Do matrices have a "to the power of" operator?

Question

Well I was sure that saying "$A^3$" (where $A$ is an $n\times n$ matrix) is nonsense. Sure one could do $(A\cdot A) A$ But that contains different operators etc. So what did my prof mean by the following statement:

show that $A^{25}\mathbf{x} = \mathbf{0}$ has only the trivial solution? (We're also given the determinant of A).

I know the proof will probably end with stating: "This means that $A^{25}$ is invertible, so $A^{25}\mathbf{x} = \mathbf{0}$ has only the trivial solution. And well I could state that $\det(A^{25}) = 5^{25} \neq 0$.

But then again: I really wonder what the "to the power of" operator means? Or did my prof make a mistake here?

Answer 1

If you want a formal definition of matrix exponentiation for non-negative integer values, just define $A^n = A^{n-1}\cdot A$ and $A^0 = I$. Since matrix multiplication is associative, we won't have any ambiguity there.

Edit: As Tobias Kildetoft points out below, it might be wiser to define the base case as $A^1=A$ instead of $A^0=I$, so as to not have to worry about how $\det(A^n)=\det(A)^n$ for $\det(A)=0$ would imply $0^0=1$. Which isn't false, depending on how we want to define it, but is something we might not want to worry about for the purposes of defining matrix powers.

Answer 2

As for numbers, if you know how to multiply square matrices together, the "power operation" is then just iteration of multiplication (this works, because your matrix is square). Therefore, what you thought was nonsense is very much "sensical".

Also, since $\det(AB)=\det(A)\det(B)$, you have $\det(A^n)=\det(A)^n$, which is what you need to prove what you want to prove.

Answer 3

$$A^3=A\cdot A\cdot A$$

$$A^{25}=\underset{25\text{ of these}}{\underbrace{A\cdot A\cdots A\cdot A}}$$

Since matrix multiplication is associative, this is completely natural and acceptable.

Answer 4

Just an add-on: the multiplication of matrices you refer to is not commutative (in general) but associative. This important fact allows you to define in an unambiguous way the powers $A^n$, for all $n>2$.

Answer 5

The "to the power of" operation with matrices requires square matrices for the typical interpretation (the "dot" matrix multiplication method), but it could be redefined in some way so that an $m$ by $n$ matrix could have an exponent applied to it. In many programming packages, $A$^$n$ means typical matrix multiplication, while $A$.^$n$ means element-by-element exponentiation.

Answer 6

The square matrix is a representation of an endomorphism in a given basis of the vector space and the product of matices is defined to be the representation of the composition of endomorphisms so if a matrix $A$ represents the endomorphism $f$ in a basis $\mathcal B$ then $A^2$ represents $f^2:=f\circ f$

Answer 7

But matrix multiplication is not commutative, $A^{2}\cdot A^{3} \neq A^{3}\cdot A^{2}$ except in some particular cases. $$ A = \pmatrix{0 & 3 \\ 1&2}; A^2 = \pmatrix{3 & 6 \\ 2 & 7}; A^3 = \pmatrix{6& 21 \\ 9 & 20} $$ $$A^2·A^3 = \pmatrix{72 & 183\\ 75 & 182}; A^3·A^2 = \pmatrix{60 & 183 \\ 61 & 182} $$ So $A^5$ has more than 1 solution, at least in this case.

Answer 8

As the currently-accepted answer has mentioned, the elementary “repeated multiplication” of exponentiation to integer exponents applies just as well to square matrices as it does to $\mathbb{R}$ or $\mathbb{C}$.

But matrix exponentiation can be generalized. Suppose that the matrix $A$ has the factorization $A = BDB^{-1}$, where $D$ is a diagonal matrix. Then, because of associativity of multiplication $A^2 = (BDB^{-1})(BDB^{-1}) = BD(B^{-1}B)DB^{-1} = BDDB^{-1} = BD^2B^{-1}$, and in general, $A^n = BD^nB^{-1}$.

Raising a diagonal matrix to a power is equivalent to raising its diagonal elements to that power, which can be extended to non-integer exponents.