So, I was reading the book Nielsen and Chuang and it's introductory chapter on Quantum Mechanics and it had a theorem called the "Spectral Decomposition Theorem" which states that an Operator $M$ is a normal operator if and only if it can be diagonalized in the orthonormal basis (which turn out to be the eigen-vectors).

Now, I have some trouble understanding the forward proof, which was to prove that if I have a Normal operator, it can be diagonalized. I am attaching the proof that is written in the book. Ultimately, it uses Projectors onto eigenspaces of a particular eigen value $P_\lambda=\sum_i |\lambda ; i\rangle \langle\lambda;i |$ and it's Orthogonal complement $Q_\lambda \equiv I-P_\lambda$ and rewrite $M=(P_\lambda+Q_\lambda)M(P_\lambda+Q_\lambda)$ which then simplifies to $M=P_\lambda MP_\lambda +Q_\lambda M Q_\lambda$ and using principle of mathematical induction, one can indeed prove that M can be diagonalized with respect to some orthonormal basis.

The part of the proof that I don't understand is that somehow, it implies that the operator $M$ can be diagonalized with respect to eigenvectors of this Normal Operator and that they are necessarily orthogonal. Also, if I try to write down $M$ in outer-product representation, it somehow simplifies as $M=\sum_i \lambda_i |e_i\rangle\langle e_i|$ where $\lambda_i$'s are the eigen-values and $|e_i\rangle$'s are the eigenvectors. I have literally no idea how this decomposition was implied by the above arguments. Any sort of help in the understanding of this is appreciated.

P.S.- I would really appreciate if you could explain the arguments with respect to the way that it is given in the proof in this book.

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1 Answer 1


$P$ and $Q$ are filtering down to complementary subspaces of $V$, so a matrix representation of $PMP+QMQ$ with respect to a basis that's made up of the union of two bases (or more precisely, concatenating the two ordered bases) of the pieces is going to look like the block matrix with nonzero entries in the upper left block and lower right block, and zeros elsewhere.

Since there is an orthonormal basis for the eigenspace and for its complement, the concatenated basis is still an orthonormal basis for the whole space. The matrix representation of both pieces is therefore going to be a block matrix of two diagonal matrices, which is again a diagonal matrix.

I believe your second question (about the outer product representation) was covered in this earlier post.

  • $\begingroup$ This still doesn't answer OP's question about why the matrix is diagonal in the "eigenvectors" basis and the representation in outer product notation. $\endgroup$
    – Tachyon209
    Aug 13, 2020 at 20:06
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    $\begingroup$ @Tachyon209 I think anyone who understands diagonalization understands that the diagonalizing matrix is made up of eigenvectors. I will take a closer look at the other part you mentioned. My reading was that the thrust of the question was why the inductive step works. $\endgroup$
    – rschwieb
    Aug 13, 2020 at 21:04
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    $\begingroup$ @Tachyon209 looks like the answer to the outer product representation part is already well covered on this site, so I included a reference. $\endgroup$
    – rschwieb
    Aug 15, 2020 at 15:16

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