Proving that every diagonalizable operator is normal

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Solution 1

Remember that, in this case, $P^{-1}=P^\dagger$ (or, to put it better, you are assuming that there is an orthogonal matrix $P$ such that $PDP^{-1}=M$; since $P$ is orthogonal, $P^{-1}=P^\dagger$).

So,\begin{align}MM^\dagger&=PDP^\dagger(PDP^\dagger)^\dagger\\&=PDP^\dagger PD^\dagger P^\dagger\\&=PDD^\dagger P^\dagger\\&=PD^\dagger DP^\dagger\\&=PD^\dagger P^\dagger PDP^\dagger\\&=M^\dagger M.\end{align}

Solution 2

In that book, "diagonalizable" means "unitarily diagonalizable". In any case, their formal definition is that if $M$ is diagonalizable, then $M=\sum_j \lambda_j x_jx_j^*$. Now all you need to do is check that $M$ is normal: \begin{align} M^*M&=\left(\sum_{j=1}^n\lambda_jx_jx_j^*\right)^*\sum_{j=1}^n\lambda_jx_jx_j^* =\sum_{j=1}^n\overline{\lambda_j}(x_jx_j^*)^*\sum_{j=1}^n\lambda_jx_jx_j^*\\ \ \\ &=\sum_{k,j}\overline{\lambda_j}\lambda_kx_jx_j^*x_kx_k^* =\sum_{k=1}^n|\lambda_k|^2\,x_kx_k^*, \end{align} since $x_j^*x_k=\delta_{k,j}$. If you now do the same for $MM^*$, you will get the same result. So $M^*M=MM^*$, and $M$ is normal.

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Updated on April 12, 2020

Comments

  • NNN
    NNN over 3 years

    This question is insipired by the proof of the Spectral Theorem in Nielsen and Chuang's book (page 72). It says:

    Theorem: Any normal operator $M$ on a vector space $V$ is diagonal with respect to some orthonormal basis for $V$. Conversely, any diagonalizable operator is normal.

    They then go on to prove only the forward implication, and say that the converse is simple. My question is: how do I prove the converse?

    So far, I've tried writing $M = PDP^{-1}$, which would mean

    \begin{align} MM^{\dagger} &= PDP^{-1}(PDP^{-1})^\dagger\\ &= PDP^{-1}(P^{-1})^\dagger D^\dagger P^\dagger\\ M^\dagger M &= (P^{-1})^\dagger D^\dagger P^\dagger PDP^{-1} \end{align}

    I know that $D^\dagger = D$, so to show that both expressions are equal, it would be enough to show that $P^{-1} = P^\dagger$, but I don't see how to do that.

    For what it's worth, all that the book has said thus far about diagonalization is that an operator is diagonalizable if it can be written in the form $\sum_i \lambda_i\mid i \rangle \langle i \mid$. So far, all I know about normal matrices is that they satisfy $M^\dagger M = M M^\dagger$.

  • NNN
    NNN about 5 years
    Why is it true that $P$ is orthogonal? All I know is that I can write $P = \sum \lambda_i \mid i \rangle \langle i \mid$, where $\{\mid i\rangle\}_i$ form an orthonormal basis.
  • José Carlos Santos
    José Carlos Santos about 5 years
    The theore states this: “Any normal operator $M$ on a vector space $V$ is diagonal with respect to some orthonormal basis for $V$”. So, the converse is: “Any operator that is diagonalizable with respect to some orthogonal bases for $V$ is normal”. Os course, not all diagonalizable operators are normal. For instance,$$\begin{array}{rccc}M\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}^2\\&(x,y)&\mapsto&(y,y)\end{array}$$is diagonalizable, but it is not normal.
  • NNN
    NNN about 5 years
    Ok, I understand that the statement of the converse is not totally correct. But this still does not explain why $P$ is orthogonal. All I know right now is that I can write $P = \sum_i \lambda_i \mid i \rangle\langle i \mid$, where the set of $\mid i \rangle$ form an orthonormal basis. I don't understand how this implies orthogonality of $P$.
  • José Carlos Santos
    José Carlos Santos about 5 years
    Because if $M$ is diagonalizable with respect to an orthogonal basis, if $P$ is the matrix whose columns are the vectors of that basis, then $P$ is an orthogonal matrix and $P^\dagger MP=D$, for some diagonal matrix $D$.