In this case
det (B - I) = det (T^{ - 1}AT - T^{ - 1}IT) |
= 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^{ - 1}AT), 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^{ - 1}AT | |
=> | B^{2} = (T^{ - 1}AT) (T^{ - 1}AT) = T^{ - 1}ATT^{ - 1}AT = T^{ - 1}A^{2}T |
=> | B^{3} = B^{2}B = (T^{ - 1}A^{2}T) (T^{ - 1}AT) = T^{ - 1}A^{3}T |
B^{k} = T^{ - 1}A^{k}T, k = 1, 2, .... | |
=> | TB^{k} = A^{k}T |
=> | A^{k} = TB^{k}T^{ - 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 A^{k} 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^{ - 1}AT, where D = diag (d_{1 }, ..., d_{n })
=> The eigenvalues of D are d_{1 }, ..., d_{n } with corresponding eigenvectors _{1 } = (1, 0, ..., 0), _{2 }, ..., _{n }, (among others) which are linearly independent. By Proposition 7, the eigenvalues of A are thus d_{1 }, ..., d_{n } and they have corresponding eigenvectors T_{1 }, ..., T_{n } which are also linearly independent(why?).
2) Assume A has n linearly independent eigenvectors _{1 }, ..., _{n }, A_{i } = _{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 }) |
= | (A_{1 } A_{2 }... A_{n }) |
= | A (_{1 }... _{n }) |
= | AT |
=> | diag (_{1 }, ..., _{n }) = T^{ - 1}AT |
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.