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I've been struggling with this problem from Axler's Linear Algebra Done Right (this is problem 11 from section 7F) and was looking for some hints.

Problem: Suppose $\mathbb{F}=\mathbb{C}$ and $T \in \mathcal{L}(V)$. Prove that for every $\epsilon > 0$ there exists a diagonalizable operator $S \in \mathcal{L}(V)$ such that $0 < \| − \| < \epsilon$.

I was thinking of maybe making use of the SVD of $T$ and then constructing $S$ using the same singular vectors and close singular values but I cant seem to quite find a construction that gives normality (which would give diagonalizability by the spectral theorem).

brianjsl
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1 Answers1

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You can make use of the following fact: every linear endomorphism of $\mathbb{C}^n$ with characteristic polynomial having only simple roots (distinct roots) is diagonalizable. The problem now turned into finding such an operator arbitrarily close to the starting one. There are various ways to approach this problem but they all come down to the fact that the roots of a polynomial depend continously on the coefficients. I propose the following even if it is not the most elementary way to phrase the solution: consider $f(t, x_{11}, \dots, x_{nn})$ to be the characteristic polynomial of a matrix with entries $x_{11}, \dots, x_{nn}$, thought as a complex function. Having only simple root (in $t$) is equivalent (by the derivative criterion) to having no common factor (in $\mathbb{C}[t]$), or saying that the resultant (the determinant of the Sylvester matrix of $f$ and $f'$) is not zero. Every open set of the coefficients $\mathbb{C}^{n^2}$ contains such point: polynomials are holomorphic functions so if they are zero on an open set they are always zero.

Alessandro Fenu
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