1
$\begingroup$

I'm reading Lawrence Perko's Differential Equations and Dynamical Systems, and he writes the following :

Theorem. If the eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$ of an $n \times n$ matrix $A$ are real and distinct, then any set of corresponding eigenvectors $\{v_1, v_2, \ldots, v_n\}$ forms a basis for $\mathbb{R}^n$, the matrix $P := [v_1 \space v_2 \space \cdots \space v_n]$ is invertible and $$ P^{-1}AP = \text{diag}[\lambda_1, \ldots, \lambda_n]. $$ This theorem says that if a linear transformation $T : \mathbb{R}^n \to \mathbb{R}^n$ is represented by the $n \times n$ matrix $A$ with respect to the standard basis $\{e_1, e_2, \ldots, e_n\}$ for $\mathbb{R}^n$, then with respect to any basis of eigenvectors $\{v_1, v_2, \ldots, v_n\}$, $T$ is represented by the diagonal matrix of eigenvalues, $\text{diag}[\lambda_1, \ldots, \lambda_n]$.

My question is : Would it be ok to write, more generally, that

"This theorem says that if a linear transformation $T : V \to V$, where $V$ is an n-dimensional real (or complex) vector space, is represented by the $n \times n$ matrix $A$ with respect to a given basis $C := \{c_1, c_2, \ldots, c_n\}$ for $V$, then with respect to any basis of eigenvectors $\{v_1, v_2, \ldots, v_n\}$, $T$ is represented by the diagonal matrix of eigenvalues, $\text{diag}[\lambda_1, \ldots, \lambda_n]$."

if we consider that the vectors $v \in V$ are identified to $\mathbb{R}^n$ via their components in $C$ ?

That is, is the author only giving the particular interpretation where $V = \mathbb{R}^n$ and $C$ is the standard basis for $\mathbb{R}^n$, in which case the identification to $\mathbb{R}^n$ of the vectors $v \in \mathbb{R}^n$ via their components in $C$ coincides with the vectors $v$ themselves ? Is it only a conventional interpretation ?

$\endgroup$
  • $\begingroup$ A general linear transformation does not necessarily have a basis of eigenvectors. The best we can do is represent $T$ in Jordan normal form (or rational canonical form). $\endgroup$ – D Wiggles Jan 29 '14 at 3:50
  • $\begingroup$ @IBWiglin Yes, but the hypothesis implies that a basis of eigenvectors exists ? $\endgroup$ – Amateur Jan 29 '14 at 4:05
  • $\begingroup$ In that case, your restatement is correct. $\endgroup$ – D Wiggles Jan 29 '14 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.