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)=[1a−1a201a]

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:

Why is this happening?

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*