# $e^{tA}$ of A1=$\begin{pmatrix} 1 & 0 \\ -3 & 1 \\ \end{pmatrix}$

A1= $\begin{pmatrix} 1 & 0 \\ -3 & 1 \\ \end{pmatrix}$

I found out that it is a double root case where an eigenvalue is 1 and tried to find the other eigenvector $\vec{d}$ by writing $C1e^t\begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} +C2(te^t\vec{d}+e^t\vec{d})$ which does not yield a linearly indepedent vector since $\vec{d} =\begin{pmatrix} 0 & 0 \\ -3 & 0\\ \end{pmatrix}\begin{pmatrix} a \\ b \\ \end{pmatrix}$ . thanks

In class I learnt how to diagonalize a matrix and using the fundamental matrix to find the exponent. So I am assuming I need to solve it through its fundamental matrix

• it is not a diagonalizable matrix... – Exodd Nov 8 '17 at 1:34
• $\exp(tA) = \exp\left(tI_2 + \begin{pmatrix}0 & 0\\-3t & 0\end{pmatrix}\right) = e^{tI_2}\exp\left(\begin{pmatrix}0 & 0\\-3t & 0\end{pmatrix}\right)$. The second factor can be easily computed with the exponential series. – amsmath Nov 8 '17 at 1:37
• I don`t think we learnt exponential series – Kaan Yolsever Nov 8 '17 at 1:45
• amsmath means the taylor series of the exponential available from lower level calculus $e^x = \sum \frac{x^n}{n!}$ – Triatticus Nov 8 '17 at 2:06

Note that $A=I+N$ with $$N=\pmatrix{0&0\\-3&0}$$ a nilpotent matrix (since $N^2=0$). It follows that for $t\in\mathbb R$, \begin{align} e^{tA} &= e^{t(I+N)}\\ &= e^{tI}e^{tN}\\ &= \pmatrix{e^t&0\\0&e^t}\left(\pmatrix{1&0\\0&1} + \pmatrix{0&0\\-3t&0}\right)\\ &= \pmatrix{e^t&0\\-3te^t&e^t}. \end{align}