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Proof that eigenvectors for distinct eigenvalues are linearly independent

Distinct eigenvalues do more than label eigenspaces. They force eigenvectors to be independent, which is exactly why diagonalization becomes possible so often in practice.

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The proof is short but easy to blur when written as one paragraph. In the Proof Builder, the cancellation step and the inductive step stay visually separate, which is the whole argument.

What the theorem says

If are eigenvectors of a linear map and their eigenvalues are all distinct, then those eigenvectors are automatically linearly independent.

That fact is one of the main reasons distinct eigenvalues are so useful: once the eigenvalues separate, the eigenvectors stop interfering with one another.

Why distinct eigenvalues matter

Suppose there were a linear relation . Applying does not scramble the vectors; it only multiplies each one by its own eigenvalue.

That is the opening the proof needs. Once you subtract times the original relation, the term disappears and the relation shrinks to one involving only .

The proof strategy

Everything turns on the same move: create two equations from the same relation, then subtract them in a way that kills one eigenvector.

  1. 1Assume there is a linear relation among the eigenvectors and apply to it.
  2. 2Use to rewrite the new equation with eigenvalue coefficients attached.
  3. 3Subtract times the original relation so that the term cancels.
  4. 4Notice that the remaining coefficients involve , which are all nonzero because the eigenvalues are distinct.
  5. 5Apply the inductive hypothesis to the shortened relation to force to vanish.
  6. 6Return to the original relation to conclude as well.

Common Pitfall

Distinct eigenvalues are a sufficient condition for independence, not a necessary one. Eigenvectors can still be independent even when some eigenvalues repeat, but then you need extra work to prove it.

Try a Variation

Look at a matrix with a repeated eigenvalue, such as . Does the repeated eigenvalue still give two independent eigenvectors, or does the argument break?

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