`AepsiM_(n,n)(F)`, `F` is a field. `B` is a matrix containing the basis of `F^n` consisting of eigenvectors of `A` as column vectors. Can we show that `A` can be diagonalized?

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txmedteach eNotes educator| Certified Educator

This is a theorem I remember from linear algebra. The theorem is as follows:

Given an nxn matrix A and a matrix B consisting of the eigenvectors of A as column vectors, one can construct a diagonal matrix containing the eigenvalues of A, matrix D, with the following calculation (see the first link):

`D =B^-1AB`

This calculation is refered to as "diagonalizing" matrix A.

To show this is possible, we'll first consider some quick facts about B.

B must have rank "n" because it contains the linearly independent n eigenvectors of A.

Because B has a rank "n," we also know that it is invertible:

`B^-1` must exist, in other words.

Now, we consider the result of multiplying matrices A and B:

`AB = A[vec(b_(1)), vec(b_(2)), ... ,vec(b_(n))]`

Here, b_(1), b_(2), etc. are the eigenvectors of A. Continuing:

`A[vec(b_(1)), vec(b_(2)), ..., vec(b_(n))] = [Avec(b_(1)), Avec(b_(2)), ..., Avec(b_(n))]`

Now, we have to go back to the definition of an eigenvector (see the second link). Given a matrix A with eigenvalues `lambda_(1), lambda_(2),...,lambda_(n)` and eigenvectors `vec(x_(1)), vec(x_(2)),...,vec(x_(n))`, we know the following for any `iepsi{1, 2, ..., n}:`

`Avec(x_(i)) = lambda_(i)vec(x_(i))`

Now, given this information and our parameters that every `b_(i)` is an eigenvector of M, we can rewrite what we have above in the following way:

`[Avec(b_(1)), Avec(b_(2)), ..., Avec(b_(n))] = [lambda_(1)vec(b_(1)), lambda_(2)vec(b_(2)),...,lambda_(n)vec(b_(n))]`

This last part should strike you as very interesting, and here's why. This is the same result as if we multiply B with D as we defined above:

`BD = [lambda_(1)vec(b_(1)), lambda_(2)vec(b_(2)),...,lambda_(n)vec(b_(n))]`

This means, we have shown that BD = AB. Recall, we mentioned before that the inverse of B must exist, so now we can isolate D:

`B^-1BD = B^-1AB`


`D =B^-1AB`

And there you have it. We can find a diagonal matrix D simply with the above calculation given the conditions in the problem.