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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!

AnInquiringMind
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  • 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. – zahbaz Aug 20 '18 at 02:31
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    @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

1 Answers1

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

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.

for instance, I believe that V needs to be an inner product space for the spectral decomposition to be formulated)

Spectral Theorem states that Every Normal matrix can be diagonalized by a complete set of orthonormal eigenvectors. So I don't think any other condition needs to be satisfied other than the normality condition.

artha
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  • 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