# The relationship between spectral decomposition / eigendecomposition and projection operators

I am trying to clarify the relationship between the spectral decomposition / eigendecomposition of a matrix and projection operators.

I understand that there is a connection between diagonalizability of a linear operator / matrix and projection operators in the following sense:

Given a finite-dimensional vector space $V$ ($\dim V = n$) and a linear operator $T \in L(V)$ that has $k \leq n$ distinct eigenvalues $\lambda_1, \dots, \lambda_k$, if $T$ is diagonalizable, then there exist $k$ linear operators $E_1, \dots, E_k$ on $V$ such that the $E_i$ are projection operators, $I = \sum_i E_i$, and $T = \sum_i \lambda_i E_i$, etc. (and an analogous converse also holds).

I also understand that the spectral decomposition / eigendecomposition of an $n \times n$ matrix $\mathbf{A}$ with $n$ linearly independent eigenvectors can be written as $\mathbf{A = Q \boldsymbol{\Lambda} Q^{-1}}$, where $\mathbf{Q}$ is the matrix whose $i$th column is the eigenvector $q_i$ of $\mathbf{A}$, and $\Lambda_{ii} = \lambda_i$.

If $\mathbf{A}$ is a normal matrix, then $\mathbf{Q}$ is unitary, so that $$\mathbf{A} = \sum_{i=1}^n\lambda_iq_i q_i^*,$$ where $q_i^*$ is the adjoint of $q_i$.

Does this mean, then, that the projection operator associated with $\lambda_i$ can be related to the sum of outer products of eigenvectors with the same eigenvalue:

$$E_i = \sum_{j=1}^{r_i} q_j q_j^*,$$

where $r_i$ is the algebraic multiplicity of $\lambda_i$? And then all of the associated properties of the projection operators hold?

Or are there additional assumptions/conditions (other than $\mathbf{A}$ being normal) that need to be taken to be true for this relationship to hold (for instance, I believe that $V$ needs to be an inner product space for the spectral decomposition to be formulated)? Any additional information on the relationship between these two paradigms would be greatly appreciated. Thank you!

• That's true of the Hermitian matrices that represent linear operators on the complex Hilbert spaces of quantum mechanics. I believe, as you said, the result extends to any normal matrix acting on an inner-product space, but I'm not confident enough to make this an answer. You might find the wiki on the spectral theorem in Hilbert spaces relevant. – suneater Aug 20 '18 at 2:31
• @zahbaz I would really like to see an online resource that explores this connection between projection operators/matrices and the $q_j q_j^∗$ form (in particular when a projection matrix is a sum of more than one eigenvector outer product) in more explicit detail. But, unless I'm missing something, it seems hard to come by. Any suggestions? Thanks! – AnInquiringMind Aug 21 '18 at 14:07

Yes. For a normal matrix, when $T$ is diagonalizable, it can be decomposed into: $$T = \lambda _1P_1 + \lambda _2P_2 + ...$$
The Projection matrices $P_i$ or $q_j q_j^*$ form eigenspaces. For a repeated eigenvalue, the corresponding eigenvectors form the basis of an eigenspace. These eigenspaces are orthogonal to each other.
• Thanks for the response. From my searching online, it seems to me that this connection between projection operators/matrices and the $q_j q_j^*$ form isn't typically explicitly explored (in particular when a projection matrix is a sum of more than one eigenvector outer product). Do you know of any good resources where they do explore this relationship in more detail? Thanks! – AnInquiringMind Aug 21 '18 at 14:04