When we introduced eigenvalues and eigenvectors, we wondered when a square matrix is similarly equivalent to a diagonal matrix? In other words, given a square matrix *A*, does a diagonal matrix *D* exist such that ? (i.e. there exists an invertible matrix *P* such that
*A* = *P*^{-1}*DP*)

In general, some matrices are not similar to diagonal matrices. For example, consider the matrix

Assume there exists a diagonal matrix

i.e is similar to . So they have the same characteristic equation. Hence

In this case, we must have

**Definition.** A matrix is **diagonalizable** if it is similar to a diagonal matrix.

**Remark.** In a previous page, we have seen that the matrix

has three different eigenvalues. We also showed that

**Theorem.** Let *A* be a square matrix of order n. Assume that *A* has n distinct eigenvalues. Then *A* is diagonalizable. Moreover, if *P* is the matrix with the columns *C*_{1}, *C*_{2}, ..., and *C*_{n} the n eigenvectors of *A*, then the matrix *P*^{-1}*AP* is a diagonal matrix. In other words, the matrix *A* is diagonalizable.

**Problem:** What happened to square matrices of order n with less than n eigenvalues?

We have a partial answer to this problem.

**Theorem.** Let *A* be a square matrix of order n. In order to find out whether *A* is diagonalizable, we do the following steps:

**1.**- Write down the characteristic polynomial

**2.**- Factorize
.
In this step, we should be able to get

where the , , may be real or complex. For every*i*, the powers*n*_{i}is called the**(algebraic) multiplicity**of the eigenvalue . **3.**- For every eigenvalue, find the associated eigenvectors. For example, for the eigenvalue ,
the eigenvectors are given by the linear system

Then solve it. We should find the unknown vector*X*as a linear combination of vectors, i.e.

where , are arbitrary numbers. The integer*m*_{i}is called the**geometric multiplicity**of . **4.**- If for every eigenvalue the algebraic multiplicity is equal to the geometric multiplicity, then we have

which implies that if we put the eigenvectors*C*_{j}, we obtained in**3.**for all the eigenvalues, we get exactly n vectors. Set*P*to be the square matrix of order n for which the column vectors are the eigenvectors*C*_{j}. Then*P*is invertible and

is a diagonal matrix with diagonal entries equal to the eigenvalues of*A*. The position of the vectors*C*_{j}in*P*is identical to the position of the associated eigenvalue on the diagonal of*D*. This identity implies that*A*is similar to*D*. Therefore,*A*is diagonalizable.

**Remark.**If the algebraic multiplicity*n*_{i}of the eigenvalue is equal to 1, then obviously we have*m*_{i}= 1. In other words,*n*_{i}=*m*_{i}.

**5.**- If for some eigenvalue the algebraic multiplicity is not equal to the geometric multiplicity, then
*A*is not diagonalizable.

**Example.** Consider the matrix

In order to find out whether

**1.**- The polynomial characteristic of
*A*is

So -1 is an eigenvalue with multiplicity 2 and -2 with multiplicity 1. **2.**- In order to find out whether
*A*is diagonalizable, we only concentrate ur attention on the eigenvalue -1. Indeed, the eigenvectors associated to -1, are given by the system

This system reduces to the equation -*y*+*z*= 0. Set and , then we have

So the geometric multiplicity of -1 is 2 the same as its algebraic multiplicity. Therefore, the matrix*A*is diagonalizable. In order to find the matrix*P*we need to find an eigenvector associated to -2. The associated system is

which reduces to the system

Set , then we have

Set

Then

But if we set

then

We have seen that if *A* and *B* are similar, then *A*^{n} can be expressed easily in terms of *B*^{n}. Indeed, if we have
*A* = *P*^{-1}*BP*, then we have
*A*^{n} = *P*^{-1}*B*^{n}*P*. In particular, if *D* is a diagonal matrix, *D*^{n} is easy to evaluate. This is one application of the diagonalization. In fact, the above procedure may be used to find the square root and cubic root of a matrix. Indeed, consider the matrix above

Set

then

Hence

Then we have

In other words,

**
**

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**Author**: M.A. Khamsi

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