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November 28, 2012

An easier way to square some 2×2 matrices

Filed under: mathematics — ckrao @ 11:16 am

The Cayley-Hamilton theorem states that a square matrix satisfies its characteristic equation. That is, if A is a matrix with characteristic equation p(x) = \det (x I - A) then p(A) = 0. (Here I is the identity matrix of the same dimension of A.)

The characteristic equation of a 2×2 matrix is a quadratic of the form p(x) = (x - \lambda_1)(x - \lambda_2) = x^2 - a x+ b, where a = \lambda_1 + \lambda_2and b = \lambda_1 \lambda_2, where \lambda_1 and \lambda_2 are the eigenvalues of the matrix A. Since the sum of the eigenvalues of a matrix is the sum of its diagonal elements (the trace) and the product of the eigenvalues is the determinant, we can rewrite this as

\displaystyle p(x) = x^2 - \text{tr} (A) x + \det (A).

By the Cayley-Hamilton theorem,

\displaystyle 0_{2 \times 2} = A^2 - \text{tr}(A) A + \det (A) I,

from which

\displaystyle A^2 = \text{tr} (A) A - \det (A) I.

In certain cases this formula might be an easier way to calculate the square of a 2×2 matrix than matrix multiplication, especially if the determinant is a nice number.

For example, if A = \left[ \begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array} \right], we easily see that its trace is 3 + 1 = 4 and its determinant is 3\times 1 - 2 \times 1 = 1, so

\displaystyle A^2 = 4A - I = \left[ \begin{array}{cc} 4\times 3 - 1 & 4 \times 2 \\ 4 \times 1 & 4 \times 1 - 1\end{array} \right] = \left[ \begin{array}{cc} 11 & 8 \\ 4 & 3\end{array} \right].

The same formula may be applied a second time if we wish to find A^3:

\begin{aligned} A^3 &= A. A^2\\ &= A(\text{tr} (A) A - \det (A) I)\\ &= \text{tr} (A) A^2- \det (A) A \\ &= \text{tr} (A) (\text{tr} (A) A - \det (A) I) - \det (A) A \\ &= \left( \left( \text{tr} (A) \right)^2 - \det (A) \right) A - \text{tr} (A) \det (A) I. \end{aligned}

So in our example,

\displaystyle \left[ \begin{array} {cc} 3 & 2 \\ 1 & 1 \end{array} \right] ^3 = (4^2 - 1) A - 4 \times 1 I = \left[ \begin{array}{cc} 41 & 30 \\ 15 & 14 \end{array} \right] .

Finally, in case you were wondering how the Cayley-Hamilton theorem can be proved, one way is via the equation

\displaystyle X (\text{adj} (X) ) = \det (X) I.

The adjugate \text{adj} (X) of an n \times n matrix is a matrix of (n-1) \times (n-1) determinants up to sign. The equation may be proved by considering the (i,j) entry of each side and using the alternating sum formula for the determinant of a matrix. All that we need to be concerned with here is that if X = xI - A then \text{adj} (X) is a polynomial of degree n-1. We can then write the above equation as

\displaystyle (x I - A) (x^{n-1} B_{n-1} + x^{n-2} B_{n-2} + \ldots + B_0) = (x^n + a_{n-1} x^{n-1} + \ldots + a_0)I,

where B_0, B_1, \ldots B_{n-1} are constant matrices and a_0, a_1, \ldots a_{n-1} are constant scalars.

From this we simply match coefficients of powers of x of each side. (For convenience set B_{-1} = B_n = 0.)

\displaystyle B_{k-1} - A B_k = a_k I, \quad k = 0, 1, \ldots n.

Multiplying both sides of this equation on the left by A^k and then summing from k =0 to n gives

\displaystyle \sum_{k=0}^n a_k A^k = \sum_{k=0}^n (A^k B_{k-1} - A^{k+1} B_k) = A^0 B_{-1} - A^{n+1} B_n = 0.

But the left side of this equation is the characteristic equation \sum_{k=0}^n a_k x^k evaluated at A, and we are done.

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