Linear Algebra 54 | Characteristic Polynomial

TL;DR

An eigenvalue λ exists exactly when the matrix A − λI has a nontrivial kernel.

Briefing Cornell Notes

Briefing

Eigenvalues can be found by turning a matrix problem into a single-variable polynomial: the characteristic polynomial. For a square matrix A, an eigenvalue λ exists exactly when the matrix A − λI becomes singular, meaning its determinant drops to zero. That link—nontrivial kernel ⇔ singular matrix ⇔ determinant equals zero—provides a systematic way to locate all eigenvalues at once, rather than hunting for eigenvectors one direction at a time.

The starting point is the eigenvalue equation Ax = λx for a nonzero vector x. This condition is equivalent to saying x lies in the kernel of A − λI. A kernel is nontrivial precisely when A − λI is singular, and a matrix is singular precisely when its determinant is 0. So λ is an eigenvalue if and only if det(A − λI) = 0. Once that equivalence is in place, the task becomes: compute det(A − λI) and solve for the values of λ that make it vanish.

A 2×2 example makes the method concrete. Take A = [[3, 2], [1, 4]]. Form A − λI by subtracting λ from the diagonal entries, giving [[3 − λ, 2], [1, 4 − λ]]. The determinant becomes a polynomial in λ (a quadratic in this case). Solving det(A − λI) = 0 yields two roots, 5 and 2, which are exactly the eigenvalues of A.

For an n×n matrix, the same construction works. The matrix A − λI subtracts λ from every diagonal entry and leaves all off-diagonal entries unchanged. Using the Leibniz formula for determinants, the result is always a polynomial in λ. The highest power term comes from the product of the diagonal entries, contributing (−λ)^n, so the characteristic polynomial always has degree n with leading coefficient (−1)^n. The remaining coefficients are collected into terms like c_{n−1}, c_{n−2}, down to a constant c_0.

The transcript then formalizes the definition: for a square matrix A, the characteristic polynomial p_A(λ) is the function λ ↦ det(A − λI). Its zeros are precisely the eigenvalues of A. In practice, finding eigenvectors becomes easier once eigenvalues are known, because eigenvectors correspond to the kernels of A − λI for each root λ. The key takeaway is therefore operational: compute the characteristic polynomial, then solve for its zeros to get the full eigenvalue set.

Cornell Notes

Eigenvalues of a square matrix A are exactly the values λ that make A − λI singular. That happens precisely when det(A − λI) = 0, because singularity is equivalent to having a nontrivial kernel. This turns the eigenvalue search into a polynomial root-finding problem: the characteristic polynomial p_A(λ) is defined as p_A(λ) = det(A − λI). For an n×n matrix, p_A(λ) is always a degree-n polynomial with leading coefficient (−1)^n. Once the zeros of p_A are found, they are the eigenvalues, and each eigenvalue can then be paired with eigenvectors from the corresponding kernel of A − λI.

Why does det(A − λI) = 0 capture the existence of an eigenvalue λ?

An eigenvalue λ means there is a nonzero vector x with Ax = λx. Rearranging gives (A − λI)x = 0, so x lies in the kernel of A − λI. A nonzero kernel exists exactly when A − λI is singular. A matrix is singular exactly when its determinant is 0, so det(A − λI) = 0 is equivalent to λ being an eigenvalue.

How is the characteristic polynomial constructed from a matrix?

For a square matrix A, form A − λI by subtracting λ from every diagonal entry and leaving off-diagonal entries unchanged. Then compute the determinant of that λ-dependent matrix. The resulting polynomial in λ is p_A(λ) = det(A − λI). Its roots are the eigenvalues.

What does the 2×2 example A = [[3,2],[1,4]] show about the method?

Compute A − λI = [[3 − λ, 2],[1, 4 − λ]]. Taking the determinant gives a quadratic polynomial in λ. Solving det(A − λI) = 0 produces roots 5 and 2, which are the eigenvalues of A. The example illustrates that eigenvalues come from solving a single polynomial equation.

Why is the characteristic polynomial always degree n for an n×n matrix?

In det(A − λI), the highest power of λ comes from the term in the Leibniz expansion that multiplies the diagonal entries. Each diagonal entry contributes a factor (−λ), so the diagonal product contributes (−λ)^n. All other determinant terms involve fewer diagonal picks and therefore lower powers of λ, so the overall degree is n.

What is the leading coefficient of the characteristic polynomial?

The leading term arises from multiplying the diagonal entries of A − λI, giving (−λ)^n. Therefore the coefficient of λ^n is (−1)^n.

Review Questions

Given a square matrix A, how do you decide whether a number λ is an eigenvalue using only determinants?
For an n×n matrix, what guarantees that det(A − λI) is a degree-n polynomial in λ?
After finding the zeros of the characteristic polynomial, what matrix’s kernel should be used to compute eigenvectors for each eigenvalue?

Key Points

1
An eigenvalue λ exists exactly when the matrix A − λI has a nontrivial kernel.
2
A − λI is singular if and only if det(A − λI) = 0, linking eigenvalues to determinants.
3
The characteristic polynomial is defined by p_A(λ) = det(A − λI).
4
The zeros of p_A(λ) are precisely the eigenvalues of A.
5
For an n×n matrix, p_A(λ) is always a degree-n polynomial with leading coefficient (−1)^n.
6
Computing det(A − λI) reduces eigenvalue finding to solving a polynomial equation in λ.
7
Once eigenvalues are known, eigenvectors come from solving (A − λI)x = 0 for each root λ.

Highlights

Eigenvalues are the λ values that make det(A − λI) = 0.

The characteristic polynomial p_A(λ) = det(A − λI) packages all eigenvalue information into one polynomial.

For n×n matrices, the characteristic polynomial always has degree n and leading coefficient (−1)^n.

In the 2×2 example A = [[3,2],[1,4]], the characteristic polynomial’s roots are 2 and 5, matching the eigenvalues. 