Why does this matrix give the derivative of a function?

I happened to stumble upon the following matrix:
A=[a10a]

And after trying a bunch of different examples, I noticed the following remarkable pattern. If P is a polynomial, then:
P(A)=[P(a)P(a)0P(a)]

Where P(a) is the derivative evaluated at a.

Futhermore, I tried extending this to other matrix functions, for example the matrix exponential, and wolfram alpha tells me:
exp(A)=[eaea0ea]
and this does in fact follow the pattern since the derivative of ex is itself!

Furthermore, I decided to look at the function P(x)=1x. If we interpret the reciprocal of a matrix to be its inverse, then we get:
P(A)=[1a1a201a]
And since f(a)=1a2, the pattern still holds!

After trying a couple more examples, it seems that this pattern holds whenever P is any rational function.

I have two questions:

  1. Why is this happening?

  2. Are there any other known matrix functions (which can also be applied to real numbers) for which this property holds?

Answer

If A = \begin{bmatrix}
a & 1 \\
0 & a
\end{bmatrix}

then by induction you can prove that
A^n = \begin{bmatrix}
a^n & n a^{n-1} \\
0 & a^n
\end{bmatrix} \tag 1

for n \ge 1 . If f can be developed into a power series

f(z) = \sum_{n=0}^\infty c_n z^n

then

f'(z) = \sum_{n=1}^\infty n c_n z^{n-1}

and it follows that

f(A) = \sum_{n=0}^\infty c_n A^n = I + \sum_{n=1}^\infty c_n
\begin{bmatrix}
a^n & n a^{n-1} \\
0 & a^n
\end{bmatrix} = \begin{bmatrix}
f(a) & f'(a) \\
0 & f(a)
\end{bmatrix} \tag 2

From (1) and

A^{-1} = \begin{bmatrix}
a^{-1} & -a^{-2} \\
0 & a^{-1}
\end{bmatrix}

one gets

A^{-n} = \begin{bmatrix}
a^{-1} & -a^{-2} \\
0 & a^{-1}
\end{bmatrix}^n =
(-a^{-2})^{n} \begin{bmatrix}
-a & 1 \\
0 & -a
\end{bmatrix}^n \\ =
(-1)^n a^{-2n} \begin{bmatrix}
(-a)^n & n (-a)^{n-1} \\
0 & (-a)^n
\end{bmatrix} =
\begin{bmatrix}
a^{-n} & -n a^{-n-1} \\
0 & a^{-n}
\end{bmatrix}

which means that (1) holds for negative exponents as well.
As a consequence, (2) can be generalized to functions
admitting a Laurent series representation:

f(z) = \sum_{n=-\infty}^\infty c_n z^n

Attribution
Source : Link , Question Author : ASKASK , Answer Author : Martin R

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