14.7 The Square Root
Now here is a uniqueness and existence theorem for the square root. It follows from this theorem that U in
the above right polar decomposition of Theorem 14.6.5 is unique.
Theorem 14.7.1 Let A be a self adjoint and nonnegative n × n matrix (all eigenvalues are
nonnegative). Then there exists a unique self adjoint nonnegative matrix B such that B2 = A.
Proof: Suppose B2 = A where B is such a Hermitian square root for A with nonnegative eigenvalues.
Then by Theorem 14.1.6, B has an orthonormal basis for Fn of eigenvectors
because both linear transformations agree on the orthonormal basis. But this implies that
Thus these are also an orthonormal basis of eigenvectors for A. Hence, letting λi = μi2
be a polynomial such that
. Say p
and so B
commutes with every matrix which commutes with A
. To see this, suppose CA
This shows that if B is such a square root, then it commutes with every matrix C which commutes with A.
It also shows, by a repeat of the argument 14.14 - 14.15 that B2 = A.
Could there be another such Hermitian square root which has all nonnegative eigenvalues? It was just
shown that any such square root commutes with every matrix which commutes with A. Suppose B1 is
another square root which is self adjoint, and has nonnegative eignevalues. Since both B,B1 are
Now, adding these together, and using the fact that the two commute because they both commute with
every matrix which commutes with A,
It follows that both inner products in 14.17 equal 0. Next use the existence part shown above to take the
square root of B and B1 which is denoted by
and so, since x is arbitrary, B1 = B. ■
Corollary 14.7.2 The U in Theorem 14.6.5 is unique.