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Questions about the properties of vector spaces and linear transformations, including linear systems in general.

15 votes
Accepted

Die hard nilpotent spaces

You conjecture is not true. Let $P$ be the $3\times 3$ matrix $$ \left( \begin{array}{ccc} 0&1&0\\ 0&0&1\\ 0&0&0\end{array} \right) $$ which is nilpotent with $P^2\neq 0$. Consider the subspace of $9 …
Lev Borisov's user avatar
  • 5,186
7 votes
Accepted

Proof for a rank-one decomposition theorem of positive (semi) definite matrices

I don't know what's going on with the paper, but here is an argument regarding existence of such decompositions. Given a rank one decomposition $$X = \sum_{i=1}^R x_ix_i^T$$ one has $\sum_{i=1}^R x_ …
Lev Borisov's user avatar
  • 5,186
2 votes
Accepted

For a set of matrices $S$, find $X$ such that the elements of $SX$ commute

This is not a complete solution by any means, but here are some ideas. If one of $A_j$ (or their linear combinations) is invertible, then one can get a necessary and sufficient condition. Namely, if …
Lev Borisov's user avatar
  • 5,186
11 votes
2 answers
818 views

Determinant and eigenvalues of a specific matrix

This came up in a conversation with an engineer friend of mine. Let $c>0$ be a constant. Let $A_{ij}$ be an $n$ by $n$ matrix with entries $$ A_{ij} = e^{-c(i-j)^2}. $$ Is there a name for this matri …
Lev Borisov's user avatar
  • 5,186
2 votes

Splitting subspaces and finite fields

Clearly, the statement is invariant under multiplication by $a\in K$, so we may assume that $W\ni 1$. This implies that $W\supseteq R$, and we want to show that $W=S$. Suppose that $t\in W$. I claim …
Lev Borisov's user avatar
  • 5,186
26 votes
1 answer
1k views

Real square roots of symmetric matrices

In joint work with Andreas Fischle (TU Dresden, Germany) and Patrizio Neff (U Essen, Germany) we needed to use the following statement: If $S$ is a real $n\times n$ matrix with $S^2$ symmetric, then t …
Lev Borisov's user avatar
  • 5,186
2 votes

When is this matrix singular?

For $\phi_j=0$ and $t_k=k$ you have (up to a constant) the matrix $A_{jk}= \lambda_j^k - \lambda_j^{-k}$ where $\lambda_j=\exp( i w_j)$. You can factor out $\lambda_j-\lambda_j^{-1}$ to get a matrix $ …
Lev Borisov's user avatar
  • 5,186