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### Why does the matrix exponential $e^A$ always exist?

It is worth to treat the matrix $A$ as a linear bounded operator $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^n)$. Therefore defining the operator $\exp A:\mathbb{R}^n\to \mathbb{R}^n$ as a limit of the ...

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### How to take the derivative of a matrix with respect to itself?

The answer is a lot easier than the previous posters are indicating $\frac{d}{dX}(A*X)=A$ define $A=I_m$ where $m$ is the number of rows of $X$ because $I_m*X=X$ this is an identity and you get your ...
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### Derivative of the inverse of a symmetric matrix w.r.t itself

In the single-variable case, we have that $$\dfrac{d}{dt}C(t)^{-1}=-C(t)^{-1}\dfrac{dC(t)}{dt}C(t)^{-1}.$$ This can obtained by differentiating the expression $C(t)C(t)^{-1}=I$ on both sides with some ...
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### Differentiating with respect to a vector transpose

The plain simple answer is that the Differentiation of $β^T$($X^T$X)B becomes ($X^T$X)$\hat \beta$ ($\hat \beta$ is the Beta estimate) $B^T$ disappears as usual. `

• 51
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### What the rank of a matrix with the elements of one column to be infinity?

The following two statements are in direct contradiction: Suppose the $m\times m$ real matrix $A$ if all the elements in the first column and first row are assigned to be infinity Infinity is not a ...
• 114k
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### Derivative of transpose of matrix

$X↦X^⊤$ is a linear map and so equal to its own derivative. A better question to ask is can we express this function in the standard form of a linear map, which is as a matrix-vector product in the ...
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### Derivative of transpose of matrix

The set of real $n \times n$ matrices forms a real vector space with respect to the usual addition of matrices. Your identification of this space with the canonical vector space $\mathbb{R}^{n^2}$ is ...
• 103
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$\def\L{{L}}\def\o{{\large\tt1}}\def\p{\partial} \def\LR#1{\left(#1\right)} \def\BR#1{\Bigl(#1\Bigr)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\Diag#1{\operatorname{Diag}\LR{#1}} \def\Lapl#1{\... • 27.5k 1 vote ### Gradient of Quadratic Function @Jake, Regarding your question, your computations are all correct. The only thing that is missing is to recognize $$\mathbf{x}_n^T \mathbf{w} \mathbf{x}_n = (\mathbf{x}_n \mathbf{x}_n^T) \mathbf{w}$$... • 1,919 2 votes ### Integration of matrix form of Vasicek variance (Python/Matlab) Assuming that$K$and$\Sigma_xare square matrices the explicit solution of your vector SDE is $$X_t=e^{-Kt}\cdot\Big\{X_0+\int_0^t e^{Ks}\cdot\mu\, ds+\Sigma_x\cdot\int_0^t e^{Ks}\cdot dZ_s\Big\}\,... • 4,274 1 vote ### Problem 2.9.6 (Perko's ODE): Show |Y(t)| \le |Y(0)|\exp\left(\int_0^t \|A(s)\|\, ds\right) for Y' = AY \def\T{{\rm tr}} \def\A{\T(A)} \def\l{\lambda} \def\qiq{\quad\implies\quad} This well known result$$\eqalign{ &\l = \log|Y| \\ &d\l = \T(Y^{-1}\,dY) \;=\; \T(Y^{-1}AY\,dt) \;=\; \A\;dt \\ &... • 27.5k 1 vote Accepted ### Problem2.9.6$(Perko's ODE): Show$|Y(t)| \le |Y(0)|\exp\left(\int_0^t \|A(s)\|\, ds\right)$for$Y' = AY$For$t \ge 0$we have$Y(t)=Y(0)+ \int_0^t A(s)Y(s)ds\$, hence $$|Y(t)| \le |Y(0)| + \int_0^t \|A(s)\|\cdot|Y(s)|ds.$$ Now (1) follows from the Grönwall-inequality: https://en.wikipedia.org/wiki/Gr%...
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