sum of all principal minors of order $i$. Since rank is one all minors of order $2$ or greater are zero. So only coefficient of $\lambda^{n-1}$ has possibility to be non zero. If it too is zero. all evalues will be zero, so rational. If it is non zero, again it has two be rational since char polynomial of ...


1

At first, to summarize this problem, let replace diagonal factor matrices as: $$ \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \lambda-mk & 1 \\ \lambda-k & 1 \end{bmatrix} \begin{bmatrix} I \\ A \end{bmatrix} $$ where $\lambda$ is eigenvalue. Therefore, we can suppose the following determinant: $$ \begin{aligned} \det \begin{bmatrix} X ...


1

You have a mistake in your calculation of the eigenvalue, but I still think the problem is not right. The eigenvalues of $M$ are $\frac a2\pm |b|$. One then finds easily that a corresponding pair of eigenvectors is given by $$ (e^{-i\phi/2},-e^{i\phi/2})^T,\ \ (e^{-i\phi/2},e^{i\phi/2})^T. $$ The eigenspaces are spanned by each of these two vectors, so ...


1

I'll just remind you of some standard facts that I hope will lead you in the right direction, but you should refresh these in the book/references. First the homogenous PDE, $ \partial_t u = \Delta u$. As mentioned, $u_{i,j}(r,t) := \psi_{i,j}(r) e^{\lambda_{i,j} t} $ solves the heat equation, with initial condition at time zero $ \psi_{i,j}(r) $. Since ...


1

The eigenfunctions will not necessarily be orthogonal with respect to an unweighted space. For the sake of discussion, assume $P$, $q$ and $r$ are real; $P$ is continuously differentiable on $[0,l]$ and strictly positive; $r$ is continuous and strictly positive; Then this problem falls into a nice classical setting where you can multiply by a factor to ...


1

Clearly (as the matrix is upper triangular) the characteristic polynomial is $\;(t-2)(t-3)^2\;$ , so the matrix is diagonizable iff $\;(t-2)(t-3)\;$ is its minimal polynomial, and indeed: ...


1

Intro I'll present an incomplete interpretation framework. Let's being by considering two points in one dimension. We can easily generalize the results to more than one dimension. Our matrix is given by, $$M=\begin{bmatrix} 0 & |x_1-x_2| \\ |x_1-x_2| & 0 \end{bmatrix}$$ The eigenvalues are given by $$\lambda_1=-\lambda_2=|x_1-x_2|$$ The ...


1

You have $$A^{20}x=\left(-\frac{1}{3}\right)^{20}V_1-\left(\frac{1}{3}\right)^{20}V_2+2V_3,$$ and this is $$\left(-\frac{1}{3}\right)^{20}\left(\begin{array}{c}1\\0\\0\end{array}\right)-\left(\frac{1}{3}\right)^{20}\left(\begin{array}{c}1\\1\\0\end{array}\right)+2\left(\begin{array}{c}1\\1\\1\end{array}\right)= ...



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