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

Question

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

Answer 1

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 $\exp(A + B) = \exp(A) \exp(B)$.

Answer 2

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

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

Answer 3

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!!!

Answer 4

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, $$\exp(D)= e^2 \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right).$$