MAT 242 Chap.9 Eigenvalues and Eigenvectors
Eigenvalues and Eigenvectors
Where the problem comes from
The course introduces eigenvalues through two masses hanging on springs. Looking for steady vibrations turns the differential equations into a homogeneous matrix system, and it has a nonzero solution only for special frequencies, the ones that make a certain matrix singular.
That is the eigenvalue problem: find the numbers λ for which A − λI is singular, and the nonzero vectors it sends to zero.
The computation
An eigenvector of A is a nonzero v with Av = λv, and λ is its eigenvalue; λ itself may be zero. The eigenvalues are the roots of the characteristic polynomial det(A − λI), which has degree n for an n × n matrix.
For each eigenvalue, solving (A − λI)v = 0 and splitting by parameter gives the eigenspace basis. If that reduction ever produces the identity matrix, something upstream is wrong, since a true eigenvalue must leave at least one free variable.
Shortcuts worth knowing
Triangular matrices show their eigenvalues on the diagonal.
The product of the eigenvalues, with multiplicity, equals the determinant, and their sum equals the trace, which gives two instant checks and shows that a matrix is invertible exactly when zero is not an eigenvalue. Real matrices can have complex eigenvalues, but the course does not treat them.
The paid chapter works a two-by-two completely, then a three-by-three with a repeated eigenvalue whose eigenspace is too small, setting up diagonalization.
Subtract λ only along the main diagonal; that single habit prevents the most common setup error.
How sample solutions grade this
Credit on a sample Test 2 eigenvalue question is split between forming A − λI, computing its determinant and solving for the eigenvalues; row reducing A itself to find eigenvalues earns only partial credit, because row operations change the eigenvalues.
For eigenvectors, each eigenvalue needs its own reduction and its own null-space basis. The course also accepts any nonzero multiple of an eigenvector, so scaling to clear fractions is allowed.
Repeated eigenvalues
A root of the characteristic polynomial can appear more than once; the number of times is its multiplicity.
A repeated eigenvalue may have an eigenspace of full size or a smaller one, and the next chapter shows that this difference decides diagonalization. For a triangular or diagonal matrix, repeated diagonal entries are repeated eigenvalues, which makes such matrices convenient practice for eigenspace dimensions. Their eigenvectors still come from null spaces, exactly as for any other matrix.
What this chapter covers
- 01
The coupled-springs motivation
- 02
Eigenvalues from det(A − λI) = 0
- 03
Eigenspaces as null spaces
- 04
Triangular matrices and repeated eigenvalues
- 05
Trace and determinant checks
Worked example · free
Eigenvalues and eigenvectors of a symmetric 2 × 2
- 3det(A − λI) = (5 − λ)(2 − λ) − 4 = λ² − 7λ + 6 = (λ − 1)(λ − 6).
- 2For λ = 6, A − 6I = [−1 2; 2 −4] reduces to [1 −2; 0 0], giving (2, 1).
- 2For λ = 1, A − I = [4 2; 2 1] reduces to [1 ½; 0 0], giving (−1, 2).
Key terms
- Eigenvalue
- An eigenvalue is a number λ for which A − λI is singular, so Av = λv has a nonzero solution.
- Characteristic Polynomial
- The characteristic polynomial is det(A − λI), whose roots are the eigenvalues of A.
- Eigenspace
- The eigenspace of λ is the null space of A − λI: every eigenvector for λ together with the zero vector.
Eigenvalues and Eigenvectors FAQ
Can zero be an eigenvalue?
Yes. Zero is an eigenvalue exactly when A itself is singular, because then A sends some nonzero vector to zero. Eigenvectors, on the other hand, are never allowed to be the zero vector.
How can I check eigenvalues without redoing the work?
Multiply them together and compare with the determinant, then add them and compare with the trace, the sum of the diagonal entries. Both must match when multiplicities are counted.
Why can I not find eigenvalues by row reducing A?
Row operations preserve solutions of linear systems but not eigenvalues, so the reduced matrix generally has different eigenvalues. The characteristic polynomial must be computed from A − λI itself.
Exam move
Drill characteristic polynomials of two-by-two matrices until the factoring is quick, then spend your time on the eigenspace reductions, where most of the credit on a test sits. For every eigenvector you find, multiply A by it once. Seeing the vector come back scaled is the fastest possible confirmation.
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