# Matrix raised to a matrix: $M^N$, is this possible? with $M,N\in M_n(\Bbb K).$

I was wondering if there is such a valid operation as raising a matrix to the power of a matrix, e.g. vaguely, if $M$ is a matrix, is $$M^N$$ valid, or is there at least something similar? Would it be the components of the matrix raised to each component of the matrix it's raised to, resulting in again, another matrix?

Thanks,

It is possible to define the exponential $$\exp(M) = \sum_{n \ge 0}^{\infty} \frac{M^n}{n!}$$

of any matrix using power series. Similarly, it is possible to define the logarithm $$\log(I + M) = \sum_{n \ge 1} \frac{(-1)^{n-1} M^n}{n}$$

when the above series converges (this is guaranteed for example if the largest singular value of $M$ is less than $1$). We can therefore define $$M^N = \exp(N \log M)$$

by imitating the identity $a^b = e^{b \log a}$ for, say, positive reals, but this won't have good properties unless $N$ and $M$ commute, I think. It's better to consider the exponential and logarithm separately.

As I have discussed elsewhere on math.SE, the fact that the ordinary exponential takes two inputs which are the same type is misleading. Most (but not all) "exponential-type" operations in mathematics take two inputs which are different types.

• Actually $\log(M)$ can be defined whenever $M$ is invertible, but there are many possible values (there is a choice of branch of the logarithm for each eigenvalue of $M$). Jun 29, 2012 at 1:45
• @Robert: that is not so clear to me if $M$ has nontrivial Jordan blocks. Do you want $M$ to be diagonalizable? Jun 29, 2012 at 1:54
• More generally, the holomorphic functional calculus defines $f(M)$ for any bounded linear operator $M$ on a Banach space and any function $f$ analytic in a neighbourhood of $\sigma(M)$. Jun 29, 2012 at 2:12
• In the case of a $d \times d$ Jordan block for eigenvalue $\lambda$, $f(M)_{i,i+k} = f^{(k)}(\lambda)/k!$ for $1 \le i \le i+k \le d$. Jun 29, 2012 at 2:17
• See e.g. Rudin, "Functional Analysis", sections 10.21 to 10.30. Jun 29, 2012 at 21:26

Please See first some quick explicit computations below:

I think the situation for to $$2\times 2$$ is more comprehensible. In oder to have a glimpse for the general case as mentioned by @Qiaochu Yuan We know that the complex plan $$\Bbb C$$, $$\Bbb C= \{a+ib: a,b\in\Bbb R\}$$ is isomorphic (as a field) to the to fields

$$G_2(\Bbb R) =\left\{\left(\begin{smallmatrix} a &b\\ -b&a\end{smallmatrix}\right): a,b\in\Bbb R\right\}$$

where one merely identify

$$1 \equiv \left(\begin{matrix} 1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv \left(\begin{matrix} 0& 1\\-1&0 \end{matrix}\right)$$

Now notion of exponential makes senses for complex numbers up to branch cut of the logarithm. Namely is not surprising to write $$z^w ~~\text{for}~~~z = a+ib, w =x+iy\in\Bbb C\setminus\{0\}.$$ which from the identification above readily represent the expression, $$\color{blue}{z^w\equiv \left(\begin{smallmatrix} a &b\\ -b&a\end{smallmatrix}\right)^\left(\begin{smallmatrix} x &y\\ -y&x\end{smallmatrix}\right)~\text{for}~~~a,b, x, y\in\Bbb R\setminus\{0\}}$$This expression is less abstract(but more particular and restricted) than, using the Taylor expansion for exponential and logarithm.

Another advantage of this expression is that, as opposed to the general cases where the expression of the $$\log(M+I)$$ $$\log(I + M) = \sum_{n \ge 1} \frac{(-1)^{n-1} M^n}{n}$$ converges only when $$\|M\|< 1$$, for the matrices taken in the fields $$G_2(\Bbb R)$$ such restriction is not require.

The above identification can be useful also to quickly compute the $$n^{th}$$ power of matrices(see here).

Application 1: Using the above identification we have,
$$\displaystyle \left(\begin{smallmatrix} 0 &1\\ -1&0\end{smallmatrix}\right)^\left(\begin{smallmatrix} 0 &1\\ -1&0\end{smallmatrix}\right)\equiv i^i = e^{-\frac{\pi}{2}}\equiv \left(\begin{smallmatrix} e^{-\frac{\pi}{2}} &0\\ 0&e^{-\frac{\pi}{2}}\end{smallmatrix}\right)$$

\begin{align}\displaystyle \lim_{n \to \infty}\left(\begin{matrix} 1&\dfrac{a}{n}\\\dfrac{-a}{n}&1\end{matrix}\right)^{\left(\begin{smallmatrix} n &0\\ 0&n \end{smallmatrix}\right)} &\equiv \displaystyle \lim_{n \to \infty}\color{red}{\left(1+\dfrac{ai}{n}\right)^n} \\&= \left(\begin{matrix} \cos a&\sin a\\-\sin a&\cos a\end{matrix}\right).\end{align}

Application 3:

$$\displaystyle\left(\begin{smallmatrix} 1 &1\\ -1&1\end{smallmatrix}\right)^\left(\begin{smallmatrix} 0 &1\\ -1&0 \end{smallmatrix}\right)\equiv (1+i)^i = \left(\sqrt{2}e^{i\frac{\pi}{4}}\right)^i = (\sqrt{2})^ie^{-\frac{\pi}{4}}\\\equiv \displaystyle\left(\begin{smallmatrix} e^{-\frac{\pi}{4}} \cos\log \sqrt{2} &e^{-\frac{\pi}{4}}\sin\log \sqrt{2}\\ -e^{-\frac{\pi}{4}}\sin\log \sqrt{2}&e^{-\frac{\pi}{4}}\cos\log \sqrt{2}&\end{smallmatrix}\right)$$

Application4: with help of Application 3: compute $$\displaystyle\left(\begin{smallmatrix} 1 &1\\ -1&1\end{smallmatrix}\right)^\left(\begin{smallmatrix} 2 &1\\ -1&2 \end{smallmatrix}\right)$$

This last one is left as an exercise to make sure the OP and future readers understood the rule of computation going on.

• Nicely thought out answer. Thanks for the heads up. Nov 19, 2017 at 23:07
• @DanielV Actually I think this can enlarge to the class of Quaternion which is also a field of dimension 3 . but the situation is more complex . Nov 20, 2017 at 4:25
• As multiplying by $i$ rotates the complex plane counterclockwise by a quarter turn, it's more natural to associate $i$ with the quarter-counterclockwise-turn matrix: $$\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$ Feb 6, 2019 at 19:43