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Mathematics · lesson 2 of 3

Eigenvectors and diagonalization

Av=λv requires a nonzero vector v.

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Learn the ideas

Prerequisites: Matrix multiplication and characteristic polynomials.

Eigenpair

Av=λv requires a nonzero vector v.

Diagonalization

A=PDP⁻¹ exists when there is a basis of eigenvectors over the chosen field.

Repeated eigenvalues

Algebraic multiplicity does not guarantee enough independent eigenvectors.

Symmetric matrices

A real symmetric matrix has an orthonormal eigenbasis over the reals.

Worked examples

Example 1

For A=[[3,1],[1,3]], find the eigenvalue of (1,1).

A(1,1)=(4,4)=4(1,1).

4

Example 2

For A=[[3,1],[1,3]], find the eigenvalue corresponding to (1,−1).

A(1,−1)=(2,−2)=2(1,−1).

2

Check your understanding

8 questions

Decimals or fractions are accepted for numeric answers. Proofs and written responses use sample answers and require human marking. This is practice, not a protected examination.

1 For A=[[3,1],[1,3]], find the eigenvalue of (1,1).

Show answer and explanation

4

A(1,1)=(4,4)=4(1,1).

2 For A=[[3,1],[1,3]], find the eigenvalue corresponding to (1,−1).

Show answer and explanation

2

A(1,−1)=(2,−2)=2(1,−1).

3 Why is the zero vector excluded from eigenvectors?

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It satisfies Av=λv for every λ and gives no eigendirection

The definition needs a nonzero vector to encode a meaningful eigenvalue relation.

4 Is [[1,1],[0,1]] diagonalizable over the reals?

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No; its eigenspace has dimension 1

The only eigenvalue is 1, and eigenvectors have second component zero.

5 A diagonal matrix has diagonal entries 2,3,5. Find the trace of its square.

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38

Squaring gives diagonal entries 4,9,25; their sum is 38.

6 Show that eigenvectors for two distinct eigenvalues are linearly independent.

Sample answer and marking guidance

If v₂=cv₁ with c≠0, applying A gives λ₂cv₁=cλ₁v₁, forcing λ₂=λ₁, a contradiction.

Credit use of nonzero vectors and the distinct-eigenvalue contradiction.

7 A real symmetric matrix always admits what decomposition?

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Orthogonal diagonalization

The real spectral theorem provides an orthonormal basis of eigenvectors.

8 Explain why diagonalization helps calculate A^n.

Sample answer and marking guidance

From A=PDP⁻¹, adjacent P⁻¹P factors cancel so A^n=PD^nP⁻¹. Powers of a diagonal matrix are computed entrywise.

Credit the cancellation argument and the correct order of factors.

Further reading and review references

These are reference links, not copied exercises, institutional endorsement or proof of independent review.