Editing Companion matrix (section)

==Diagonalizability==
The roots of the characteristic polynomial <math>p(x)</math> are the [[eigenvalue]]s of <math> C(p) </math>. If there are ''n'' distinct eigenvalues <math>\lambda_1,\ldots,\lambda_n</math>, then <math> C(p) </math> is [[diagonalizable]] as <math> C(p)  = V^{-1}\!D V</math>, where ''D'' is the diagonal matrix and ''V'' is the [[Vandermonde matrix]] corresponding to the {{mvar|λ}}'s: 
<math display="block"> D = \begin{bmatrix}
\lambda_1 & 0 & \!\!\!\cdots\!\!\! & 0\\
0 & \lambda_2 & \!\!\!\cdots\!\!\! & 0\\
0 & 0 & \!\!\!\cdots\!\!\! & \lambda_n
\end{bmatrix}, 
\qquad 
V = \begin{bmatrix}
1 & \lambda_1 & \lambda_1^2 & \!\!\!\cdots\!\!\! & \lambda_1^{n-1}\\
1 & \lambda_2 & \lambda_2^2 & \!\!\!\cdots\!\!\! & \lambda_2^{n-1}\\[-1em]
\vdots & \vdots & \vdots & \!\!\!\ddots\!\!\! &\vdots \\
1 & \lambda_n & \lambda_n^2 & \!\!\!\cdots\!\!\! & \lambda_n^{n-1}
\end{bmatrix}.
</math>
Indeed, a reasonably hard computation shows that the transpose <math> C(p)^T </math> has eigenvectors <math> v_i=(1,\lambda_i,\ldots,\lambda_i ^{n-1}) </math> with <math> C(p)^T\!(v_i) = \lambda_i v_i </math>, which follows from <math> p(\lambda_i)=c_0+c_1\lambda_i+\cdots+c_{n-1}\lambda_i^{n-1}+\lambda_i  ^n = 0 </math>. Thus, its diagonalizing [[change of basis]] matrix is <math> V^T = [v_1^T \ldots v_n^T] </math>, meaning <math> C(p)^T = V^T D\, (V^T) ^{-1} </math>, and taking the transpose of both sides gives <math> C(p)  = V^{-1}\!D V</math>.

We can read the eigenvectors of <math> C(p) </math> with <math> C(p)(w_i) = \lambda_i w_i </math> from the equation <math> C(p)  = V^{-1}\!D V</math>: they are the column vectors of the [[Vandermonde matrix#Inverse Vandermonde Matrix|inverse Vandermonde matrix]] <math> V^{-1} = [w_1^T \cdots w_n^T] </math>. This matrix is known explicitly, giving the eigenvectors <math> w_i = (L_{0i}, \ldots, L_{(n-1)i}) </math>, with coordinates equal to the coefficients of the [[Lagrange polynomial|Lagrange polynomials]]
<math display="block">
L_i(x) = L_{0i} + L_{1i}x + \cdots + L_{(n-1)i}x^{n-1} = \prod_{j\neq i} \frac{x-\lambda_j}{\lambda_j-\lambda_i} 
= \frac{p(x)}{(x-\lambda_i)\,p'(\lambda_i)}. 
</math>
Alternatively, the scaled eigenvectors <math> \tilde w_i = p'\!(\lambda_i)\,w_i </math> have simpler coefficients.

If <math> p(x) </math> has multiple roots, then <math> C(p) </math> is not diagonalizable. Rather, the [[Jordan canonical form]] of <math> C(p) </math> contains one [[Jordan block]] for each distinct root; if the multiplicity of the root is ''m'', then the block is an ''m'' × ''m'' matrix with  <math> \lambda </math> on the diagonal and 1 in the entries just above the diagonal. in this case, ''V'' becomes a [[Vandermonde matrix#Confluent Vandermonde matrices|confluent Vandermonde matrix]].<ref> {{cite book
	| last1 = Turnbull
	| first1 =  H. W.
	| last2 = Aitken
	| first2 = A. C.
	| date = 1961
	| title = An Introduction to the Theory of Canonical Matrices
	| location =  New York
	| publisher = Dover
	| page = 60
	| isbn = 978-0486441689
}}
</ref>