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7.3: Diagonal matrices

( \newcommand{\kernel}{\mathrm{null}\,}\)

Note that if T has n=dim(V) distinct eigenvalues, then there exists a basis (v1,,vn) of Vsuch that
Tvj=λjvj,for all j=1,2,,n.

Then any vV can be written as a linear combination v=a1v1++anvnof v1,,vn. Applying T to this, we obtain Tv=λ1a1v1++λnanvn.

Hence the vector M(v)=[a1an]

is mapped to M(Tv)=[λ1a1λnan].

This means that the matrix M(T)for T with respect to the basis of eigenvectors (v1,,vn) is diagonal, and so we call T diagonalizable: M(T)=[λ100λn].

We summarize the results of the above discussion in the following Proposition.

Proposition 7.3.1. If TL(V,V) has dim(V) distinct eigenvalues, then M(T) is diagonal with respect to some basis of V. Moreover, V has a basis consisting of eigenvectors of T. ​


This page titled 7.3: Diagonal matrices is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling.

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