Tag Info

$Sv_i$ 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 $c_i$.
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.