What Distinct Eigenvalues Can Tell You
When a list consists of eigenvectors, their eigenvalues can provide a quick test for linear independence. If every vector in the list has a different eigenvalue, the list cannot satisfy a nontrivial linear relation. If an eigenvalue repeats, that test no longer decides the question: vectors with the same eigenvalue may be dependent, but they may also be independent. The useful task is to distinguish what the eigenvalue information proves from what still needs to be checked.
Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). An eigenvector of \(\mathbf{A}\) is a nonzero vector \(\mathbf{v}\in\mathbb{R}^n\) for which \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for some scalar \(\lambda\in\mathbb{R}\); \(\lambda\) is its eigenvalue. A list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is linearly dependent if there are scalars \(c_1,\dots,c_p\), not all zero, such that \(\sum_{i=1}^p c_i\mathbf{v}_i=\mathbf{0}\). The coefficients matter: a nonzero coefficient is what makes the relation nontrivial.
The theorem is a one-way test: distinct eigenvalues guarantee independence. It does not say that vectors with a repeated eigenvalue are dependent, nor does it say that every dependent list of eigenvectors consists of vectors with one common eigenvalue. The next result states the conclusion you can draw when dependence does occur.
Proof. Suppose instead that every vector in the list has a different eigenvalue. The distinct-eigenvalue independence theorem then says that the list is linearly independent. This contradicts the assumption that it is dependent. Therefore at least two vectors must have the same eigenvalue. \(\square\)
This proposition is often most useful in contrapositive form: if the eigenvalues attached to the vectors are all different, a nontrivial linear relation is impossible. It does not identify a dependence relation when eigenvalues repeat; that requires further information about the vectors.
Using the Test on a Specified List
Worked Example: Three Different Eigenvalues
Let \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&3&0\\0&0&5\end{bmatrix}\), with \(\mathbf{v}_1=(1,0,0)^{\top}\), \(\mathbf{v}_2=(0,1,0)^{\top}\), and \(\mathbf{v}_3=(0,0,1)^{\top}\). Decide whether a nontrivial linear relation among these vectors is possible.
First verify the eigenvalue assigned to each vector by multiplying:
The eigenvalues \(2,3,5\) are pairwise distinct. By the distinct-eigenvalue independence theorem, the vectors are independent, so there is no nontrivial relation. The coefficient equations also verify the conclusion directly: if \(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}\), then
which forces \(c_1=c_2=c_3=0\). Thus any relation is trivial, as required.
In practice, apply the theorem only after checking the eigenvector and eigenvalue information. In particular, each listed vector must be nonzero and must satisfy the eigenvector equation for the matrix in question. If the problem gives eigenvalue labels but not the vectors, those labels are enough for the theorem; if a label or vector is uncertain, verify it before using the test.
When an Eigenvalue Repeats
If two or more vectors have the same eigenvalue, the distinct-eigenvalue theorem cannot be applied to the entire list. It is still possible to find a dependence relation by expressing one vector using others. It is also possible for the vectors to remain independent. These two cases are important: repetition removes the automatic guarantee of independence, but does not itself prove dependence.
Worked Example: Repeated Eigenvalue with a Dependence Relation
Let \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&5\end{bmatrix}\), and take \(\mathbf{u}_1=(1,0,0)^{\top}\), \(\mathbf{u}_2=(0,1,0)^{\top}\), and \(\mathbf{u}_3=(1,1,0)^{\top}\). Determine whether this list is dependent.
Multiplication verifies that all three vectors have eigenvalue \(2\):
Here the theorem does not settle independence because the eigenvalue is repeated. Instead, compare the vectors directly: \(\mathbf{u}_1+\mathbf{u}_2-\mathbf{u}_3=(1+0-1,\;0+1-1,\;0+0-0)^{\top}=(0,0,0)^{\top}\). The coefficients \(1,1,-1\) are not all zero, so this is a nontrivial relation and the list is dependent.
Check: the relation is exactly \((1,0,0)^{\top}+(0,1,0)^{\top}-(1,1,0)^{\top}=(0,0,0)^{\top}\). Substitution verifies every coordinate, and the nonzero coefficients verify that the relation is not trivial.
When eigenvalues repeat, it can help to sort the list into groups: one group for each eigenvalue. The next criterion explains what this grouping accomplishes. Within a group, ordinary linear-independence checks are still needed; between groups, distinct eigenvalues prevent a nontrivial relation from cancelling one group against another.
Proof. First suppose the full list is independent. Any relation among the vectors in one group can be extended to a relation among the full list by assigning coefficient \(0\) to every vector outside that group. Independence of the full list forces every coefficient in the group relation to be \(0\). Thus every group is independent.
For the reverse direction, suppose every eigenvalue group is independent and that a relation among the full list is given. For each distinct eigenvalue \(\lambda_j\), let \(\mathbf{y}_j\) be the sum of the terms in the relation belonging to that group, including their coefficients. The relation becomes \(\sum_j\mathbf{y}_j=\mathbf{0}\). Because each vector in that group satisfies \(\mathbf{A}\mathbf{v}=\lambda_j\mathbf{v}\), linearity gives \(\mathbf{A}\mathbf{y}_j=\lambda_j\mathbf{y}_j\). If \(\mathbf{y}_j\ne\mathbf{0}\), it is therefore an eigenvector with eigenvalue \(\lambda_j\). The nonzero \(\mathbf{y}_j\)'s have pairwise distinct eigenvalues, so the distinct-eigenvalue independence theorem applies to them. Their sum can be zero only if there are no nonzero \(\mathbf{y}_j\)'s. Hence every \(\mathbf{y}_j=\mathbf{0}\). Since each group is independent, all the coefficients in each group are zero. Thus the original relation is trivial, and the full list is independent. \(\square\)
Applying the Grouped Criterion
Worked Example: A Repeated Eigenvalue Does Not Force Dependence
Use \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&5\end{bmatrix}\) and consider \(\mathbf{w}_1=(1,1,0)^{\top}\), \(\mathbf{w}_2=(1,-1,0)^{\top}\), and \(\mathbf{w}_3=(0,0,1)^{\top}\). Test the list.
Check the eigenvector equations:
The eigenvalue \(2\) occurs twice, so the distinct-eigenvalue theorem alone does not settle the full list. Check the group for eigenvalue \(2\): if \(a\mathbf{w}_1+b\mathbf{w}_2=\mathbf{0}\), then \((a+b,\;a-b,\;0)^{\top}=\mathbf{0}\). Thus \(a+b=0\) and \(a-b=0\). Adding gives \(2a=0\), so \(a=0\), and then \(b=0\). This group is independent. The group for eigenvalue \(5\) contains one nonzero vector and is independent as well.
By the grouped independence criterion, the full list is independent. Direct substitution checks the conclusion: if \(a\mathbf{w}_1+b\mathbf{w}_2+c\mathbf{w}_3=\mathbf{0}\), its coordinates give \(a+b=0\), \(a-b=0\), and \(c=0\). The first two equations imply \(a=b=0\), so all three coefficients vanish.
The grouped criterion gives a complete way to test a list of eigenvectors when the eigenvalues are known. Check independence within each group. If any one group is dependent, then the full list is dependent, because its nontrivial relation remains a relation when the other groups receive coefficient \(0\). If every group is independent, the theorem proves the whole list independent, even when some eigenvalues repeat.
Common Mistakes and a Reliable Checklist
Mistake: “A repeated eigenvalue means the eigenvectors are dependent.” In the last example, \(\mathbf{w}_1\) and \(\mathbf{w}_2\) both have eigenvalue \(2\), but the equations \(a+b=0\) and \(a-b=0\) force \(a=b=0\). They are independent. The correction is to treat repeated eigenvalues as a reason to check the vectors within that group, not as a verdict about them.
Mistake: “Different vectors must have different eigenvalues.” The second example gives three different vectors with the same eigenvalue, and they satisfy \(\mathbf{u}_1+\mathbf{u}_2-\mathbf{u}_3=\mathbf{0}\). Being different as vectors does not make the eigenvalues different and does not guarantee independence.
For a specified list, use this sequence:
- Verify that each vector is nonzero and satisfies \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\).
- If all eigenvalues are distinct, conclude that the list is independent.
- If an eigenvalue repeats, group the vectors by eigenvalue and test independence within each group.
- If you claim dependence, write a nontrivial relation and substitute the vectors to check it.
The distinct-eigenvalue test is powerful precisely because it is decisive when its condition holds. When that condition fails, the grouped criterion keeps the reasoning precise: repeated eigenvalues require an additional check, while distinct eigenvalues ensure that independent groups cannot cancel one another.
Check Your Understanding
For each question, state what the eigenvalue information proves and, where needed, show the coefficient calculation.
- Three eigenvectors have eigenvalues \(1\), \(4\), and \(7\). Can they satisfy a nontrivial linear relation? Explain.
- For \(\mathbf{A}=\begin{bmatrix}3&0\\0&3\end{bmatrix}\), test whether \((1,0)^{\top}\) and \((1,1)^{\top}\) are independent, and verify their eigenvalue equations.
- Suppose a list of eigenvectors is dependent. What must be true of the eigenvalues attached to its vectors?
- A list has two vectors with eigenvalue \(2\), and one with eigenvalue \(5\). What must be checked before the grouped independence criterion can establish that the full list is independent?
- Explain why a relation involving vectors from distinct eigenvalue groups cannot have nonzero group sums that cancel to zero.
