6
votes
Accepted
Order of eigenvectors within basis for Jordan Normal Form?
The basis can be rearranged in any way that rearranges only the order of the Jordan blocks, but does not rearrange the basis elements within the blocks themselves. That is, basis elements coming from ...
- 3,142
2
votes
Constructing an arbitrary set of real vectors whose pairwise dot products are bounded
Partial answer/strategy:
Start with any orthonormal basis, perturb the entries randomly by at most some $\epsilon$. That will yeild inner products in a range $1 \pm f(\epsilon)$ for some function $f$ ...
- 86.9k
1
vote
Accepted
If B is a $2\times3$ matrix with some condition, find a non-zero matrix $C$ such that $BCB^T=0$
Presumably the matrices have real elements. Then $\operatorname{rank}(B^TB)=\operatorname{rank}(B)=\operatorname{rank}(B^T)=\operatorname{rank}(BB^T)$. The condition $\det(B^TB)=\det(BB^T)$ thus ...
- 132k
1
vote
Find 3x3 Matrix with specific eigenvalue and eigenvector
If $(\lambda, x)$ is an eigenpair of $M$ then we indeed have $Mx = \lambda x.$ But $\lambda = 0$ in the given eigenpair, so we must have $Mx = 0.$ That is, $x \in \mathrm{Nul} \ M$ (the null space). ...
- 8,690
1
vote
Accepted
Can I decompose this matrix into separate parts?
While I don't see a way to decompose this into the form you suggest I was able to decompose the matrix to a similar form.
First, suppose $a$ and $b$ are $n$-dimensional column vectors and $c$ is an $m$...
- 760
1
vote
Can I decompose this matrix into separate parts?
While this isn't a fully rigorous non-existence proof, this argument suggests that no such decomposition exists for $n \times n$ matrices for $n>2$.
It is easy to check that $$\begin{bmatrix} 0 &...
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