Question

If \(A=\left[\begin{matrix}1&0\\0&2\end{matrix}\right]\), then \(e^{tA}=\left[\begin{matrix}e^t&0\\0&e^{2t}\end{matrix}\right]\)
Now, if \(B=\left[\begin{matrix}0&1\\0&0\end{matrix}\right]\), then \(e^{tB}=\left[\begin{matrix}1&t\\0&1\end{matrix}\right]\)
I don't get why t there. Please, help

Answer 1

@oldrin.bataku

Answer 2

Are you specifically referring to the \[e^{tB}\] matrix?

Answer 3

Yes, sir

Answer 4

OK, lemme check. I believe exponentiation of matrices is defined by its infinite series, but let me see if there is a more elegant approach.

Answer 5

OK, I got it. I'll let @oldrin.bataku reply first and see if they have any additional insights.

Answer 6

We first need that
\[e^x = 1 + x + x^2/2! + x^3/3! + \cdots \]Are you familiar and/or comfortable with that?

Answer 7

Yes, I do

Answer 8

OK, great. This definition can be extended to matrices so that
\[e^{tB} = I + tB + (tB)^2/2! + (tB)^3/3! + \cdots\]
In general,
\[e^A = \sum_{n=0}^\infty \frac{A^n}{n!}\]

Answer 9

Expanded, we have
\[e^{tB} = I + tB + t^2B^2/2! + t^3B^3/3! + \cdots\]

Answer 10

OK, so far?

Answer 11

Yes

Answer 12

Alright, awesome! Now, the annoying thing is the powers of matrices. We really hope that we have a "nice" matrix, otherwise this can get messy. Fortunately, B isn't that bad.

Answer 13

\[B = \left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right], \ \ B^2 = \left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]\]

Answer 14

The zero matrix times any matrix is still the zero matrix, so all powers of B above 1 will be zeroed out.

Answer 15

I know it

Answer 16

Hence, our problem reduces to this:
\[e^{tB} = I + tB + 0 + 0 + \cdots = I + tB\] and the result follows immediately.

Answer 17

yyyyyyyyyyyyyes!

Answer 18

\[I + tB = \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] + \left[\begin{matrix}0 & t \\ 0 & 0\end{matrix}\right] = \left[\begin{matrix}1 & t \\ 0 & 1\end{matrix}\right]\]

Answer 19

Yay!! Indeed, any questions?