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

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.

• There's a nice approach here if $u$ is diagonalizable, but otherwise it gets a bit tricky. – Omnomnomnom Nov 21 '16 at 15:49
• "Let $u$ have negative eigenvalues" means (a) it has some negative eigenvalues (b) all its (real) eigenvalues are negative, or (c) all complex roots of its characteristic polynomial are real (so they are eigenvalues) and negative. – Marc van Leeuwen Nov 21 '16 at 17:05
• Also, which are you talking about a Hermitian inner product if your vector space is real (symmetric would seem more appropriate). – Marc van Leeuwen Nov 21 '16 at 17:07
• I have updated the question according to your remark. The original question didn't actually say if $x$ lied in $\mathbb R^d$, but real $x$ was used in subsequent questions. – Groovy. Nov 21 '16 at 17:09
• @Omnomnomnom Can't one extend the approach to non-diagonalizable $u$ via density/continuity arguments? – Fimpellizieri Nov 21 '16 at 18:47

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

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.

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.