Page 192 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes basis  which consists of vectors which are simultaneously characteristic vectors for every
i
operator in  .
i
Since T is diagonalizable, the lemma before Theorem 2 of unit 12 tells us that  = ( ,...,  ) is a
1 k
basis for V. That is the basis we seek.
If we consider finite dimensional vector space V over a complex field F, then there is a basis such
that the matrix of the linear operator T is diagonal. This is due to the key fact that every complex
polynomial of positive degree has a root. This tells us that every linear operator has at least one
eigenvector.
From the theorem above we now have that every complex n × n matrix A is similar to an upper
–1
triangular matrix i.e. there is a matrix P, such that P AP is upper triangular.
Equally we also state that for a linear operator T on a finite dimensional complex vector space V,
there is a basis of V such that the matrix of T with respect to that basis is upper triangular.
Let V contain an eigenvector of A, call it v . Let be its eigenvalue. We extend (v ) to a Basis
1 1
= (v , v , …, v ) for V. There will be a matrix P for which the new matrix A = P A P has the block
–1
1 2 n
form
  * 
A    
  O D  
where D is an (n – 1) × (n – 1) matrix, is a 1 × 1 matrix of the restriction of T to W (v ). Here O
1
denotes n – 1 zeros below in the first column. By induction on n, we may assume that there
exists a matrix Q such that Q D Q is upper triangular. If we denote Q by the relation
–1
1
 1 O 
Q 1    
  O Q  

then

  * 
1
A   Q A Q 1   1 
1
  O Q DQ  
is the upper triangular and thus
–1
A = (P Q ) A (P Q ).
1 1
Knowing one vector v corresponding to the characteristic value we can find a linear operator
P and then Q to find A .
1
Self Assessment

–1
–1
1. Find an invertible real matrix P such that P AP and P BP are both diagonal, where A and
B are the real matrices
8
 1 2  3  
(a) A    , B   
 0 2   0  1 
 1 1  1 a
(b) A    , B   
 1 1   a 1 

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