Definition (T -invariant subspace) Example. Example

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1 Eigenvalues, Eigenvectors, Similarity, and Diagonalization We now turn our attention to linear transformations of the form T : V V. To better understand the effect of T on the vector space V, we begin by looking for T -invariant subspaces of V. Definition (T -invariant subspace) Let V be a vector space, let W be a subspace of V, and let T : V V be a linear transformation. We say that W is a T -invariant subspace of V if T (W ) W. In other words, we say W is T -invariant if for all w W, it must be the case that T (w) W. Example If T : P P is differentiation (where P is the vector space of all polynomials with real coefficients), then P k is T -invariant for all k. Example If T : V V is a linear transformation, and W is a one-dimensional T -invariant subspace of V, then there exists a scalar λ such that T (w) = λw for all w W.

2 Definition (direct sum) Let W 1,..., W k be subspaces of a vector space V. If every vector v V can be written as v = w w k where w i W i, then we say V is a sum of the subspaces W 1,..., W k and write V = W W k. Furthermore, if W i W j = {0} when i j, then we say V is the direct sum of W 1,..., W k and write V = W 1 W k. In this case, the expression v = w w k is unique.

3 Example R 3 is a direct sum of the xy-plane and the z-axis. We understand a linear transformation T : V V if we understand what T does to all of the T -invariant subspaces of V. As such, we might ask for a direct sum decomposition V = W 1 W k of V into T -invariant subspaces. Knowing what T does to each W i will completely determine what T does to V. The simplest T -invariant subspaces are the one-dimensional T -invariant subspaces. As we saw in the example above, the vectors in these spaces are simply scaled when T acts on them. Our study of eigenvalues and eigenvectors will help us better understand such T -invariant subspaces.

4 Definition (eigenvalue, eigenvector) Let V be a vector space over F, and let T : V V be a linear transformation. An eigenvalue of T is a scalar λ F such that there is a nonzero vector v V such that T (v) = λv. In this case, we say v is an eigenvector of T associated to λ. Also, we define the eigenspace of λ to be E λ = {v V T (v) = λv} = {v V (T λi )v = 0} = ker(t λi ). Note Suppose V is finite-dimensional. If B is a basis for V, then T (v) = λv [T (v)] B = [λv] B [T ] B [v] B = λ[v] B. In other words, λ is an eigenvalue of T if and only if it is an eigenvalue of any matrix representation [T ] B of T. We can use this fact when actually trying to find the eigenvalues of T!

5 Example Example

6 Because the action of T on each of its eigenspaces is so straightforward, it s tempting to ask if V can be written as a direct sum of its eigenspaces: V = E λ1 E λk. If this is indeed the case, we say that T is diagonalizable. Definition (diagonalizable) Let V be a finite-dimensional vector space, and let T : V V be a linear transformation. Then T is diagonalizable if there is a basis C for V such that the matrix [T ] C is a diagonal matrix. Note that if [T ] C is diagonal, then the vectors in C are eigenvectors for T, and V is a direct sum of the eigenspaces of T.

7 Note If V is finite-dimensional with basis B, then by Theorem 6.29 ([T ] C = P 1 [T ] B P), T : V V is diagonalizable if and only if [T ] B is similar to a diagonal matrix. Theorem Let V be a finite-dimensional vector space. A linear transformation T : V V is diagonalizable if and only if V has a basis of eigenvectors of T.

8 Remember, in Math 12 we learned that... If T : V V is a linear transformation, then if λ 1,..., λ m are distinct eigenvalues for T with corresponding eigenvectors v 1,..., v m, then {v 1,..., v m } is linearly independent if V is n-dimensional, and T has n distinct eigenvalues, then T is diagonalizable (but the converse is false!) T is invertible if and only if 0 is not an eigenvalue of T T is diagonalizable if and only if the algebraic and geometric multiplicity of every eigenvalue is equal the eigenvalues of a triangular matrix are on the diagonal of the matrix similar matrices have the same characteristic polynomial (and thus the same eigenvalues) not all linear transformations are diagonalizable being diagonalizable has nothing to do with being invertible, and vice versa!

9 Jordan Canonical Form Even though not all linear transformations are diagonalizable, it turns out we can get pretty close. More specifically, let V be a finite-dimensional complex vector space, and let T : V V be a linear transformation. Then there is a basis B (of generalized eigenvectors ) of V such that [T ] B is block-diagonal and the blocks are all Jordan blocks :

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