Prove that matrix equation $AX-XA=I$ doesn't have a solution for any $A\in M_n(\mathbb{R})$

Note that $X\in\mathcal{M}_n(\mathbb{R}),$ else doing operations like $AX-XA$ have no sense. Use the trace and note that $$\mathrm{Tr}(AX-XA)=\mathrm{Tr}(AX)-\mathrm{Tr}(XA)=\mathrm{Tr}(AX)-\mathrm{Tr}(AX)=0$$ whereas $$\mathrm{Tr}(I)\geq 2.$$

Tags:

Linear Algebra

Matrices

Related

Intuitive or visual understanding of the real projective plane Show that a generalized knight can return to its original position only after an even number of moves How many possible phone words exist for a phone number of length N when also counting words less than length N within that phone number? If a matrix is triangular, is there a quicker way to tell if it is can be diagonalized? Number of $n^2\times n^2$ permutation matrices with a 1 in each $n\times n$ subgrid Why is $e$ the number that it is? Does there exist an $n$ such that all groups of order $n$ are Abelian? Is $(3^p-1)/2$ always squarefree? Proof of $\sum\limits_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim x \log(x) e^x$ as $x \to \infty$ What does the case of $\operatorname{Spec}C^{\infty}(M)$ tell us about the relavance of scheme theory to general rings? The "inverse" of $\nabla\times$ operator What are the applications of functional analysis?

Recent Posts