math111_logo Properties

1. About single matrix

In this part, we compare the eigenvalues and eigenvectors of a matrix and its modification.

Proof Since the determinant is not changed by transpose, we have

det(AT - λI) = det((A - λI)T) = det(A - λI).

By the definition of diagonalization, we have

A = PDP-1AT = (P-1)T DT PT = QDTQ-1,

where Q = (P-1)T is still invertible and DT is still diagonal.

The relation between eigenvectors of A and AT is rather complicated in general: The eigenvectors of A are columns of P. The eigenvectors of AT are columns of Q = (P-1)T.

Proof The first part follows from (also see this discussion)

det(PAP-1 - λI) = det(P(A - λI)P-1) = det(P-1P(A - λI)) = det(A - λI).

The second part follows from

Av = λvPAP-1(Pv) = λ(Pv).

For the third part, we note that since P is invertible, we have

{v1, v2, ..., vn} is a basis of eigenvectors for A ⇔ {Pv1, Pv2, ..., Pvn} is a basis of eigenvectors for PA.

Then we apply the definition of diagonalizability in terms of bases of eigenvectors.

Proof In the discussion leading to the computation of eigenvalues, we showed

λ is an eigenvalue of AA - λI is not invertible.

In particular, taking λ = 0 gives us

0 is an eigenvalue of AA is not invertible.

This is the same as the first part.

Now assume A is invertible and v is an eigenvector of A with eigenvalue λ. Then λ ≠ 0 and Av = λv. Applying λ-1A-1 to both sides, we get λ-1v = A-1v. This is the second part.

The third part can be proved by using either definition of diagonalizability, similar to the proof of the previous two problems.

The properties were stated in terms of matrices. They also hold for linear transformations. For example, T: VV and its dual T*: V*V* have the same eigenvectors and T is diagonalizable if and only if T* is diagonalizable.

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