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4.2. Diagonalizable matrices

Two matrices of dimension n x n, say, A and B, are said to be similar if there exists a matrix T (similarity transformation matrix), such that

B = T - 1AT.

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

A (T ) = T (T - 1AT) = T (B) = T () = (T )

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 0 2 0 ... 0 : : 0 ... n
 = (1 1 2 2 ...n n ) = (A1 A2 ... An ) = A (1 ... n ) = AT => diag (1 , ..., n ) = T - 1AT

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.

Note. If the order of eigenvectors in T is changed, the same change of order happens in the resulting diagonal matrix.

Exercises: E37, E38, E39, E40, E41
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