A linear operator T on a finite-dimensional vector space

[jeff1evesque]

Definitions: A linear operator T on a finite-dimensional vector space V is called diagonalizable if there is an ordered basis B for V such that [T]_B is a diagonal matrix. A square matrix A is called diagonalizable if L_A is diagonalizable.

We want to determine when a linear operator T on a finite-dimensional vector space V is diagonalizable and, if so, how to obtain an ordered basis B = {v_1, v_2, ... , v_n} for V such that [T]_B is a diagonal matrix. Note that, if D = [T]_B is a diagonal matrix, then for each vector v_j in B, we have

T(v_j) = [SUMMATION: from i = 1 to n]D_i_jv_i = D_j_jv_j = (lambda_j)v_j
where (lambda_j) = Djj.

Questions: Could someone explain the following:
1. T(v_j) = [SUMMATION: from i = 1 to n]D_i_jv_i

2. And maybe touch upon the other two equality relation in the line above.



Thanks,


JL

 

[jbunniii]

Definitions: A linear operator T on a finite-dimensional vector space V is called diagonalizable if there is an ordered basis B for V such that [T]_B is a diagonal matrix. A square matrix A is called diagonalizable if L_A is diagonalizable.

We want to determine when a linear operator T on a finite-dimensional vector space V is diagonalizable and, if so, how to obtain an ordered basis B = {v_1, v_2, ... , v_n} for V such that [T]_B is a diagonal matrix. Note that, if D = [T]_B is a diagonal matrix, then for each vector v_j in B, we have

T(v_j) = [SUMMATION: from i = 1 to n]D_i_jv_i = D_j_jv_j = (lambda_j)v_j
where (lambda_j) = Djj.

Questions: Could someone explain the following:
1. T(v_j) = [SUMMATION: from i = 1 to n]D_i_jv_i

2. And maybe touch upon the other two equality relation in the line above.

T is a linear operator mapping from V to V. So for each element v_j of the basis, you can represent T(v_j) uniquely as a linear combination of \{v_1,\ldots,v_n\}. The coefficients of that linear combination are precisely the elements of the j'th column of [T]_B.

T(v_j) = \sum_{i=1}^{n} D_{ij} v_i

expresses exactly what I wrote above: on the right hand side we are holding j (the column index) fixed, and summing over i (the row index).

Now, if the matrix is diagonal, only one of the D_{ij}'s in a given column is nonzero, which is why the next equality is true:

\sum_{i=1}^{n} D_{ij} v_i = D_{jj}v_j

So now you have the fact that

T(v_j) = D_{jj} v_j

Since v_j \neq 0 (because it is an element of a basis), this equality says precisely that D_{jj} is an eigenvalue of T, and v_j is a corresponding eigenvector. Thus the suggestive \lambda_j in place of D_{jj} in the last equality.

 

[HallsofIvy]

A linear operator, L, (and its associated matrix) is diagonalizable if and only if there is a basis for the space consisting entirely of eigenvectors of L. In that case we can 'diagonalize' L by D= P^-1LP where P is the matrix having the eigenvectors of L as columns and D is the diagonal matrix having the eigenvalues of L on its main diagonal.

 

[jeff1evesque]

Thanks for the clarification. Now I have to try to commit this to memory.

JL