# Find $\exp(D)$ where $D = \begin{bmatrix}2& 1 \\ 0& 2\end{bmatrix}.$ [closed]

This one I cannot just use the diagonal method right? Could anyone give any hint on solving this?

• Can you show what work you've done so far? – user18862 Jan 30 '14 at 5:52
• Weird that you can deal with this but not with the present question. – Git Gud Jan 30 '14 at 7:40
• The big motivator for matrix exponentials is their use in solving systems of differential equations. This $2\times2$ case is small enough that it'd be just as easy to solve the corresponding system in order to compute the matrix exponential! Given the tag (differential-equations) though, one might suspect that this is a pedagogically undesirable way of doing it. – anon Jan 30 '14 at 8:05
• Does this answer your question? Find $\exp(D)$ where $D = \begin{bmatrix}2&amp; -1 \\ 1 &amp; 2\end{bmatrix}.$ – Keen-ameteur Apr 2 at 9:42

Set

$\Lambda = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \tag{1}$

and

$P = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \tag{2}$

so that $D = \Lambda + P$. Furthermore, since $\Lambda = 2I$, $\Lambda$ commutes with any matrix $X$, i.e. $\Lambda X = X \Lambda$. Thus we may conclude, based upon the discussion/proofs given in my answer to this question, that $\exp (D) = \exp(\Lambda + P) = \exp(\Lambda) \exp(P)$. $\Lambda = 2I$ implies $\exp(\Lambda) = e^2I$, easy to see by taking the power series for $\exp(\Lambda)$ and noting that $\Lambda = 2I$ implies $\Lambda^k = 2^kI$ for all integral $k \ge 0$, whence

$\exp(\Lambda) = \sum_0^\infty \dfrac{2^k I}{k!} = e^2 I; \tag{3}$

noting that $P^2 = 0$, we see that the series for $\exp(P)$ is truncated after the linear term, so that

$\exp(P) = I + P = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \tag{4}$

and finally

$\exp(D) = \exp(\Lambda) \exp(P) = e^2I (I + P) = \begin{bmatrix} e^2 & e^2 \\ 0 & e^2 \end{bmatrix}. \tag{5}$

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

• what is the phase portrait is going to be? – afsdf dfsaf Jan 31 '14 at 1:18

Hint: $$D = \begin{bmatrix}2& 0 \\ 0& 2\end{bmatrix} + \begin{bmatrix}0& 1 \\ 0& 0\end{bmatrix}.$$

The two matrices in the sum commute, so you can use $\exp(A + B) = \exp(A) \exp(B)$.

$D=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}+\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$

These two matrices commute, and the square of the second matrix is $0$ matrix.

• Can anyone explain to me how is it that voldemort and arsmath give the same answer (apart from an important detail that it's OK not to mention so some work is left for the OP), voldemort answers first, but gets less up votes? – Git Gud Jan 30 '14 at 7:47

Clearly, $$D^n=\left( \begin{matrix} 2 & 1 \\ 0 & 2 \end{matrix} \right)^n = \left( \begin{matrix} 2^n & a_n \\ 0 & 2^n \end{matrix} \right).$$ Because $$\left( \begin{matrix} 2^{n+1} & a_{n+1} \\ 0 & 2^{n+1} \end{matrix} \right) = \left( \begin{matrix} 2^n & a_n \\ 0 & 2^n \end{matrix} \right) \left( \begin{matrix} 2 & 1 \\ 0 & 2 \end{matrix} \right) = \left( \begin{matrix} 2^{n+1} & 2^n+2a_n \\ 0 & 2^{n+1} \end{matrix} \right),$$ we deduce that $a_{n+1}= 2^n+2a_n$. By induction, it is easy to prove that $a_n=n2^{n-1}$ for $n \geq 1$. Therefore, $$\exp(D)= \sum\limits_{n \geq 0} \frac{D^n}{n!} = \left( \begin{matrix} \sum\limits_{n \geq 0} \frac{2^n}{n!} & \sum\limits_{n \geq 0} \frac{n \cdot 2^n}{2 \cdot n!} \\ 0 & \sum\limits_{n \geq 0} \frac{2^n}{n!} \end{matrix} \right).$$ But $\displaystyle \sum\limits_{n \geq 0} \frac{2^n}{n!}=e^2$ and $\displaystyle \sum\limits_{n \geq 0} \frac{n \cdot 2^n}{2 \cdot n!}= \sum\limits_{n \geq 1} \frac{2^{n-1}}{(n-1)!} = e^2$. Finally, $$\exp(D)= e^2 \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right).$$

• Does $\text{Id}$ in your last line mean the identity map? If so, then how can $\exp(D) = e^2 \text{Id}$ in light of your preceding equation for $\exp(D)$? – Robert Lewis Jan 30 '14 at 8:07
• Thank you, I edited my answer. – Seirios Jan 30 '14 at 11:02
• what is the phase portrait is going to be? – afsdf dfsaf Jan 31 '14 at 1:18