# Spectral Decomposition and Polynomials

• Nov 17th 2011, 06:35 AM
H12504106
Spectral Decomposition and Polynomials
Let T be a normal operator on a finite-dimensional complex inner product space V. Use the spectral decomposition to show that If g is a polynomial, then
g(T) = g(m_1)T_1 + g(m_2)T_2 + ... + g(m_k)T_k, where the m_i's are the eigenvalues of T and the T_i's are the orthogonal projection of T on the eigenspace, W_i corresponding to the eigenvalue m_i.

I know that from the spectral decomposition, we have T = (m_1)(T_1) + (m_2)(T_2) + ... + (m_k)(T_k). Hence, g(T) = g((m_1)(T_1)) + g((m_2)(T_2)) + ... + g((m_k)(T_k)). But how do i pull out the T_i's to obtain the above result?

Thank You.
• Nov 17th 2011, 01:39 PM
Opalg
Re: Spectral Decomposition and Polynomials
Quote:

Originally Posted by H12504106
Let T be a normal operator on a finite-dimensional complex inner product space V. Use the spectral decomposition to show that If g is a polynomial, then
g(T) = g(m_1)T_1 + g(m_2)T_2 + ... + g(m_k)T_k, where the m_i's are the eigenvalues of T and the T_i's are the orthogonal projection of T on the eigenspace, W_i corresponding to the eigenvalue m_i.

I know that from the spectral decomposition, we have T = (m_1)(T_1) + (m_2)(T_2) + ... + (m_k)(T_k).

Start by computing monomials (powers of T):

$T^2 = (m_1T_1 + m_2T_2+\ldots+m_kT_k)(m_1T_1 + m_2T_2+\ldots+m_kT_k).$

Multiply out the brackets, remembering that the $T_i$ are pairwise orthogonal projections so that $T_i^2=T_i$ and $T_iT_j=0$ if $j\ne i.$ Thus

$T^2 = m_1^2T_1^2 + m_2^2T_2^2 + \ldots+m_k^2T_k^2.$

Use induction to prove the analogous result for each power of T. Finally, use linearity to deduce the result for a polynomial.