In this case
|det (B - I) = det (T - 1AT - T - 1IT)|| = det [ T - 1 (A - I)T ] = det T - 1·det (A - I)· det T|| = det (A - I)|
And if is an eigenvalue of B with a corresponding eigenvector , then
or is an eigenvalue of A with T as a corresponding eigenvector. We have just proved
Proposition 7. If A and B similar (B = T - 1AT), they have the same characteristic polynomial, the same eigenvalues and if is an eigenvector of B, then T is the corresponding eigenvector of A.
We can use similarity, for example, in the following way:
|B = T - 1AT||=> ||B2 = (T - 1AT) (T - 1AT) = T - 1ATT - 1AT = T - 1A2T||=> ||B3 = B2B = (T - 1A2T) (T - 1AT) = T - 1A3T||Bk = T - 1AkT, k = 1, 2, ....||=> ||TBk = AkT||=> ||Ak = TBkT - 1, k = 1, 2, ....|
In particular, if B is a diagonal matrix and if T can easily be computed, it is then easy to compute Ak or determine the eigenvalues of A, and so on.
A is diagonalizable if it is similar to a diagonal matrix B.
Proposition 8. An n x n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors.
Proof. 1) Assume A is diagonalizable, i.e. D = T - 1AT, where D = diag (d1 , ..., dn )
=> The eigenvalues of D are d1 , ..., dn with corresponding eigenvectors 1 = (1, 0, ..., 0), 2 , ..., n , (among others) which are linearly independent. By Proposition 7, the eigenvalues of A are thus d1 , ..., dn and they have corresponding eigenvectors T1 , ..., Tn which are also linearly independent(why?).
2) Assume A has n linearly independent eigenvectors 1 , ..., n , Ai = i i . Denote T = (1 , 2 ...n ), i 's are columns. Then the rank of T n, so that T - 1 exists.
|=>||(1 2 ...n )||1||0||...||0|
|=||(1 1 2 2 ...n n )|
|=||(A1 A2 ... An )|
|=||A (1 ... n )|
|=||AT||=> ||diag (1 , ..., n ) = T - 1AT|
Example 1: Diagonalization of a matrix
Note. If the eigenvalues of A are all distinct, their corresponding eigenvectors are linearly independent and therefore A is diagonalizable. It is possible for a matrix A to have n linearly independent eigenvectors while it has eigenvalues with multiplicities grater than one.
Example 2: An application of diagonalization
Example 3: A non-diagonalizable matrix
Note. If the order of eigenvectors in T is changed, the same change of order happens in the resulting diagonal matrix.