Tag Info

Hot answers tagged

4

$Sv_i$ is an eigenvector for $T$ with eigenvalue $\lambda_i$. The dimension of each eigenspace of $T$ is 1 since the eigenvalues are distinct, so $Sv_i = c_iv_i$ for some $c_i$.


3

You have stated that $\{ v_1,v_2,\cdots,v_n\}$ is a basis of $V$ consisting of eigenvectors of $T$ with distinct eigenvalues $\{\lambda_1,\lambda_2,\cdots,\lambda_n\}$. And you have assumed that $S$ is another linear operator on $V$ that commutes with $T$. Because $TS=ST$, then $$ TSv_j = STv_j = \lambda_j Sv_j. $$ Because $\{ v_1,v_2,\cdots,v_n ...


3

Note that $x^3-x^2-16x-20=(x-5)(x^2+4x+4)=(x-5)(x+2)^2$.


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

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

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

V isn’t just any arbitrary matrix. They're trying to take you step by step through the diagonalization process in this exercise, so the eigenvectors that you computed in part b are going to come into play here.


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 $A$.


1

You started with a system of linear equations. In matrix form this can be written as: $$\left(\begin{array}{cc|c} 0 & -1 & 0 \\ 0 & -1 & 0 \end{array}\right)$$ from which you have gotten to $$\left(\begin{array}{cc|c} 0 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right).$$ As we have a complete row of $0$'s, there might be infinitely ...


1

From your equations, you have $y=0$ and $x$ does not feature in the equations, so can be any value.



Only top voted, non community-wiki answers of a minimum length are eligible