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For a real square matrix to have a nontrivial null space, it must be singular. The matrix in your example isn’t, so it’s not surprising that your algorithm didn’t product a null vector for it. In fact, it’s possible to show directly that, with $v$ constructed as in your question, the last element of $Av$ vanishes iff $\lvert A\rvert = 0$.
First, observe ...
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PART 1.
Firstly, your matrix is only non-negative; consequently, $\lambda=\rho(A)$ is an eigenvalue of $A$ associated to a non-negative eigenvector $u$; yet, $\rho(A)$ is not necessarily simple and there may be other eigenvalues of A of same modulus. See for example $A=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}$. Anyway, the ...
2
When $N=4$, your unitary matrix $G$ has four eigenvalues $1,1,-1,i$. Since $1$ is a repeated eigenvalue, numpy.linalg.eig will produce two linearly independent eigenvectors in the corresponding eigenspace, but these two eigenvectors are not guaranteed to be mutually orthogonal.
In general, given a normal matrix $G$, its Schur triangulation $G=URU^\ast$ (...
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Having distinct eigenvalues is a sufficient condition for a matrix to be diagonalizable, but not a necessary one. For example, any diagonal matrix is diagonalizable.
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What can I say about: 1) the convergence?
It converges when there is a dominant eigenvalue. It has trouble converging when they are very close together, which isn't the case here.
2) the absolute value
greatest eigenvalue?
The matrix is small so you can check it by hand.
$$ A = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} $$
then
$$ \det(A- ...
1
I'm no expert with Matlab but when you call upon the inv() function it applies different methods depending upon the structure of your Matrix. I'm guessing that you're looking to form an Eigenvalue Decomposition for your Markov Chain Matrix. I would recommend that you first check the Eigenvalues for $W_2$, following that it's non-degenerate that you check ...
1
If you need to show that $T_1$ is a tridiagonal matrix in an iterative QR process, you put $T_0 = Q_0R_0 \Longrightarrow R_0 = Q_0^{T}T_0$. So,
$$
T_1 = R_0Q_0 \Longrightarrow T_1 = Q_0^{T}T_0Q_0.
$$
Now you calculate $T_1^{T}$ and conclude that $T_1$ is a symmetric matrix. I think that you can use that $T_0$ is symmetric by first part of your question.
Then,...
1
This is an old question, but I recently worked through this problem, and would like to share my insight... so to make this work as a saxpy formulation, we require some column vector $C$ that will be updated with an assignment $$C := b_kA(:,k)+C$$ where we are looping over the $k$ columns of $A$, and $b_k$ is a scalar.
The way I tackled this, was to first ...
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