Distinct Eigenvalues Give Independent Eigenvectors

A proof that eigenvectors associated with distinct eigenvalues are linearly independent, which is the main doorway to diagonalization.
Modified

April 26, 2026

Keywords

proof, eigenvalues, eigenvectors, diagonalization

1 Claim

The key theorem behind diagonalization is:

different eigenvalues force genuinely different directions.

Theorem 1 (Theorem) Let \(A \in \mathbb{R}^{n \times n}\), and let \(v_1,\dots,v_k\) be eigenvectors of \(A\) with distinct eigenvalues \(\lambda_1,\dots,\lambda_k\).

Then \(v_1,\dots,v_k\) are linearly independent.

2 Why This Proof Matters

This is the theorem that starts turning eigenvalue computation into structural understanding.

It matters because:

  • it explains why distinct eigenvalues automatically help diagonalization
  • it lets you count independent modes in a dynamical system
  • it keeps reappearing in graph and spectral arguments

3 Dependencies

  • \(Av_i = \lambda_i v_i\) for each eigenpair
  • a linearly independent list is the same as a list with only the trivial linear relation

4 Strategy Before Details

Use induction on the number of eigenvectors.

The main trick is to start from a hypothetical linear relation and then apply \((A-\lambda_k I)\) to kill the last vector while keeping the others alive with shifted eigenvalues.

5 Full Proof

Proof. We prove the statement by induction on \(k\).

For \(k=1\), the statement is immediate because an eigenvector is nonzero.

Now assume the theorem holds for \(k-1\) distinct eigenvalues, and suppose

\[ c_1 v_1 + \cdots + c_k v_k = 0 \]

for eigenvectors \(v_1,\dots,v_k\) with distinct eigenvalues \(\lambda_1,\dots,\lambda_k\).

Apply \(A-\lambda_k I\) to both sides:

\[ (A-\lambda_k I)(c_1 v_1 + \cdots + c_k v_k) = 0. \]

Because \((A-\lambda_k I)v_k = 0\), this becomes

\[ c_1(\lambda_1-\lambda_k)v_1 + \cdots + c_{k-1}(\lambda_{k-1}-\lambda_k)v_{k-1} = 0. \]

The eigenvalues are distinct, so each coefficient \((\lambda_j-\lambda_k)\) is nonzero for \(j<k\).

Therefore the vectors \(v_1,\dots,v_{k-1}\) satisfy a linear relation. By the induction hypothesis, they are linearly independent, so

\[ c_1(\lambda_1-\lambda_k)=\cdots=c_{k-1}(\lambda_{k-1}-\lambda_k)=0. \]

Hence

\[ c_1=\cdots=c_{k-1}=0. \]

Returning to the original relation, we get

\[ c_k v_k = 0. \]

Since \(v_k \neq 0\), it follows that \(c_k=0\).

So all coefficients are zero, which proves that \(v_1,\dots,v_k\) are linearly independent.

6 Step Annotations

  1. The operator \(A-\lambda_k I\) is chosen to kill one selected eigenvector exactly.
  2. The other eigenvectors survive, but with nonzero scaling factors because their eigenvalues are different.
  3. Distinctness of eigenvalues is what prevents those scaling factors from vanishing.

7 Why The Assumptions Matter

  • The eigenvalues must be distinct. Repeated eigenvalues do not automatically give dependence, but they do not guarantee independence either.
  • The theorem does not say every matrix with repeated eigenvalues fails to diagonalize. It says distinct eigenvalues are a sufficient source of independence.

8 Common Failure Modes

  • forgetting to use the distinctness assumption
  • applying \(A\) instead of \(A-\lambda_k I\), which does not isolate the last vector
  • concluding “repeated eigenvalues imply dependence,” which is false

9 Reusable Pattern

The proof teaches a durable move:

apply an operator that annihilates one chosen mode and rescales the others.

This same pattern appears in spectral decomposition arguments and minimal-polynomial reasoning.

10 Where This Shows Up Again

  • diagonalization tests
  • spectral methods for repeated matrix powers
  • graph operators where eigenvectors define independent modes

11 Exercises

  1. Adapt the proof to the case \(k=2\) without induction.
  2. Give an example of a matrix with repeated eigenvalue that still has two independent eigenvectors.
  3. Explain why the theorem implies that a matrix with \(n\) distinct eigenvalues is diagonalizable.

12 Sources and Further Reading

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