I am reading Hoffman and Kunze linear algebra book. In chapter $8$ there is a theorem that states as " Let $V$ be a finite-dimensional complex inner product space and $T$ a normal operator on $V.$ Then $V$ has an orthonormal basis consisting of characteristic vectors for $T".$ Which also says that every normal matrix is unitarily diagonalizable. The proof is totally depends on the fact that matrix or operator over complex field have characteristic vectors. Now my question is that, if $A$ is real normal matrix with real eigen values then, is it true that $A$ is diagonalizable over $\mathbb{R}?$ Please suggest me. Thanks in advance.
[Math] Diagonalizability of real Normal matrix.
linear algebra
Related Solutions
Being diagonal is not a property of an operator but of a matrix. That there exists an orthonormal basis $\beta$ in which $T$ is represented by a diagonal matrix doesn't imply that it is represented by a diagonal matrix in all orthonormal bases, so there's no contradiction here.
With respect to any orthonormal basis a self-adjoint operator is represented by a Hermitian (or self-adjoint) matrix, and the fact that there exists a basis $\beta$ in which it is represented by a diagonal matrix corresponds to the fact that every Hermitian matrix is diagonalizable by a unitary matrix.
Note: we will prove that the commutant of $T$ is equal to $K[T]$. This property is actually equivalent to the fact that the characteristic and minimal polynomials are equal. That's also equivalent to similarity to a companion matrix.
Let $\lambda_j$ be the $n$ pairwise distinct eigenvalues of $T$. Such an operator must be diagonalizable, and its eigenspaces are all one-dimensional. Since $U$ commute with $T$, it leaves the eigenspaces of $T$ invariant. Since they are one-dimensional, this implies that every eigenvector of $T$ is an eigenvector for $U$. So $T$ and $U$ are simultaneously diagonalizable.
In a basis of simultaneous diagonalization, $T=\mbox{diag}(\lambda_1,\ldots,\lambda_n)$ and $U=\mbox{diag}(\mu_1,\ldots,\mu_n)$.
Now we will do Lagrange interpolation. Since the $\lambda_j$ are pairwise distinct, we can consider the degree $n$ polynomials $$ L_j(x)=\prod_{i\neq j}\frac{x-\lambda_i}{\lambda_j-\lambda_i}\qquad L_j(\lambda_i)=\delta_{ij}. $$ Now $$p(x)=\sum_{j=1}^n\mu_jL_j(x)\qquad\mbox{satisfies }\quad p(\lambda_j)=\mu_j \;\forall j. $$ Therefore $p(T)=\mbox{diag}(p(\lambda_1),\ldots,p(\lambda_n))=U$ belongs to $K[T]$. The converse is clear, so the commutant of $T$ is $$ \{T\}'=\{U\in L(V)\,;\,UT=TU\}=K[T] $$ the subagebra of all polynomials in $T$.
Note: for a less explicit argument, just compare the dimensions of $K[T]$ and the subspace of all diagonal operators in this basis. The latter is clearly $n$. The former is $n$ by minimal polynomial consideration and Euclidean division. And clearly $K[T]$ is contained in the commutant, so they must be equal. Oh...but that's TTS argument...
Best Answer
If $A$ is normal, it is diagonalizable over $\mathbb C$. Hence its minimal polynomial $m(x)$ is a product of distinct linear factors over $\mathbb C$. Yet by assumption, all eigenvalues of $A$ are real. Therefore $m(x)$ is also a product of distinct linear factors over $\mathbb R$. Consequently $A$ is diagonalisable over $\mathbb R$.
(In fact, if $A$ is a normal matrix with a real spectrum, it must be real orthogonally diagonalisable and hence real symmetric, but that's another story.)