# Normal matrix over real inner product space with real eigenvalues is Hermitian

Let $V$ be a real inner product space. Let $T$ be a normal linear operator on V having only real eigenvalues. Then how can we prove that $T$ is also Hermitian?

If $V$ were a complex inner product space, the reasoning is provided in A normal matrix with real eigenvalues is Hermitian

However, I am not sure if such an argument holds if $V$ is restricted to a inner product space over the field $\mathbb{R}$. After all, in the equation $A = U^{*}DU$, the similarity matrix $U$ could have complex-valued entries which would render the argument inadmissible in the case where $V$ is a real inner product space.

• I must be missing sometuhng, because if it was real, it's just $U^* = U^t$. Mar 10, 2017 at 23:41
• That is correct, but the matrix could still have complex entries. Mar 10, 2017 at 23:46
• If the field is $\mathbb{R}$, why would it have complex entries in the first place? Mar 10, 2017 at 23:49
• Which is exactly my point, the argument doesn't hold for a real vector space. So what is an alternative way to prove the result? Mar 11, 2017 at 0:43
• @user308485, what exactly do you mean by 'let $T$ be a normal linear operator on $V$'? Are you assuming that there is a complex vector space structure on $V$? Mar 11, 2017 at 7:46

## 1 Answer

If $V$ is a real vector space with an inner product, we can use the Gram-Schmidt process to obtain an othonormal basis. Under this basis, the matrix $A$ associated to the linear operator $T$ will be a matrix with real entries such that $A=A^{\mathrm{T}}$. The point is that to study $T$, we only need to study $A$. So, we have transformed our question into a question regarding matrices. But from the answer to the question you cited, viewing $A$ as a complex matrix (then $A$ as a complex matrix is normal having only real eigenvalues), we conclude that $A^{*}=A$. This means that $A$ is Hermitian ($A^\mathrm{T}=A$) and thus $T$ is Hermitian.

In other words, your question is really only a question about matrices. Although if you want a more abstract point of view, what we are doing here is in fact working with the complexification of the inner product space to obtain a result about the original operator on the real inner product space.

Addendum:

If you really feel uneasy about treating real matrices as complex matrices, which you shouldn't feel uneasy about, and insist that everything be done in the real domain. Here is another way to prove it without using complex numbers.

The basic strategy is to use Schur's Decomposition Theorem for real matrices with real eigenvalues.

Suppose that $A$ is a real matrix with all of its eigenvalues being real numbers, then there is an orthogonal matrix $Q$, and an upper triangular real matrix $U$ such that $A=QUQ^{\mathrm{T}}$.

Proof. By induction on the size of $A$ (Cf. here). Let $\lambda_1$ be an eigenvalue of $A$, and $\pmb{x}_1\in \mathbf{R}^n$ an eigenvector corresponding to $\lambda_1$. Extend $\pmb{x}_1$ to an orthonormal basis $\pmb{x}_1, \pmb{x}_2, \ldots, \pmb{x}_n$. Let $Q_1$ be the matrix whose $i$th column is the vector $\pmb{x}_i$. Then we have $$AQ_1=A(\pmb{x}_1, \pmb{x}_2, \ldots, \pmb{x}_n) = (\pmb{x}_1, \pmb{x}_2,\ldots, \pmb{x}_n) \begin{pmatrix} \lambda_1 & \pmb{*} \\ \pmb{0} & A_1 \end{pmatrix} = Q_1 \begin{pmatrix} \lambda_1 & \pmb{*} \\ \pmb{0} & A_1 \end{pmatrix} .$$ Since $A_1$ has a smaller size, by induction, there exists an $(n-1)\times (n-1)$ orthogonal matrix $Q_2$ such that $A=Q_2U_2Q_2^\mathrm{T}$ where $U_2$ is an upper triangular matrix. Therefore we have $$AQ_1 =Q_1 \begin{pmatrix} \lambda_1 & \pmb{0} \\ \pmb{0} & Q_2U_2Q_2^\mathrm{T} \end{pmatrix}=Q_1 \begin{pmatrix} 1 & \pmb{0} \\ \pmb{0} & Q_2 \end{pmatrix} \begin{pmatrix} \lambda_1 & \pmb{0} \\ \pmb{0} & U_2 \end{pmatrix} \begin{pmatrix} 1 & \pmb{0} \\ \pmb{0} & Q_2 \end{pmatrix}^\mathrm{T}.$$ Now let $$Q=Q_1 \begin{pmatrix} 1 & \pmb{0} \\ \pmb{0} & Q_2 \end{pmatrix}\quad \text{and}\quad U= \begin{pmatrix} \lambda_1 & \pmb{0} \\ \pmb{0} & U_2 \end{pmatrix}.$$ Then $U$ is upper triangular and $Q$ is orthogonal and $A=QUQ^\mathrm{T}$.

After you prove the decomposition theorem, we can then show that if $A$ corresponds to your linear operator $T$, then $A=QUQ^{\mathrm{T}}$ and so using $AA^\mathrm{T}=A^\mathrm{T}A$, we get that $UU^{\mathrm{T}}=U^\mathrm{T}U$.

Suppose that $U$ is an upper triangle matrix with $UU^\mathrm{T}=U^\mathrm{T}U$, then $U$ is diagonal.

Proof. Denote the $ij$-entry of $U$ by $u_{ij}$. Then the for any $i$, we have $$\sum_{j=1}^n u_{ij}u_{ij} = \sum_{j=1} u_{ji}u_{ji}\implies \sum_{j=1}u_{ij}^2 =\sum_{j=1}^n u_{ji}^2. \tag1$$ But because $U$ is upper triangular, Let $i$ be $1$ in Eq.(1), then we get that on the first row of $U$, every entry except the one on the first column is zero. Now, let $i$ be $2$ in Eq. (1), we immediately get that on the second row of $U$, every entry except the one on the second column is zero. This carries on. So $U$ has to be diagonal.

Now, since $U$ is diagonal, it follows that $A^\mathrm{T}=QU^\mathrm{T}Q^\mathrm{T}= QUQ^\mathrm{T}=A$. This shows that $A$ is indeed Hermitian.