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4

No. The eigenvalues of $A$ are the zeroes of the characteristic polynomial. It doesn't matter whether you take the characteristic polynomial to be $\det(A -\lambda I)$ or $\det(\lambda I - A)$; they differ by at most a factor of $-1$, so their zeroes are the same.


3

Hint: If $\lambda$ is an eigenvalue of $A$, let $x$ be the associated eigenvector, and consider $x'Ax$.


2

Suppose our matrix $A$ has eigenvalue $\lambda$. If $\lambda = 0$, then there is some eigenvector $x$ so that $Ax = 0$. But then $x^T A x = 0$, and so $A$ is not positive definite. If $\lambda < 0$, then there is some eigenvector $x$ so that $Ax = \lambda x$. But then $x^T A x = \lambda \lvert x \rvert^2$, which is negative since $\lvert x \rvert^2 ...


2

Let $$ C=\pmatrix{I&0\\0&D}, $$ where the identity $I$ and an arbitrary symmetric $D$ have the same dimension greater than one. For $v_A=[1,0]^T$ and any nonzero $v_B$ (of the same dimension as $I$ and $D$), $v=v_A\otimes v_B$ is an eigenvector of $C$. The matrix $C$ does not need to have a Kronecker product form since $D$ is arbitrary symmetric.


2

This is in general false. Consider $v_A = v_B = (1, 1)^t$ and $C = \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1\end{pmatrix}$. Then $v_A \otimes v_B = (1, 1, 1, 1)^t$ and $C (v_A \otimes v_B) = 10 (v_A \otimes v_B)$, but clearly $C$ is not of the form $A \otimes I_2$. In ...


2

Since $A$ is real symmetric, there exists a matrix $P$ such that $$PAP^{-1} = D =\verb/diag/(\lambda_1,\lambda_2,\ldots,\lambda_n)$$ is diagonal with eigenvalues $\lambda_1,\ldots,\lambda_n$. This leads to $$P (I_n+A)^{-2}P^{-1} = (I_n + PAP^{-1})^{-2} = \verb/diag/\left(\frac{1}{(1+\lambda_1)^2},\ldots,\frac{1}{(1+\lambda_n)^2}\right)\\ \implies ...


2

Note that the trace of any permutation is its number of fixed points (that is, the number of $1$-cycles). As in my previous answer, let $x_{ij}$ denote $x_ix_j = x_{ji}$. In order to have $A^{(2)}(x_{ij}) = x_{ij}$, we must have $$ (Ax_i)(Ax_j) = x_{ij} \DeclareMathOperator{\tr}{Tr} $$ There are precisely two ways in which this can occur: $x_i$ and $x_j$ ...


2

You can easily see that $$A^nx^{(0)} = \sum_{i=1}^N\alpha_i \lambda_i^n x_i$$ You can rewrite that as $$\lambda_1^n\sum_{i=1}^N \alpha_i\left(\frac{\lambda_i}{\lambda_1}\right)^nx_i$$ and then study what happens to $\frac{\lambda_i}{\lambda_1}$ if $i=1$ or of $i\neq 1$.


2

Hint: In a basis of eigenvectors, the matrix of the endomorphism is diagonal, and the correspondence between the matrix $A$ in the original basis and the matrix $D$ in the basis of eigenvectors is given by: $$D=P^{-1}AP.$$


2

The eigenvalue algorithm produces a diagonal matrix containing the eigenvalues of $A$. That matrix is positive definite if and only if the eigenvalues of $A$ are all positive. So, generally we will not get a positive definite eigenvalue matrix.


1

We can always do so if $D$ is diagonalizable invertible. Otherwise, we have counterexamples like $$ D=\pmatrix{0&1\\0&0} $$ Note that $D^2$ as the eigenvector $(0,1)$.


1

By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e $$ P^{T}AP=\pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n} $$ where $\lambda_i$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$. Since $$ \pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n}=Q\pmatrix{I_r\\ &-I_s\\ && 0}Q $$ where $r$ is number of ...


1

Finally, uranix hint with preconditioning lead me to a solution. The key performance problem comes from solving lots of systems of the form $Ax = b$ with our $A$. Fortunately, $A$ has so many nice properties, that the PCG algorithm works well when using ichol as a preconditioner. Thus using eigs' capability to take $x\mapsto A^{-1}x$ as a function leads to ...


1

Your questions regarding the usefulness of the index and finding a Jordan basis I will answer together: as with diagonalization one usually finds the Jordan form first and then determine the basis of vectors that transforms the given matrix to Jordan form. One "useful" way to determine the Jordan form of a matrix $A$ is to find the Segre characteristics, ...


1

If $A$ acts on this basis than the trace is simply going to be the size $N$ of the set of basis vectors that it fixes (i.e. the number of $i$ such that $A(x_i) = x_i$) and as such we have that $Tr(A) = \sum_{i = 1}^n \lambda_i = N$. Then the number of basis vectors fixed by $A^{(2)}$ is simply $\frac{\text{Tr}(A^2) + \text{Tr}(A)^2}{2} - N$ which is ...


1

An eigenbasis is a basis in which every vector is an eigenvector. In your case, $$ \left\{ \pmatrix{-1\\1\\0}, \pmatrix{-1\\0\\1}, \pmatrix{1\\1\\1} \right\} $$ is an eigenbasis for your matrix $A$.



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