Given $u$ with negative eigenvalues, find a Hermitian inner product s.t. $\operatorname{Re}\langle u(x),x\rangle\leq 0$

Question

Let $u\in \mathrm{End}(\mathbb R^d)$ have all eigenvalues with negative real part. I need to show that there exists a Hermitian inner product $\langle\,\cdot\,|\,\cdot\,\rangle$ such that

$$ \operatorname{Re}\langle u(x)|x\rangle \leq 0\quad \forall x\in\mathbb C^d. $$

Edit : I already did the case where $u$ is Hermitian: I just take the standard inner product.

Answer 1

Let $\mathcal E =(e_i)$ be an upper trigonalisation basis for $u$. We thus have that $u$'s matrix in that basis is

$$ A = \operatorname{Mat}_\mathcal{E}(u) = \begin{pmatrix} \mu_1 & \cdots & \cdots & (\star) \\ & \mu_2 & & \vdots \\ & & \ddots & \vdots \\ (0) & & & \mu_d \end{pmatrix} $$

Let $P:(0,+\infty)\longrightarrow\mathrm{GL}_d(\mathbb C),\; \varepsilon\longmapsto \operatorname{diag}(\varepsilon^{-1},\ldots,\varepsilon^{-n})$. We have that for all $\varepsilon>0$

$$ P(\varepsilon)AP(\varepsilon)^{-1} = (\varepsilon^{j-i}a_{i,j})_{1\leqslant i,j\leqslant d}. $$

We call $\mathcal B_\varepsilon=(b_i(\varepsilon))$ be the new basis. We define the hermitian product $$ \langle x|y \rangle_\varepsilon := x^i\overline{y^j} $$ for $x=x^ie_i$, $y=y^ie_i$.

Thus, for all nonzero $x=x^ie_i\in\mathbb R^d$, \begin{align*} \langle u(x)|x \rangle_\varepsilon = \left\langle x^i\varepsilon^{k-i}a_{i,k}e_k\mid x^je_j \right\rangle_\varepsilon &= x^i\overline{x^j}\varepsilon^{k-i}a_{i,k}\langle e_k|e_j\rangle_\varepsilon = x^i\overline{x^j}\varepsilon^{j-i}a_{i,j}\\ &= \sum_{i=1}^d \mu_i|x_i|^2 + \underbrace{\sum_{\substack{1\leqslant i,j \leqslant d\\ i\neq j}}\epsilon^{j-i}a_{i,j}x^i\overline{x^j}}\limits_{=:\eta(\varepsilon)}. \end{align*}

So $$ \operatorname{Re}\langle u(x)|x\rangle_\varepsilon = \sum_{i=1}\operatorname{Re}(\mu_i)|x_i|^2+\operatorname{Re}\eta(\varepsilon) $$

We have that $\eta(\varepsilon) = \mathcal O(\varepsilon)$ as $\varepsilon\to 0$. The first term is negative. We thus take $\varepsilon>0$ such that $\operatorname{Re}\langle u(x)|x\rangle_\varepsilon$ is nonpositive, and take the inner product $\langle\,\cdot|\cdot\,\rangle_\varepsilon$ as the answer.

Answer 2

Note: The below no longer works since the question was updated.

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Begin by selecting any basis $\mathcal A = \{v_1,\dots,v_d\}$ that upper-triangularizes $u$.

Now, for a sufficiently small $\epsilon>0$, define the new basis $\mathcal B = \{w_1,\dots,w_n\}$ by $w_i = \epsilon^{i-1}v_i$. In particular, we'll have $$ [u]_{\mathcal A} = \pmatrix{\lambda_1 & a_{12}&\cdots & a_{1n}\\ &\lambda_2 & \ddots & \vdots\\ &&\ddots&a_{(n-1)n}\\ &&& \lambda_{n}}, \quad [u]_{\mathcal B} = \pmatrix{\lambda_1 & \epsilon \,a_{12}&\cdots & \epsilon^{n-1} a_{1n}\\ &\lambda_2 & \ddots & \vdots\\ &&\ddots& \epsilon \,a_{(n-1)n}\\ &&& \lambda_{n}} $$ and define your inner product relative to the basis $\mathcal B$. In particular, take $$ \langle w_i,w_j \rangle = \begin{cases} 1&{i=j}\\0&\text{otherwise}\end{cases} $$ and extend the definition by linearity.

Choose $\epsilon$ small enough so that by the Gershgorin cricle theorem, $[u]_{\mathcal B} + [u]_{\mathcal B}^T$ has negative eigenvalues.