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The study of the properties of real and complex matrices that are more close to analysis and operator theory. For instance: the properties of positive definite matrices, matrix inequalities, perturbation analysis, matrix functions, inequalities between eigenvectors and singular values, majorization.
17
votes
Accepted
Closed form solution for $XAX^{T}=B$
$B^{-1/2}XAX^TB^{-1/2}=I$, so $B^{-1/2}XA^{1/2}=Q$ must be orthogonal. On the other hand, for any orthogonal $Q$, it is simple to verify that $X = B^{1/2}QA^{-1/2}$ solves the equation, so this is a c …
14
votes
Accepted
Are Diagonally dominant Tridiagonal matrices diagonalizable?
Counterexample:
$$
\begin{bmatrix}
-1 & 1 & 0 & 0\\
0 & -1 & 1 & 0\\
0 & 0 & -2 & 2\\
0 & 0 & 2 & -2
\end{bmatrix}
$$
is defective: its eigenvalues are $-1,-1, 0, -4$ (it is block triangular, so its …
14
votes
Accepted
Is every real matrix conjugate to a semi antisymmetric matrix?
Yes. Every matrix can be written as the sum of a symmetric plus an antisymmetric one: $A = \frac{A+A^T}{2}+\frac{A-A^T}{2}$. Now change basis such that the symmetric part is diagonal.
12
votes
Matrix elements of exponential of tridiagonal matrices
Yes! Most methods to compute exponentials of large sparse matrices are based on computing directly $\exp(A)b$ for a given vector $b$ rather than the full matrix $\exp(A)$. Just take $b$ as a vector of …
11
votes
Accepted
When the sum of positive definite matrices converges, does the sum of the norm of the associ...
You can bound $\|A_k\| \leq C(n)\max_{i,j} |(A_k)_{ij}|$ for some function of the dimension only $C(n)$, because all norms are equivalent in finite dimension. If I am not mistaken $C(n)=\sqrt{n}$, but …
9
votes
Accepted
Source for roots of matrix polynomials?
As Geoff Robinson says, a Jordan form takes you quite far. Evaluating a scalar polynomial (or an analytic function) $f(x)$ at a Jordan block $J_{\lambda,t}$ of size $t$ and eigenvalue $\lambda$ gives …
8
votes
Accepted
How to solve a non-homogeneous quadratic matrix equation?
(commenting about the equation with the plus sign, I hope that the correction was right).
This is one of the few quadratic matrix equations that have a closed form solution. Set $A=-H^{-1}$; then $G …
8
votes
About Sylvester's determinant
The common ground of those two formulas is related to the Woodbury matrix identity. This relation is a useful statement that shows what happens to the inverse when one "updates" a matrix $A\in\mathbb{ …
7
votes
Is this inequality involving the Frobenius norm right?
For a short fat matrix $G$ (more columns than rows), $\|AG\|_F \geq \sigma_{\min}(G)\|A\|_F \geq n \sigma_{\min}(G) \|A\|$, where $\sigma_{\min}(G)$ is the least singular value of $G$. This follows fr …
7
votes
Accepted
upper bounds on a certain matrix norm
You can use the surprising identity $(A^{-1}+B^{-1})^{-1}=A(A+B)^{-1}B$, and take the norms of the three factors separately.
5
votes
How bad could $\|A^k\|$ be when $\rho(A) < 1-\delta$
To get a hang of the behaviour of matrix powers, you should consider powers of Jordan blocks:
$$
J_k(\lambda)^n = \begin{bmatrix}
\lambda^n & \binom{n}{1}\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} & \c …
5
votes
Accepted
Solving a vector of quadratic equations
Shameless advertisement to a paper of mine: http://www.sciencedirect.com/science/article/pii/S0024379511004484 Quadratic vector equations, in Linear Algebra and its Applications, volume 438, 2013. Ar …
5
votes
Optimizing the condition number
I have been working recently on a similar problem, namely, maximizing the absolute value of the determinant of a chosen $m\times m$ submatrix $S$ of a $m\times N$ submatrix $V$ (think to the columns o …
5
votes
Fast Upper Triangular Matrix Exponentiation
You probably want the Schur-Parlett method for computing matrix functions. It is a method to compute a generic function of a triangular matrix. Essentially, you apply the function to its diagonal elem …
5
votes
Accepted
Norm numerical range
If $A=R^*R$ is the Cholesky factorization of $A$, then $$\frac{(ABv,v)}{(Av,v)} = \frac{(R^*RBv,v)}{(R^*Rv,v)} = \frac{(RBv,Rv)}{(Rv,Rv)} = \frac{(RBR^{-1}w,w)}{(w,w)}$$ for $w=Rv$, hence $W_A(B) = W( …