Educational tool; results are estimates for learning, verify with your textbook or instructor.
This tool finds the eigenvalues of one $3\times 3$ real matrix and writes the characteristic polynomial $\det(A-\lambda I)$ with the steps. It also gives an eigenvector for each real eigenvalue. Enter integers, decimals, or fractions such as $3/4$. A terminating decimal is read as an exact fraction of a power of ten, so $0.5$ is $1/2$.
The polynomial is built from three numbers. The trace is the sum of the diagonal entries. The next coefficient is the sum of the three principal $2\times 2$ minors: delete one row and the matching column, then take that determinant, and add the three results. The constant term is $\det A$. For a $3\times 3$ matrix,
$$\det(A-\lambda I)=-\lambda^{3}+(\operatorname{tr} A)\lambda^{2}-\sigma\lambda+\det A,$$
where $\sigma$ is that sum of principal minors. The page prints the trace, $\sigma$, and the determinant before it writes the polynomial. The eigenvalues are the roots. A rational root is reported exactly. Any other real root is numeric. Nonreal roots come in a conjugate pair, and neither one has a real eigenvector.
For each real eigenvalue $\lambda$, the page builds $A-\lambda I$ and row reduces it. A nonzero vector in the null space is an eigenvector. The page then multiplies $A$ by that vector and checks that the result is $\lambda$ times the vector. If the same eigenvalue is repeated and there are not enough independent eigenvectors, the steps say so. That is the difference between a diagonalizable matrix and one that is not.
This page does not invert $A$, and it does not solve a linear system you typed yourself. The determinant and the inverse, with every minor written out, are on the 3 by 3 matrix calculator. A system of your own is on the reduced row echelon form calculator, which uses the same elimination. A $2\times 2$ characteristic polynomial, without eigenvectors, is on the 2 by 2 eigenvalue calculator.
How to read the characteristic polynomial
The three coefficients are the audit line. Trace, sum of principal minors, determinant. If you only stare at the list of $\lambda$ values, you miss the arithmetic that produced them. A triangular matrix is the easy check: the eigenvalues are the diagonal entries, the trace is their sum, and the determinant is their product.
A repeated real root is one eigenvalue with algebraic multiplicity greater than one. Row reduction of $A-\lambda I$ then tells you the geometric multiplicity, which is the dimension of the eigenspace. If those two numbers disagree, there is no basis of eigenvectors and the matrix is not diagonalizable. A negative discriminant in a leftover quadratic means a conjugate pair. Over the reals, picture a rotation and a scale in a plane, plus the real eigenvalue on the remaining line.
Check. Counted with multiplicity, the eigenvalues sum to the trace and multiply to the determinant. On the example below, $1+2+11=14$ and $1\cdot 2\cdot 11=22$. If either check fails, do not trust the list.
A worked example
Let $A$ have rows $(2,0,0)$, $(0,3,4)$, and $(0,4,9)$. That is the matrix already filled in above. The trace is $2+3+9=14$. The principal $2\times 2$ minors are $11$, $18$, and $6$, so they sum to $35$. The determinant is $22$. Therefore
$$\det(A-\lambda I)=-\lambda^{3}+14\lambda^{2}-35\lambda+22,$$
and the monic equation $\lambda^{3}-14\lambda^{2}+35\lambda-22=0$ has roots $1$, $2$, and $11$. An eigenvector for $1$ is $(0,2,-1)$, for $2$ is $(1,0,0)$, and for $11$ is $(0,1,2)$. The sum of the roots is the trace, and the product is the determinant. The steps below are written in the page, so you can read them with scripts turned off.
Characteristic polynomial det(A − λI).
Trace = a₁₁ + a₂₂ + a₃₃ = 2 + 3 + 9 = 14.
Principal 2 by 2 minor M₁₁ deletes row 1 and column 1, leaving 3, 4; 4, 9.
M₁₁ = (3)(9) − (4)(4) = 27 − 16 = 11.
Principal 2 by 2 minor M₂₂ deletes row 2 and column 2, leaving 2, 0; 0, 9.
M₂₂ = (2)(9) − (0)(0) = 18 − 0 = 18.
Principal 2 by 2 minor M₃₃ deletes row 3 and column 3, leaving 2, 0; 0, 3.
M₃₃ = (2)(3) − (0)(0) = 6 − 0 = 6.
Sum of the principal 2 by 2 minors = 11 + 18 + 6 = 35.
Determinant by cofactor expansion along row 1.
The minor for column 1 deletes row 1 and column 1, leaving 3, 4; 4, 9.
That minor is (3)(9) − (4)(4) = 11.
a₁₁ times that minor is (2)(11) = 22. The cofactor sign is plus, so this term is 22.
The minor for column 2 deletes row 1 and column 2, leaving 0, 4; 0, 9.
That minor is (0)(9) − (4)(0) = 0.
a₁₂ times that minor is (0)(0) = 0. The cofactor sign is minus, so this term is 0.
The minor for column 3 deletes row 1 and column 3, leaving 0, 3; 0, 4.
That minor is (0)(4) − (3)(0) = 0.
a₁₃ times that minor is (0)(0) = 0. The cofactor sign is plus, so this term is 0.
det(A) = 22 + 0 + 0 = 22.
det(A − λI) = −λ³ + (trace)λ² − (sum of the principal 2 by 2 minors)λ + det(A).
det(A − λI) = −λ³ + 14λ² − 35λ + 22.
The eigenvalues are the roots. They also solve the monic equation λ³ − 14λ² + 35λ − 22 = 0, which is λ³ − (trace)λ² + (sum of the principal 2 by 2 minors)λ − det(A) = 0.
λ = 1 is a rational root. Substitution into the monic polynomial gives 0.
Dividing by λ − 1 leaves λ² − 13λ + 22.
Discriminant of λ² − 13λ + 22 = (−13)² − 4(22) = 81.
The discriminant is a perfect square, so the other two eigenvalues are rational.
Eigenvalues: 1, 2, and 11.
Eigenvector for λ = 1.
A − λI with λ = 1
Row 1: 1, 0, 0
Row 2: 0, 2, 4
Row 3: 0, 4, 8
1. Multiply row 2 by 1/2 so the pivot in column 2 is 1.
Row 1: 1, 0, 0
Row 2: 0, 1, 2
Row 3: 0, 4, 8
2. Add −4 times row 2 to row 3 to clear the entry in column 2.
Row 1: 1, 0, 0
Row 2: 0, 1, 2
Row 3: 0, 0, 0
Reduced row echelon form
Row 1: 1, 0, 0
Row 2: 0, 1, 2
Row 3: 0, 0, 0
Free variable: x₃. The other variables are determined by it.
Set x₃ = 1.
One null vector is (0, −2, 1).
Multiply by −1 so the first nonzero entry is positive.
An eigenvector is (0, 2, −1).
Check. A times (0, 2, −1) = (0, 2, −1), and 1 times the vector is (0, 2, −1). They agree.
Eigenvector for λ = 2.
A − λI with λ = 2
Row 1: 0, 0, 0
Row 2: 0, 1, 4
Row 3: 0, 4, 7
1. Swap row 1 and row 2, because row 1 has a zero in column 2 and row 2 does not.
Row 1: 0, 1, 4
Row 2: 0, 0, 0
Row 3: 0, 4, 7
2. Add −4 times row 1 to row 3 to clear the entry in column 2.
Row 1: 0, 1, 4
Row 2: 0, 0, 0
Row 3: 0, 0, −9
3. Swap row 2 and row 3, because row 2 has a zero in column 3 and row 3 does not.
Row 1: 0, 1, 4
Row 2: 0, 0, −9
Row 3: 0, 0, 0
4. Multiply row 2 by −1/9 so the pivot in column 3 is 1.
Row 1: 0, 1, 4
Row 2: 0, 0, 1
Row 3: 0, 0, 0
5. Add −4 times row 2 to row 1 to clear the entry in column 3.
Row 1: 0, 1, 0
Row 2: 0, 0, 1
Row 3: 0, 0, 0
Reduced row echelon form
Row 1: 0, 1, 0
Row 2: 0, 0, 1
Row 3: 0, 0, 0
Free variable: x₁. The other variables are determined by it.
Set x₁ = 1.
One null vector is (1, 0, 0).
An eigenvector is (1, 0, 0).
Check. A times (1, 0, 0) = (2, 0, 0), and 2 times the vector is (2, 0, 0). They agree.
Eigenvector for λ = 11.
A − λI with λ = 11
Row 1: −9, 0, 0
Row 2: 0, −8, 4
Row 3: 0, 4, −2
1. Multiply row 1 by −1/9 so the pivot in column 1 is 1.
Row 1: 1, 0, 0
Row 2: 0, −8, 4
Row 3: 0, 4, −2
2. Multiply row 2 by −1/8 so the pivot in column 2 is 1.
Row 1: 1, 0, 0
Row 2: 0, 1, −1/2
Row 3: 0, 4, −2
3. Add −4 times row 2 to row 3 to clear the entry in column 2.
Row 1: 1, 0, 0
Row 2: 0, 1, −1/2
Row 3: 0, 0, 0
Reduced row echelon form
Row 1: 1, 0, 0
Row 2: 0, 1, −1/2
Row 3: 0, 0, 0
Free variable: x₃. The other variables are determined by it.
Set x₃ = 1.
One null vector is (0, 1/2, 1).
Multiply by 2 to clear the denominator.
An eigenvector is (0, 1, 2).
Check. A times (0, 1, 2) = (0, 11, 22), and 11 times the vector is (0, 11, 22). They agree.
Check. Sum of the eigenvalues = 14, which equals the trace 14.
Product of the eigenvalues = 22, which equals the determinant 22.
There are 3 independent eigenvectors, so A is diagonalizable.
Common pitfalls
The diagonal entries are the eigenvalues only when the matrix is triangular. On a full matrix, do not read $a_{11}$ and $a_{33}$ as the spectrum.
A repeated root that is rational is exact on this page. A numeric root is an approximation. If a homework answer is supposed to be a simple fraction, type the fraction. The decimal $0.333$ is $333/1000$, which is not $1/3$.
An eigenvector is never the zero vector. If row reduction leaves a free variable, set that variable to $1$ and read the other entries. Then multiply $A$ by the vector yourself. The check on this page is the same multiplication.
Related reading
The geometry of an invariant line is in eigenvalues in plain language, and the linear algebra topic puts that picture next to bases and determinants. A study order is the linear algebra first course. Glossary lines: eigenvalue, eigenvector, and eigenspace.
Build the determinant first on the 3 by 3 matrix calculator if the cofactor expansion is the part you want to practice. Use the reduced row echelon form calculator when you want every elimination step of a system that is not $A-\lambda I$. The quadratic case, without an eigenvector, stays on the 2 by 2 eigenvalue calculator. An eigenbasis is not the output of the Gram Schmidt calculator unless you were building an orthonormal eigenbasis by hand.
