Spectral theorem:

"Every self adjoint operator $T$ of a hermitian space $V$ is diagonalisable and its eigenvalues are real"

The proof consists of 3 parts:

  1. Proof that all the eigenvalues of $T$ are real.

  2. Induction proof that there exists a basis of eigenvectors of $V$ for any dimension of $V$.

  3. Proof that all the eigen-subspaces are orthogonal.

Why is that enough to say that T is diagonalisable

I understand that this means that we created a diagonal matrix who's coefficients are $T$'s eigenvalues and that we created an orthogonal matrix who's component are T's eigenvectors but how is that enough to show that T is indeed diagonalisable?

  • $\begingroup$ Every operator can be cast into triangular form using some basis transformation. If this operator is further self adjoint, it must be diagonal and all of its diagonal entries must be real in order to satisfy $[T]_D=[T]_D^\dagger$ where the subscript $D$ indicates diagonal form. $\endgroup$
    – nougako
    Jun 8, 2016 at 5:41
  • $\begingroup$ matrix or operator : you are considering finite vector spaces, not Hilbert spaces ? in that case you can prove only the singular value decomposition of matrices $\endgroup$
    – reuns
    Jun 8, 2016 at 10:02

1 Answer 1


An operator $T \colon V \rightarrow V$ is said to be diagonalizable if there exists a basis for $V$ consisting of eigenvectors of $T$. Part 2 shows precisely that.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .