Similar Matrices, Real/Imaginary Eigen Values

Real/Imaginary Eigen Values

In this section we will show that a hermitian operator has orthogonal eigen vectors and real eigen values. We already know that a hermitian matrix is normal, and has orthogonal eigen vectors, but let's prove this for general hermitian operators, even those that cannot be implemented by a matrix.

Let t be a hermitian operator, x an eigen vector, and s its associated eigen value. Thus t(x) = sx. Rewrite this as t(x).x over x.x = s.

Remember that in the world of complex numbers, the dot product is not commutative. In fact, u.v is the conjugate of v.u. Also, x.x is real, and is not affected by conjugation. Therefore the conjugate of s is x.t(x) over x.x. Yet x.t(x) is the same as t(x).x, since t is hermitian, hence the conjugate of s equals s, and s is real. All eigen values are real.

An analogous proof shows the eigen values of a skew hermitian operator are pure imaginary.

Next let t be hermitian, and let x and y be eigen vectors with distinct eigen values j and k. We already showed j and k are real. Remember that t(x).y = x.t(y). Rewrite this as j×x.y = k×x.y. With j and k distinct, x.y = 0, and x and y are orthogonal.

If t is skew hermitian than j and k are imaginary. Write t(x).y = -x.t(y), hence j×x.y = -l×x.y, where l is the conjugate of k. Since k is imaginary l is simply -k, and once again x.y = 0 whenever j and k are distinct. Different eigen values always lead to orthogonal vectors.

Next consider a hermitian operator in finite dimensional space. It can be implemented by a matrix M, which is normal, and has an orthogonal eigen basis. By the above, all its eigen values are real. Similarly, a finite dimensional skew hermitian operator has an orthogonal eigen basis with imaginary eigen values. Let's see if we can prove the converse.

Let M be a matrix with orthogonal eigen vectors, hence it is normal. Also, assume its eigen values are all real. Write M = Q*D/Q, where Q is unitary and D is diagonal. We can do this because M is normal. Now the entries of D are the eigen values of D, and of M, so D is a real matrix. Also, the tranjugate of Q is its inverse. Take the tranjugate of Q*D/Q and get Q*D/Q back again. Thus M equals its tranjugate, and M is hermitian. A finite dimensional linear operator is hermitian iff it has an orthogonal eigen basis with real eigen values.

An analogous Q*D/Q proof shows M is skew hermitian iff it has an orthogonal eigen basis and imaginary eigen values.

Symmetric Matrices and Real Transformations

An important corollary arises when M is symmetric, i.e. hermitian and real. We know M has orthogonal eigen vectors, and their eigen values are real. Use the last variant of Schur's theorem to find a real, orthonormal matrix P, such that PM/P is upper triangular. In this case the upper triangular matrix must be diagonal. Therefore PM/P = D, and all entries are real numbers.

The Inverse of a Symmetric Matrix

Let M be a hermitian invertible matrix. Write M as QD/Q, where D is diagonal, and holds the eigen values. If E is the inverse of D, then M inverse is QE/Q. Like M, the eigen values of M inverse are real, and its eigen vectors are orthogonal. This makes geometric sense. If M expands the x y and z axes by real eigen values, the inverse of M will contract these very same axes by the same eigen values. Therefore M inverse has real eigen values and orthogonal eigen vectors, and is hermitian. The inverse of a hermitian matrix is hermitian.

A similar proof shows the inverse of a skew hermitian matrix is skew hermitian.

Assume M is symmetric, whence it is hermitian and real. We showed above that Q is real, hence QE/Q is real. Thus M inverse is hermitian and real, and the inverse of a symmetric matrix is symmetric.

Of course there is a simpler proof involving cofactors.

Skew Hermitian plus Identity

This theorem has no applications, but it's interesting. If you add the identity matrix to a skew hermitian matrix, the result is always nonsingular.

Let S be a skew hermitian matrix. Such a matrix has n independent eigen vectors, and their eigen values are all pure imaginary.

Let v be such a vector. Thus sv = cv, where c is an imaginary scaling factor. What happens to Sv if we add ones down the main diagonal of S? Well we have cv as before, plus something else. The first element of v is added to the first component of the product. The second element of v is added to the second component, and so on. In other words, v is added to the result.

With sv = (1+c)v, v remains an eigen vector. Since c is imaginary, 1+c is nonzero. In other words, v is not squashed down to zero; it is a true eigen vector. This holds for all n linearly independent eigen vectors. The new matrix S+I has n linearly independent eigen vectors, and is nonsingular.