# How to understand that a normal matrix commutes with its adjoint, but a diagonalizable matrix may not?

According to the answer to this question, the geometrical meanings of normal and diagonalizable matrices are: normal matrices have orthogonal eigenvectors whiles diagonalizable have independent eigenvectors (not necessarily orthogonal).

And according to its definition, a normal matrix shares the same eigenvectors with its adjoint matrix (conjugate transpose).

My question is: how to connect the geometrical meanings of normal matrices to the definition? In other words, how to understand the fact that a matrix shares the same eigenvectors with its adjoint if it has orthogonal eigenvectors, but it may not be so if it only has independent but NOT orthogonal eigenvectors?

On the flip side, how to understand, if a matrix shares the same eigenvectors with its adjoint, then it has orthogonal eigenvectors?

• That's not quite stated properly. An $n \times n$ matrix is diagonalizable iff it has $n$ linearly independent eigenvectors. An $n \times n$ matrix is normal iff it has $n$ orthogonal eigenvectors. – Robert Israel Jul 19 '18 at 18:30
• The heart of the matter is having a geometrical meaning of the adjoint of a matrix. Would that be helpful? – Christopher A. Wong Jul 19 '18 at 19:32
• @ChristopherA.Wong I agree. Could you please elaborate the geometrical meaning of the adjoint of a matrix? – chaohuang Jul 20 '18 at 1:07
• @ChristopherA.Wong And how is the geometrical meaning of the adjoint of a matrix related to the question above? Thanks. – chaohuang Jul 20 '18 at 1:14
• @chaohuang A geometric interpretation of the adjoint is not crystal clear to me, actually. The symmetric transpose (e.g. for real matrices) is much easier compared to the complex case, so I'll have to think about how to explain it clearly. If you have this geometrical meaning then it'll be easier to understand why "matrix is normal" is equivalent to "orthogonal eigenvectors". – Christopher A. Wong Jul 25 '18 at 19:59

Every diagonalizable matrix $N$ on a finite-dimensional vector space $X$ is normal with respect to some inner product on $X$. Indeed, if $B=\{ b_1,b_2,\cdots,b_n \}$ is a basis for $X$ such that $Nb_k=\lambda_k b_k$, then you can define an inner product $\langle \cdot,\cdot\rangle_{B}$ by
$$\langle \alpha_1b_1+\cdots+\alpha_n b_n,\beta_1 b_1+\cdots+\beta_n b_n\rangle_{B}=\sum_{j=1}^{n} \alpha_j\beta_j^*,$$ and you have the adjoint $N^*$ of $N$ with respect to this inner product given by $$N^*(\alpha_1 b_1+\cdots+\alpha_n b_n)=\alpha_1^*b_1+\cdots+\alpha_n^*b_n.$$ So $N^*N=NN^*$ because $N^*,N$ share eigenvectors. Furthermore, the basis of eigenvectors is orthonormal with respect to $\langle\cdot,\cdot\rangle_B$. So, normality is not more or less general than diagonalizability; it's a matter of choosing the right inner product in order to make a diagonalizable $N$ normal.
For a linear operator $$A:V\to V$$ on a finite dimensional vector space, to be diagonalizable (also called semisimple) is an intrinsic property; i.e., it does not need any further structure.
The adjoint, however, needs an inner product. Call it $$g:V\times V\to \mathbb R$$. We also view it as an isomorphism $$\check g:V\to V^*$$. The adjoint $$A^{*,g}$$ of $$A$$ with respect to $$g$$ is given by either $$g(Av,w)=g(v,A^{*,g}w)$$ or by $$A^{*,g} = \check g^{-1}.A^*.\check g: V\to V^*\to V^*\to V$$, where $$A^*:V^*\to V^*$$ is the "dual operator" between the dual spaces. Running through all inner products $$g$$ then gives you all possible adjoints.