# A linear operator T on a finite-dimensional vector space

1. May 16, 2009

### 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

2. May 16, 2009

### jbunniii

Re: Diagonalization

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.

3. May 16, 2009

### HallsofIvy

Re: Diagonalization

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.

4. May 16, 2009

### jeff1evesque

Re: Diagonalization

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

JL