What Eigenvalues Are
An eigenvalue of a square matrix $A$ is a scalar $\lambda$ for which there is a non-zero vector $\mathbf{v}$ satisfying
$$A\mathbf{v} = \lambda\mathbf{v}$$
The vector $\mathbf{v}$ is the corresponding eigenvector. In words: most vectors get knocked off their line when a matrix multiplies them, but an eigenvector keeps its direction — the matrix only stretches it (or shrinks or flips it) by the factor $\lambda$. That factor is the eigenvalue. Eigenvalues are also called characteristic roots, characteristic values, or latent roots.
The eigenvalue is the how much and the eigenvector is the which direction. They always come in pairs, and this article focuses on the eigenvalue — the scaling number itself. For the partner concept, the direction that stays fixed, see the companion article on eigenvectors; this page is the degree-first, "find the number" view of the same equation.
How Do You Find Eigenvalues?
Eigenvalues come from the characteristic equation. Rewrite $A\mathbf{v} = \lambda\mathbf{v}$ as
$$A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$$
$$(A - \lambda I)\mathbf{v} = \mathbf{0}$$
where $I$ is the identity matrix of the same order as $A$. For a non-zero $\mathbf{v}$ to exist, the matrix $A - \lambda I$ must be singular, which means its determinant is zero:
$$\det(A - \lambda I) = 0$$
This is the characteristic equation. Expanding the determinant gives a polynomial in $\lambda$ — the characteristic polynomial — whose roots are the eigenvalues. The steps:
Form $A - \lambda I$ by subtracting $\lambda$ from each diagonal entry of $A$.
Compute $\det(A - \lambda I)$.
Set that determinant to 0 and solve for $\lambda$.
An $n \times n$ matrix gives a degree-$n$ polynomial, so it has at most $n$ eigenvalues.
Variable glossary. $A$ is the square matrix; $\lambda$ (lambda) is an eigenvalue; $\mathbf{v}$ is the eigenvector; $I$ is the identity matrix; $\det(\cdot)$ is the determinant; the characteristic polynomial is $\det(A - \lambda I)$ written as a polynomial in $\lambda$.
What Are the Properties of Eigenvalues?
A few properties let you check your work and shortcut the algebra.
Sum equals the trace. The eigenvalues add up to the trace of $A$ (the sum of its diagonal entries). If the diagonal sums to 7, so do the eigenvalues.
Product equals the determinant. Multiply the eigenvalues together and you get $\det A$. A zero among the eigenvalues means $\det A = 0$, so the matrix is not invertible.
Triangular matrices give their diagonal free. For an upper- or lower-triangular matrix, the eigenvalues are simply the diagonal entries — no characteristic equation needed.
Symmetric matrices have real eigenvalues. A symmetric matrix always has real eigenvalues, which is why they appear so often in physics and statistics.
Transpose shares eigenvalues. $A$ and $A^{\mathsf{T}}$ have the same eigenvalues.
The trace and determinant checks are the fastest sanity tests you have: after solving, add your eigenvalues and multiply them, then compare with the trace and determinant of the original matrix.
Examples of Eigenvalues
Example 1
Find the eigenvalues of $A = \begin{bmatrix} 2 & 0 \ 0 & 3 \end{bmatrix}$.
The matrix is diagonal (a special triangular case), so the eigenvalues are the diagonal entries. Final answer: $\lambda = 2$ and $\lambda = 3$. (Check: trace $= 5 = 2 + 3$ ✓; determinant $= 6 = 2 \times 3$ ✓.)
Example 2
Find the eigenvalues of $A = \begin{bmatrix} 5 & 4 \ 1 & 2 \end{bmatrix}$.
Form $A - \lambda I$:
$$A - \lambda I = \begin{bmatrix} 5 - \lambda & 4 \ 1 & 2 - \lambda \end{bmatrix}$$
Set the determinant to zero:
$$(5 - \lambda)(2 - \lambda) - (4)(1) = 0$$
$$\lambda^2 - 7\lambda + 10 - 4 = 0$$
$$\lambda^2 - 7\lambda + 6 = 0$$
$$(\lambda - 6)(\lambda - 1) = 0$$
Final answer: $\lambda = 6$ and $\lambda = 1$. (Check: trace $= 7 = 6 + 1$ ✓; determinant $= 10 - 4 = 6 = 6 \times 1$ ✓.)
Example 3 (the mistake worth making once)
Find the eigenvalues of $A = \begin{bmatrix} 4 & 1 \ 2 & 3 \end{bmatrix}$.
Wrong attempt. A student forms $A - \lambda I$ but forgets to subtract $\lambda$ from both diagonal entries, writing the determinant as $(4 - \lambda)(3) - (1)(2)$ — leaving the bottom-right entry as plain 3.
The break. That gives $12 - 3\lambda - 2 = 0$, a linear equation with a single root $\lambda = \tfrac{10}{3}$. But a $2 \times 2$ matrix must produce a quadratic with up to two eigenvalues. A linear characteristic equation is the tell that a $\lambda$ was dropped.
Correct. Subtract $\lambda$ from both diagonal entries:
$$\det\begin{bmatrix} 4 - \lambda & 1 \ 2 & 3 - \lambda \end{bmatrix} = (4 - \lambda)(3 - \lambda) - (1)(2) = 0$$
$$\lambda^2 - 7\lambda + 12 - 2 = 0$$
$$\lambda^2 - 7\lambda + 10 = 0$$
$$(\lambda - 5)(\lambda - 2) = 0$$
Final answer: $\lambda = 5$ and $\lambda = 2$. (Check: trace $= 7$ ✓; determinant $= 12 - 2 = 10 = 5 \times 2$ ✓.)
Example 4
Find the eigenvalues of the triangular matrix $A = \begin{bmatrix} 3 & 7 & -1 \ 0 & 2 & 4 \ 0 & 0 & -3 \end{bmatrix}$.
The matrix is upper-triangular, so the eigenvalues are the diagonal entries directly, no expansion needed. Final answer: $\lambda = 3$, $\lambda = 2$, $\lambda = -3$.
Example 5
Find the eigenvalues of $A = \begin{bmatrix} 2 & -1 \ 1 & 2 \end{bmatrix}$.
$$\det\begin{bmatrix} 2 - \lambda & -1 \ 1 & 2 - \lambda \end{bmatrix} = (2 - \lambda)^2 + 1 = 0$$
$$(2 - \lambda)^2 = -1$$
$$2 - \lambda = \pm i$$
$$\lambda = 2 \mp i$$
Final answer: $\lambda = 2 + i$ and $\lambda = 2 - i$ — a complex conjugate pair. (This is exactly what a rotation-and-scaling matrix produces; a real matrix can have complex eigenvalues.)
Example 6
Find the eigenvalues of $A = \begin{bmatrix} 3 & 1 & 1 \ 2 & 4 & 2 \ 1 & 1 & 3 \end{bmatrix}$, given that one eigenvalue is $\lambda = 2$.
The characteristic polynomial of this $3 \times 3$ matrix factors as $(\lambda - 2)^2(\lambda - 6) = 0$. Using the property checks: trace $= 3 + 4 + 3 = 10$, and $2 + 2 + 6 = 10$ ✓; determinant $= 2 \times 2 \times 6 = 24$. Final answer: $\lambda = 2$ (repeated) and $\lambda = 6$.
Why Eigenvalues Matter — "the few numbers that define a matrix"
Eigenvalues were not invented as abstract algebra. They surfaced in the 1700s and 1800s in the study of rotating bodies and vibrating systems — the principal axes of a spinning top, the natural frequencies of an oscillating string. The question was always the same: as a system transforms, what stays aligned, and by how much does it grow? That is the eigenvalue question.
PageRank. Google's original ranking was the dominant eigenvalue problem for the web's link matrix — the eigenvector of that matrix is the ranking.
Vibration and resonance. The eigenvalues of a structure's stiffness matrix are its natural frequencies. The Tacoma Narrows Bridge collapse in 1940 is the textbook warning about what happens when a driving frequency meets a structure's natural one.
Quantum mechanics. The allowed energy levels of a system are the eigenvalues of its energy operator — discrete, fixed numbers, exactly like the eigenvalues of a matrix.
Data science. Principal component analysis ranks the directions of greatest variance in data by their eigenvalues, the basis of dimensionality reduction.
Stability analysis. Whether a system settles or blows up is read from the signs of its eigenvalues.
A connection worth carrying: a rotation matrix has complex eigenvalues (as in Example 5), because a pure rotation leaves no real direction unchanged — there is nothing for a real eigenvector to point along. That single fact ties the geometry of rotation to the algebra of eigenvalues.
The Mathematicians Behind Eigenvalues
Augustin-Louis Cauchy (1789–1857, France) gave the eigenvalue problem its modern footing, proving that symmetric matrices have real eigenvalues and connecting the idea to the principal axes of quadratic forms.
David Hilbert (1862–1943, Germany) extended eigenvalues from finite matrices to infinite-dimensional spaces and coined the term Eigenwert ("proper value"), the German root of the word we still use.
Common Confusions Cleared Up
Mistake 1: Subtracting λ from only one diagonal entry
Where it slips in: Forming $A - \lambda I$ for the characteristic equation.
Don't do this: Subtracting $\lambda$ from just the top-left entry. That changes the polynomial's degree and loses an eigenvalue.
The correct way: Subtract $\lambda$ from every diagonal entry — that is what $\lambda I$ does. A quick check: an $n \times n$ matrix should give a degree-$n$ characteristic polynomial, so a $2 \times 2$ that produces a linear equation means a $\lambda$ was missed. The first instinct is to treat $\lambda I$ as a single number subtracted once, when it is subtracted along the whole diagonal.
Mistake 2: Multiplying out a polynomial that is already factored
Where it slips in: After reaching something like $(\lambda - 3)(\lambda - 5) = 0$.
Don't do this: Expanding it back to $\lambda^2 - 8\lambda + 15 = 0$ and then re-factoring. That is wasted work and a fresh chance for an arithmetic slip.
The correct way: If the characteristic equation arrives partly factored, read the roots straight off. The habit that fixes this is pausing to ask "is this already factored?" before expanding anything.
Mistake 3: Forgetting that eigenvalues can be complex
Where it slips in: Matrices that rotate, like Example 5.
Don't do this: Assuming a real matrix must have real eigenvalues and treating $(2 - \lambda)^2 = -1$ as having no answer.
The correct way: A real matrix can have complex eigenvalues, appearing in conjugate pairs. Solve over the complex numbers rather than declaring "no solution."
The Short Version
An eigenvalue is the scalar $\lambda$ by which a matrix scales its eigenvector, defined by $A\mathbf{v} = \lambda\mathbf{v}$.
Find eigenvalues by solving the characteristic equation $\det(A - \lambda I) = 0$.
The eigenvalues sum to the trace and multiply to the determinant — two quick checks.
The most common mistake is subtracting $\lambda$ from only one diagonal entry, which drops an eigenvalue.
Real matrices can have complex eigenvalues, and a zero eigenvalue signals a singular matrix.
Practice Questions on Eigenvalues
Work through these, then check your answers below.
Find the eigenvalues of $\begin{bmatrix} 4 & 0 \ 0 & 7 \end{bmatrix}$.
Solve $\det(A - \lambda I) = 0$ for $\begin{bmatrix} 3 & 2 \ 1 & 4 \end{bmatrix}$, then confirm with the trace-and-determinant check.
Find the eigenvalues of the $3 \times 3$ matrix $\begin{bmatrix} 2 & 0 & 0 \ 1 & 3 & 0 \ 4 & 5 & 1 \end{bmatrix}$ from scratch.
The eigenvalues of a $2 \times 2$ matrix are $3$ and $-2$. What are its trace and determinant?
Answer to Question 1: The matrix is diagonal, so the eigenvalues are the diagonal entries: $\lambda = 4$ and $\lambda = 7$.
Answer to Question 2: $\det\begin{bmatrix} 3 - \lambda & 2 \ 1 & 4 - \lambda \end{bmatrix} = (3 - \lambda)(4 - \lambda) - (2)(1) = 0$. $$\lambda^2 - 7\lambda + 12 - 2 = 0$$ $$\lambda^2 - 7\lambda + 10 = 0$$ $$(\lambda - 5)(\lambda - 2) = 0$$ So $\lambda = 5$ and $\lambda = 2$. Check: trace $= 3 + 4 = 7 = 5 + 2$ ✓; determinant $= 12 - 2 = 10 = 5 \times 2$ ✓.
Answer to Question 3: The matrix is lower-triangular, so the eigenvalues are the diagonal entries directly: $\lambda = 2$, $\lambda = 3$, $\lambda = 1$. (Verifying from scratch, the characteristic polynomial expands along the first row to $(2 - \lambda)(3 - \lambda)(1 - \lambda) = 0$, giving the same three roots.)
Answer to Question 4: The eigenvalues sum to the trace and multiply to the determinant. So trace $= 3 + (-2) = 1$ and determinant $= (3)(-2) = -6$.
A Practical Next Step
If a sign trips you up in the characteristic equation, come back to Mistake 1 and verify you subtracted $\lambda$ down the whole diagonal. At Bhanzu, our trainers teach the trace-and-determinant check alongside the method, so students catch a wrong eigenvalue before it spreads into the eigenvector work.
Want a live Bhanzu trainer to walk through more eigenvalues problems? Book a free demo class.
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