Similar Matrices, Hermitian and Skew Hermitian

Hermitian and Skew Hermitian

If M is a matrix, the transpose of M, written M^T, is the reflection of M through the main diagonal.

If M is square, as it usually is, then the diagonal is unchanged. This means the trace is unchanged. In fact, the determinant is unchanged, hence the norm is unchanged.

Subtract s from the main diagonal and take the determinant again. The resulting polynomial is the same, for M and M^T, hence the eigen values are the same, including their multiplicities.

The conjugate of a complex matrix is the conjugate of all its entries. The tranjugate is the transpose of the conjugate. This is written M^*. Note that M^* = M^T when M is real.

A symmetric or hermitian matrix has M = M^*. (We usually use the word symmetric when M is real, hermitian when M is complex.)

A skew symmetric or skew hermitian matrix has M = -M^*.

The diagonal of a hermitian matrix is real, whereas the diagonal of a skew hermitian matrix is pure imaginary.

As an exercise, show that every matrix is a unique sum of a hermitian matrix and a skew hermitian matrix.

Show that (AB)^T = B^TA^T, and (AB)^* = B^*A^*. If A and B are inverses, write AB = 1 and take the tranjugate of everything. This shows B^* and A^* are inverses. In other words, the inverse of the tranjugate is the tranjugate of the inverse.

Hermitian Operators

In general, hermitian operators are defined in terms of the dot product.

Let the vectors x and y be drawn from an arbitrary real or complex vector space S, with a valid dot product. The function f, from S into S,is hermitian, if f(x).y = x.f(y) for all vectors x and y.

Let's prove that every hermitian operator is linear. You will need to know that the space perpendicular to the space perpendicular to S is S. This is pretty obvious, but you can review the proof here.

Let x y and z be vectors and let c be a scalar. Watch what happens to f(cx).

f(cx).y =
cx.f(y) =
c×(x.f(y)) =
c×(f(x).y) =
cf(x).y

If y is perpendicular to f(x), with a zero dot product, then y is perpendicular to f(cx). This holds for all vectors y in the subspace P perpendicular to f(x). Therefore f(cx) is perpendicular to P, and is a multiple of f(x).

For some scalar b, f(cx) = bf(x). Write f(cf(x)).x = bf(f(x)).x = bf(x).f(x). At the same time, f(cf(x)).x = cf(x).f(x). If f(x) is 0 then bf(x) = 0, so we may as well write f(cx) = cf(x). When f(x) is nonzero f(x).f(x) is also nonzero. Hence b = c, and f commutes with scaling by a constant.

Now show f commutes with addition.

f(x+y).z =
(x+y).f(z) =
x.f(z) + y.f(z) =
f(x).z + f(y).z

If d = f(x+y)-(f(x)+f(y)), then d.z = 0 for every vector z. The only vector that is perpendicular to everything is 0, hence f(x+y) = f(x)+f(y), and f is linear.

If f and g are hermitian, show that c×f is hermitian, and f+g is hermitian. They both satisfy the dot product formula. Therefore the set of hermitian operators forms a vector space.

Hermitian Operator = Hermitian Matrix

For finite dimensions a function is hermitian iff its matrix is hermitian.

Let M be a hermitian matrix and consider x*M.y, where x is a row vector and y is a column vector. Here x*M runs x through the matrix M, thus implementing f(x). This result is dotted with y. Yet this is the same as multiplying by the column vector z, where z is the conjugate of y. So f(x).y = x*M*z.

Next evaluate x.f(y). Remember that f(y) is y*M, where y has become a row vector on the left. The result, y*M, becomes the second operand of the dot product. In other words, we have x.(y*M). Using matrix notation, this is x times the conjugate of y*M. Since M is hermitian, replace the conjugate of y*M with M*z, where z is the conjugate of y. Thus we obtain the same formula as before: x*M*z. A hermitian matrix implements a hermitian operator.

For the converse, let the matrix M implement a hermitian operator, and let the i^th component of x and the j^th component of y equal 1. In other words, x and y are basis vectors. Now x*M.y pulls out M_i,j, while x.(y*M) pulls out the conjugate of M_j,i. These are equal, hence M_i,j is always the conjugate of M_j,i, and M is a hermitian matrix.

Skew Hermitian

As you might surmise, f is skew hermitian if f(x).y = -x.f(y).

Review the above proofs. A skew hermitian operator is linear, and the set of skew hermitian operators on S forms a vector space. When S is finite dimensional, f is skew hermitian iff it is implemented by a skew hermitian matrix.

An Infinite Dimensional Example

Leaving finite dimensions, let S be the space of continuous functions on [a,b], with ∫f×g as inner product. Let w be a fixed continuous curve on [a,b], and let w(f) = w×f. Note that w is a linear operator. That is, w times cf is cwf, and w times f+g is wf+wg. But is w a symmetric operator? There is no matrix to consult, so apply the dot product criterion. Both w(f).g and f.w(g) become ∫wfg, so w is indeed a symmetric operator.

Next let S be the space of differentiable functions that are equal at the endpoints of [a,b]. That is, f(a) = f(b). Note that S is still a vector space. Also, the dot product f.g = ∫f×g is still well defined.

Let the operator d(f) = f′. In other words, d(f) takes the derivative of f. This is a linear operator. But is it symmetric?

Remember that the dot product is the integral of f×g, so d(f).g = ∫f′g. Use integration by parts; this is the same as fg -∫fg′. Yet the function f×g is equal at the endpoints of the interval, so that term drops out. We are left with -∫fg′, which is the same as -f.d(g). Therefore d(f).g = -f.d(g), and differentiation is a skew symmetric operator.