I am a bit confused with your notation. Since you have used ##Q_{\bar{B}_2}## which comes out of the QR factorization as ##\bar{B}_2=Q_{\bar{B}_2}R_{m}##, and ##Q_C## in your formulation is defined as ##C=[\bar{B}_2|\textbf{v}_i|_{i=m+1:p}]=Q_CR##, shouldn't ##Q_C=Q_{\bar{B}_2}##? In that case...
Given a real-valued matrix ## \bar{B}_2=\begin{bmatrix}
\bar{B}_{21}\\
\bar{B}_{22}
\end{bmatrix}\in{R^{p \times m}}
##, I am looking for an orthogonal transformation matrix i.e., ##T^{-1}=T^T\in{R^{p \times p}}## that satisfies:
$$
\begin{bmatrix}
T_{11}^T & T_{21}^T\\
T_{12}^T...
That's a nasty mistake that I made; I think deep down, I was looking for something like SPDness of ##R## to simplify things but ##R## is not square in general.
No, it won't help really. The matrix inequality is already linear (LMI) and Schur complement in that sense will just cause a second problem that the off-diagonal terms will cause nonlinearty (later ##R## should be found by interior point method). I am expecting the answer to be in the form of a...
Hello, I want to know if there exist any result in literature that answers my question:
Under which conditions on the real valued matrix ## R ## (symmetric positive definite), the first argument results in and guarantees the second one:
1) for real valued matrices ##A, B, C,## and ## D ## with...
Sure. My confusion was because of the example in my mind that s can be a very small value which then the upper bound will be also very small and as a result, physically conservatism but again mathematically correct. Thank you very much.
Based on Schwartz inequality, I am trying to figure out why there
should/can be the "s" variable which is the lower bound of the
integration in the RHS of the following inequality:
## \left \|\int_{-s}^{0} A(t+r)Z(t+r) dr \right \|^{2} \leq s\int_{-s}^{0}\left \| A(t+r)Z(t+r) \right \|^{2} dr...