In this section we’re going to take a look at a special kind of matrix.  We’ll start out with the following definition.

 

Definition 1  Suppose that A is a square matrix and further suppose that there exists an invertible matrix P (of the same size as A of course) such that  is a diagonal matrix.  In such a case we call A diagonalizable and say that P diagonalizes A.

 

The following theorem will not only tell us when a matrix is diagonalizable, but the proof will tell us how to construct P when A is diagonalizable.

 

Theorem 1  Suppose that A is an  matrix, then the following are equivalent. (a) A is diagonalizable. (b) A has n linearly independent eigenvectors.

 

Proof : We’ll start by proving that .  So, assume that A is diagonalizable and so we know that an invertible matrix P exists so that  is a diagonal matrix.  Now, let , , … ,  be the columns of P and suppose that D is the diagonal matrix we get from , i.e. .  So, both P and D have the following forms,

Also note that because P is an invertible matrix Theorem 8 from the Fundamental Subspaces section tells us that the columns of P will form a basis for  and hence must be linearly independent.  Therefore, , , … ,  are a set of linearly independent column vectors.

 

Now, if we rewrite  we arrive at  or,

 

Theorem 1 from the Matrix Arithmetic section tell us that the j^th column of PD is P[j^th column of D] and so the j^th column of PD is nothing more than .  The same theorem tells us that j^th column of AP is A[j^th column of P] or .

 

Now, since we have  the columns of both sides must be equal and so we must have,

 

 

So, the diagonal entries from D, , , … ,  are the eigenvalues of A and their corresponding eigenvectors are the columns of P, , , … , .  Also as we noted above these are a set of linearly independent vectors which is what we were asked to prove.

 

We now need to prove  and we’ve done most of the work for this in the previous part.  Let’s start by assuming that the eigenvalues of A are , , … ,  and that their associated eigenvectors , , … ,  are linearly independent. 

 

Now, form a matrix P whose columns are , , … , .  So, P has the form,

 

 

Now, as we noted above the columns of AP are given by

However, using the fact that , , … ,  are the eigenvectors of A each of these columns can be written as,

 

Therefore, AP can be written as,

 

However, as we saw above, the matrix on the right can be written as PD where D is the following diagonal matrix,

 

So, we’ve managed to show that by defining P as above we have .  Finally, since the columns of P are n linearly independent vectors in  we know that they will form a basis for  and so by Theorem 8 from the Fundamental Subspaces section we know that P must be invertible and hence we have,

where P is an invertible matrix.  Therefore A is diagonalizable.

 

Let’s take a look at a couple of examples.

 

Example 1  Find a matrix P that will diagonalize each of the following matrices. (a)   (b)   Solution Okay, provided we can find 3 linearly independent eigenvectors for each of these we’ll have a pretty easy time of this since we know that that the columns of P will then be these three eigenvectors.   Nicely enough for us, we did exactly this in the Example 6 of the previous section.  At the time it probably seemed like there was no reason for writing down specific eigenvectors for each eigenvalue, but we did it for the problems in this section.  So, in each case we’ll just go back to Example 6 and pull the eigenvectors from that example and form up P.   (a) This was part (a) from Example 6 and so P is,                                                            We’ll leave it to you to verify that we get,               (b) This was part (c) from Example 6 so P is,                                                              Again, we’ll leave it to you to verify that,

 

Example 2  Neither of the following matrices are diagonalizable. (a)   (b)   Solution To see that neither of these are diagonalizable simply go back to Example 6 in the previous section to see that neither matrix has 3 linearly independent eigenvectors.  In both cases we have only two linearly independent eigenvectors and so neither matrix is diagonalizable.   For reference purposes.  Part (a) of this example matches part(b) of Example 6 and part (b) of this example matches part (d) of Example 6.

 

We didn’t actually do any of the work here for these problems so let’s summarize up how we need to go about finding P, provided it exists of course.  We first find the eigenvalues for the matrix A and then for each eigenvalue find a basis for the eigenspace corresponding to that eigenvalue.  The set of basis vectors will then serve as a set of linearly independent eigenvectors for the eigenvalue.  If, after we’ve done this work for all the eigenvalues we have a set of n eigenvectors then A is diagonalizable and we use the eigenvectors to form P.  If we don’t have a set of n eigenvectors then A is not diagonalizable.

 

Actually, we should be careful here.  In the above statement we assumed that if we had n eigenvectors that they would be linearly independent.  We should always verify this of course.  There is also one case where we can guarantee that we’ll have n linearly independent eigenvectors.

 

Theorem 2  If , , … ,  are eigenvectors of A corresponding to the k distinct eigenvalues , , … ,  then they form a linearly independent set of vectors.

 

Proof : We’ll prove this by assuming that , , … ,  are in fact linearly dependent and from this we’ll get a contradiction and we’ll see that , , … ,  must be linearly independent.

 

So, assume that , , … ,  form a linearly dependent set.  Now, since these are eigenvectors we know that they are all non-zero vectors.  This means that the set  must be a linearly independent set.  So, we know that there must be a linearly independent subset of , , … , .  So, let p be the largest integer such that  is a linearly independent set.  Note that we must have  because we are assuming that , , … ,  are linearly dependent.  Therefore we know that if we take the next vector   and add it to our linearly independent vectors, , the set  will be a linearly dependent set. 

 

So, if we know that  is a linearly dependent set we know that there are scalars , not all zero so that,

(1)

 

Now, multiply this by A to get,

 

We know that the  are eigenvectors of A corresponding to the eigenvalues  and so we know that .  Using this gives us,

(2)

 

Next, multiply both sides of (1) by  to get,

and subtract this from (2).  Doing this gives,

 

Now, recall that we assumed that  were a linearly independent set and so the coefficients here must all be zero.  Or,

 

However the eigenvalues are distinct and so the only way all these can be zero is if,

 

Plugging these values into (1) gives us

but,  is an eigenvector and hence is not the zero vector and so we must have

 

So, what have we shown to this point?  Well we’ve just seen that the only possible solution to

 

is

 

This however would mean that the set  is linearly independent and we assumed that at least some of the scalars were not zero.  Therefore, this contradicts the fact that we assumed that this set was linearly dependent.  Therefore our original assumption that , , … ,  form a linearly dependent set must be wrong.

 

We can then see that , , … ,  form a linearly independent set.

 

We can use this theorem to quickly identify some diagonalizable matrices.

 

Theorem 3  Suppose that A is an  matrix and that A has n distinct eigenvalues, then A is diagonalizable.

 

Proof : By Theorem 2 we know that the eigenvectors corresponding to each of the eigenvalues are a linearly independent set and then by Theorem 1 above we know that A will be diagonalizable.

 

We’ll close this section out with a nice theorem about powers of diagonalizable matrices and the inverse of an invertible diagonalizable matrix.

 

Theorem 4  Suppose that A is a diagonalizable matrix and that  then, (a) If k is any positive integer we have,                                                                 (b) If all the diagonal entries of D are non-zero then A is invertible and,

 

Proof :

(a) We’ll give the proof for  and leave it to you to generalize the proof for larger values of k.  Let’s start with the following.

So, we can see that,

We can finish this off by multiplying the left of this equation by P and the right by  to arrive at,

 

(b) First, we know that if the main diagonal entries of a diagonal matrix are non-zero then the diagonal matrix is invertible.  Now, all that we need to show is that,

This is easy enough to do.  All we need to do is plug in the fact that from part (a), using , we have,

So, let’s do the following.

 

So, we’re done.