# Tagged Questions

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### If $A$ is a $2\times 2$ matrix with a repeated eigenvalue $r$, then $\mathrm{e}^{At}=\mathrm{e}^{rt}\left[I+(A-rI)t\right]$

If $A$ is a $2\times 2$ matrix with a repeated eigenvalue $r$, show that $\mathrm{e}^{At}=\mathrm{e}^{rt}\left[I+(A-rI)t\right]$. I have already been able to show that if $A$ is a general ...
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### Changing a negative definite matrix to a positive definite matrix

Consider a negative definite matrix $X$,then $(I-e^X)$ is a positive definite matrix. What condition should matrix $X$ satisfy?
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### Commuting Exponential Matrices

Let $x(t)=\exp(tA)\exp(tB)$ and $y(t)=\exp(t(A+B))$. Show that if $AB=BA$ then $x(t)$ and $y(t)$ satisfy the same initial value problem for ODEs and therefore must be equal. $A, B$ square matrices. ...
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### Inverse function of product of exponential matrices

I am looking for the value of $\mathbf{X}$ in a function of the type \begin{align} (\mathbf{X}-\mathbf{A})e^{\mathbf{X}}e^{-\mathbf{A}} = \mathbf{B} \end{align} where ...
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### $e^{At}$ for $A = B^{-1} \lvert \cdots \rvert B$

For a homework problem, I have to compute $e^{At}$ for $$A = B^{-1} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} B$$ I know how to compute the result ...
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### Commuting in Matrix Exponential

Let $A$, and $B$ be commuting $n\times n$ matrices, i.e. $A.B = B.A$. Let $$\exp(A) = \sum_{i=0}^\infty\frac{1}{i!} A^i$$ show that $\exp(A+B) = \exp(A).\exp(B)$.
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### Definition of $\exp(A)$ in terms of spectral decomposition.

I am read this question Plugging a matrix multiplied by an imaginary number in the exponential function. Here the question'author defined $$\exp(A) := \sum_{1\le k\le n}\exp(\lambda_k) P_k$$ What ...
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### Show that $e^{A+B}=e^A e^B$

If $A$ and $B$ are $n\times n$ matrices such that $AB = BA$ (that is, $A$ and $B$ commute), show that $$e^{A+B}=e^A e^B$$ Note that $A$ and $B$ do NOT have to be diagonalizable.
I have diagonlised P to get $$P=\left(\begin{matrix} -1 &0 &0\\ 0 &0 &0\\ 0 &0 &1 \end{matrix}\right)$$ however am unsure on how to proceed, would appreciate any help! By ...