# Linear independence of eigenvectors

Eigenvectors corresponding to distinct eigenvalues are linearly independent. As a consequence, if all the eigenvalues of a matrix are distinct, then their corresponding eigenvectors span the space of column vectors to which the columns of the matrix belong.

If there are repeated eigenvalues, but they are not defective (i.e., their algebraic multiplicity equals their geometric multiplicity), the same spanning result holds.

However, if there is at least one defective repeated eigenvalue, then the spanning fails.

These results will be formally stated, proved and illustrated in detail in the remainder of this lecture.

## Independence of eigenvectors corresponding to different eigenvalues

We now deal with distinct eigenvalues.

Proposition Let be a matrix. Let ( ) be eigenvalues of and choose associated eigenvectors. If there are no repeated eigenvalues (i.e., are distinct), then the eigenvectors are linearly independent.

Proof

The proof is by contradiction. Suppose that are not linearly independent. Denote by the largest number of linearly independent eigenvectors. If necessary, re-number eigenvalues and eigenvectors, so that are linearly independent. Note that because a single vector trivially forms by itself a set of linearly independent vectors. Moreover, because otherwise would be linearly independent, a contradiction. Now, can be written as a linear combination of :where are scalars and they are not all zero (otherwise would be zero and hence not an eigenvector). By the definition of eigenvalues and eigenvectors we have thatand that
By subtracting the second equation from the first, we obtainSince are distinct, for . Furthermore, are linearly independent, so that their only linear combination giving the zero vector has all zero coefficients. As a consequence, it must be that . But we have already explained that these coefficients cannot all be zero. Thus, we have arrived at a contradiction, starting from the initial hypothesis that are not linearly independent. Therefore, must be linearly independent.

When in the proposition above, then there are distinct eigenvalues and linearly independent eigenvectors, which span (i.e., they form a basis for) the space of -dimensional column vectors (to which the columns of belong).

Example Define the matrixIt has three eigenvalueswith associated eigenvectorswhich you can verify by checking that (for ). The three eigenvalues , and are distinct (no two of them are equal to each other). Therefore, the three corresponding eigenvectors , and are linearly independent, which you can also verify by checking that none of them can be written as a linear combination of the other two. These three eigenvectors form a basis for the space of all vectors, that is, a vectorcan be written as a linear combination of the eigenvectors , and for any choice of the entries , and .

## Independence of eigenvectors when no repeated eigenvalue is defective

We now deal with the case in which some of the eigenvalues are repeated.

Proposition Let be a matrix. If has some repeated eigenvalues, but they are not defective (i.e., their geometric multiplicity equals their algebraic multiplicity), then there exists a set of linearly independent eigenvectors of .

Proof

Denote by the eigenvalues of and by a list of corresponding eigenvectors chosen in such a way that is linearly independent of whenever there is a repeated eigenvalue . The choice of eigenvectors can be performed in this manner because the repeated eigenvalues are not defective by assumption. Now, by contradiction, suppose that are not linearly independent. Then, there exist scalars not all equal to zero such thatDenote by the number of distinct eigenvalues. Without loss of generality (i.e., after re-numbering the eigenvalues if necessary), we can assume that the first eigenvalues are distinct. For , define the sets of indices corresponding to groups of equal eigenvaluesand the vectorsThen, equation (1) becomesDenote by the following set of indices:The set must be non-empty because are not all equal to zero and the previous choice of linearly independent eigenvectors corresponding to a repeated eigenvalue implies that the vectors in equation (2) cannot be made equal to zero by appropriately choosing positive coefficients . Then, we haveBut, for any , is an eigenvector (because eigenspaces are closed with respect to linear combinations). This means that a linear combination (with coefficients all equal to ) of eigenvectors corresponding to distinct eigenvalues is equal to . Hence, those eigenvectors are linearly dependent. But this contradicts the fact, proved previously, that eigenvectors corresponding to different eigenvalues are linearly independent. Thus, we have arrived at a contradiction. Hence, the initial claim that are not linearly independent must be wrong. As a consequence, are linearly independent.

Thus, when there are repeated eigenvalues, but none of them is defective, we can choose linearly independent eigenvectors, which span the space of -dimensional column vectors (to which the columns of belong).

Example Define the matrixIt has three eigenvalueswith associated eigenvectorswhich you can verify by checking that (for ). The three eigenvalues are not distinct because there is a repeated eigenvalue whose algebraic multiplicity equals two. However, the two eigenvectors and associated to the repeated eigenvalue are linearly independent because they are not a multiple of each other. As a consequence, also the geometric multiplicity equals two. Thus, the repeated eigenvalue is not defective. Therefore, the three eigenvectors , and are linearly independent, which you can also verify by checking that none of them can be written as a linear combination of the other two. These three eigenvectors form a basis for the space of all vectors.

## Defective matrices do not have a complete basis of eigenvectors

The last proposition concerns defective matrices, that is, matrices that have at least one defective eigenvalue.

Proposition Let be a matrix. If has at least one defective eigenvalue (whose geometric multiplicity is strictly less than its algebraic multiplicity), then there does not exist a set of linearly independent eigenvectors of .

Proof

Remember that the geometric multiplicity of an eigenvalue cannot exceed its algebraic multiplicity. As a consequence, even if we choose the maximum number of independent eigenvectors associated to each eigenvalue, we can find at most of them because there is at least one defective eigenvalue.

Thus, in the unlucky case in which is a defective matrix, there is no way to form a basis of eigenvectors of for the space of -dimensional column vectors to which the columns of belong.

Example Consider the matrixThe characteristic polynomial isand its roots areThus, there is a repeated eigenvalue () with algebraic multiplicity equal to 2. Its associated eigenvectors solve the equationorwhich is satisfied for and any value of . Hence, the eigenspace of is the linear space that contains all vectors of the formwhere can be any scalar. In other words, the eigenspace of is generated by a single vectorHence, it has dimension 1 and the geometric multiplicity of is 1, less than its algebraic multiplicity, which is equal to 2. This implies that there is no way of forming a basis of eigenvectors of for the space of two-dimensional column vectors. For example, the vectorcannot be written as a multiple of the eigenvector . Thus, there is at least one two-dimensional vector that cannot be written as a linear combination of the eigenvectors of .

## Solved exercises

Below you can find some exercises with explained solutions.

### Exercise 1

Consider the matrix

Try to find a set of eigenvectors of that spans the set of all vectors.

Solution

The characteristic polynomial isand its roots areSince there are two distinct eigenvalues, we already know that we will be able to find two linearly independent eigenvectors. Let's find them. The eigenvector associated to solves the equationorwhich is satisfied for any couple of values such that or For example, we can choose , so that and the eigenvector associated to isThe eigenvector associated to solves the equationorwhich is satisfied for any couple of values such that or For example, we can choose , so that and the eigenvector associated to isThus, and form the basis of eigenvectors we were searching for.

### Exercise 2

Define

Try to find a set of eigenvectors of that spans the set of all column vectors having the same dimension as the columns of .

Solution

The characteristic polynomial iswhere in step we have used the Laplace expansion along the third row. The roots of the polynomial areHence, is a repeated eigenvalue with algebraic multiplicity equal to 2. Its associated eigenvectors solve the equationorThis system of equations is satisfied for any value of and . As a consequence, the eigenspace of contains all the vectors that can be written aswhere the scalar can be arbitrarily chosen. Thus, the eigenspace of is generated by a single vectorHence, the eigenspace has dimension and the geometric multiplicity of is 1, less than its algebraic multiplicity, which is equal to 2. It follows that the matrix is defective and we cannot construct a basis of eigenvectors of that spans the space of vectors.