Dedicated to My Favorite Martian, a classic TV show from the 60's.

Multiplying matrices is a bit like multiplying numbers, with the identity matrix "I" replacing the number 1 in many ways. We've seen some differences, such as AB ¹ BA, and nonzero matrices A, B such that AB = O. This page looks at some odd things that can happen with powers of a matrix.

The examples are simple, and I think you can read most of this page any time after Ch 1 - just skip the remarks on Chs 4 and 6. At this time, it's not required reading, but it may help you with

• motivation to learn about transformations, similar matrices and diagonalization
• answering bonus questions on exams
• doing a project related to these topics

Example 1. Let's try to solve A² = I for A (a 2x2 matrix, let's say). We know x² = 1 implies x = 1 or x = -1. And we find that A = I, or A = - I solves the matrix equation. But there are other answers! We could choose the entries on the diagonal to be a₁₁ = 1 and a₂₂ = -1 (or vice versa). Also, if A solves this, any matrix B that is similar to A solves it too [see Ch 4.3]. This means we can pick any nonsingular 2x2 matrix S, and set B = S^-1AS, and B works. For example:

Check that A² = I and B² = I. Does this method describe all the solutions? I am not really sure! Especially for "nondiagonalizable" matrices (see "defective" matrices, Ch 6). If you are interested in this you might also read up on the Cayley-Hamilton theorem.

Example 2. I gave a bonus question on the Summer 2002 final - to find a matrix A¹ I, so that A³ = I (or to prove this is impossible). The reasoning of Example 1 doesn't help much. We know x³ = 1 implies x = 1, but this doesn't help us find an A¹ I (so most students guessed or "proved" it is impossible). There is a fairly easy solution, though. Let L be a rotation by 120 degrees (see Ch 4). So, L³ is rotation 360 degrees, which is the identity function. Let A be its matrix representation (also Ch 4). Then A³ = I. If we round off sin (2 pi / 3) = 3^1/2 / 2 = .85, an answer is:

Again, there are other solutions "similar" to this one. A completely different method is to pick a complex number, such as z = e^it, where t = 2pi/3, so that z³ = 1. And set A = z I. We'll cover complex numbers in Ch 6.4, but not in this detail.

If you followed this example, you should be able to find at least one solution to A^m = I , A ¹ I, for any choice of m (perhaps with some trig in the answer). There's also a 3x3 matrix M ¹ I with M⁴ = I described in my Ch 4 page on linear transformations.

Example 3. If A^m = O (the zero matrix), then A is called nilpotent. Any "seriously upper-triangular" matrix (one with zeroes on the diagonal) is nilpotent. For a small example, with m=2, set

Any nilpotent matrix must be singular (why? see Ch. 2). In fact, all it's eigenvalues must be zero (why? see Ch 6). Can you find a nilpotent matrix such that A³ = O but A² ¹ O ? Hint: this might be easier when A is a bit larger.

A different hint: Consider the derivative transformation D on the vector space P₃ (see Chs 3-4).The third derivative D³ is zero on functions in P₃. If the matrix representation of D is A, then the matrix representation of D³ is A³ = O. Why isn't A² = O, too?

Example 4. If A² = A then A is called idempotent. These are always singular (except for A = I). Any matrix that represents a projection onto a subspace S is idempotent. [If w Î S, then L(w) = w. So, since L(v) Î S, L(L(v))=L(v). So, L² = L, and A² = A.] For example, if you change any of the 1's in I to 0's, you get a projection matrix that is idempotent.

Suppose A is idempotent. Can you show that I-A is also idempotent?

Example 5. What if A³ = A ? (I have not seen this discussed in a text). If A is nonsingular, then A² = I as in Example 1 (why?). Also, any solution for Example 4, such as a projection, works. There are other solutions to A³ = A, in which A² ¹ A and A² ¹ I. Hint: look at singular 2x2 diagonal matrices.

Example 6. Suppose that B is given and we want to find a "square root", meaning a matrix C such that C² = B. [Example 1 is the special case of B = I]. This is easy if B is a diagonal matrix - just take a square root of each entry. Try it ! If some entry is negative, you'll get complex numbers in C.

If B is not diagonal, the best solution is to find a diagonal matrix A that is similar to it, so B = S^-1AS (Ch 6 explains how to do this). Then set C = S^-1 A^1/2 S (square both sides of this to check that C² = B). For example, if B is the same matrix given in Example 1, we already know what A and S are, so finding C is pretty easy:

Check that C² = B. If all B's eigenvalues (the diagonal entries of A) are positive, then the entries of C will be real. Note: some matrices are not diagonalizable and this method won't work. The matrix in Example 3 has no square root.

Summary: For many simple matrix equations, like A² = I, there are zillions of solutions. Rather than trying to calculate the entries of A directly, we can use ideas like linear transformations and diagonalization to find answers pretty quickly. If you are still in Ch. 1, be patient - this will make more sense soon !

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