The Characteristic Polynomial

From $Av=\lambda v$A v = λ v to one equation

Trying columns one at a time is a slow way to find the eigenvectors of A = (4 −1; 2 1). The eigenvalues can be found directly, starting from the equation that defines them, $Av=\lambda v$A v = λ v.

Bring everything to one side: $Av-\lambda v=0$A v − λ v = 0. To factor out v, both terms must be a matrix times v, and $\lambda$λ is a number, so $A-\lambda$A − λ would mean taking a number from a matrix, which has no meaning. Write $\lambda v$λ v as $\lambda Iv$λI v instead, since the identity matrix I leaves v unchanged. Then $Av-\lambda Iv=0$A v − λI v = 0, and factoring out v gives $\left(A-\lambda I\right)v=0$(A − λI) v = 0.

Why the determinant must be 0

The matrix $A-\lambda I$A − λI sends the eigenvector v to the zero column, and v is not the zero column. If $A-\lambda I$A − λI had an inverse, multiplying both sides of $\left(A-\lambda I\right)v=0$(A − λI) v = 0 by it would give v = 0, which is false. So $A-\lambda I$A − λI has no inverse, and a 2 × 2 matrix has no inverse exactly when its determinant is 0.

So every eigenvalue $\lambda$λ satisfies $\det \left(A-\lambda I\right)=0$det(A − λI) = 0. It works the other way too: when the determinant is 0, the two rows of $A-\lambda I$A − λI are multiples of each other, they give one condition on the column, and that condition has non-zero solutions. The eigenvalues are exactly the numbers that make this determinant 0.

Subtracting $\lambda I$λI

$\lambda I$λI has $\lambda$λ in the two places of the main diagonal and 0 in the other two. Subtracting it from A takes $\lambda$λ off each entry of the main diagonal and leaves the other entries alone. So $A-\lambda I$A − λI has $4-\lambda$4 − λ and $1-\lambda$1 − λ on its main diagonal, and −1 and 2 where they were.

$\lambda I$λI has $\lambda$λ on the main diagonal and 0 elsewhere, so subtracting it changes only the two diagonal entries of A.

The characteristic polynomial

Take the determinant the usual way, the product down the main diagonal minus the product of the other two entries: $\det \left(A-\lambda I\right)=\left(4-\lambda \right)\left(1-\lambda \right)-\left(-1\right)\left(2\right)$det(A − λI) = (4 − λ)(1 − λ) − (−1)(2).

Expand the brackets: $\left(4-\lambda \right)\left(1-\lambda \right)=4-4\lambda -\lambda +{\lambda }^2={\lambda }^2-5\lambda +4$(4 − λ)(1 − λ) = 4 − 4λ − λ + λ² = λ² − 5λ + 4. The second product is (−1)(2) = −2, and subtracting it adds 2. So $\det \left(A-\lambda I\right)={\lambda }^2-5\lambda +4+2={\lambda }^2-5\lambda +6$det(A − λI) = λ² − 5λ + 4 + 2 = λ² − 5λ + 6.

This quadratic in $\lambda$λ is called the characteristic polynomial of A, and setting it equal to 0 gives the characteristic equation.

Its roots are the eigenvalues

Set it to zero and factor: ${\lambda }^2-5\lambda +6=\left(\lambda -2\right)\left(\lambda -3\right)=0$λ² − 5λ + 6 = (λ − 2)(λ − 3) = 0, so $\lambda =2$λ = 2 or $\lambda =3$λ = 3. These are the two eigenvalues of A, the stretches along (1, 2) and (1, 1).

Check each one by putting it back into $A-\lambda I$A − λI. At $\lambda =3$λ = 3 the matrix is (1 −1; 2 −2), with determinant 1 × (−2) − (−1) × 2 = 0. At $\lambda =2$λ = 2 it is (2 −1; 2 −1), with determinant 2 × (−1) − (−1) × 2 = 0. A number that is not a root fails: at $\lambda =1$λ = 1 the matrix is (3 −1; 2 0), with determinant 3 × 0 − (−1) × 2 = 2, so only the zero column is sent to zero, and 1 is not an eigenvalue.

The graph of ${\lambda }^2-5\lambda +6$λ² − 5λ + 6 against $\lambda$λ. It crosses the axis at $\lambda =2$λ = 2 and $\lambda =3$λ = 3, the two values that make $\det \left(A-\lambda I\right)$det(A − λI) zero.

A check: the diagonal total and the determinant

For any 2 × 2 matrix (a b; c d), $\det \left(A-\lambda I\right)=\left(a-\lambda \right)\left(d-\lambda \right)-bc={\lambda }^2-\left(a+d\right)\lambda +\left(ad-bc\right)$det(A − λI) = (a − λ)(d − λ) − b c = λ² − (a + d)λ + (a d − b c). The coefficient of $\lambda$λ is minus the total of the main diagonal, and the constant term is det A.

A quadratic ${\lambda }^2-s\lambda +p$λ² − sλ + p has roots that add to s and multiply to p. So the two eigenvalues always add to the diagonal total a + d and multiply to det A. Here 2 + 3 = 5 = 4 + 1, and 2 × 3 = 6 = 4 × 1 − (−1) × 2. This gives a quick check on any pair of eigenvalues, and a quick way to write the polynomial down.

Other ways the equation can come out

For (3 1; 0 2), $\det \left(A-\lambda I\right)=\left(3-\lambda \right)\left(2-\lambda \right)-1\times 0=\left(3-\lambda \right)\left(2-\lambda \right)$det(A − λI) = (3 − λ)(2 − λ) − 1 × 0 = (3 − λ)(2 − λ), so the eigenvalues are 3 and 2, the diagonal entries. That happens only because an entry off the diagonal is 0. For A = (4 −1; 2 1), the diagonal entries 4 and 1 are not the eigenvalues.

For (2 1; 0 2), the polynomial is ${\left(2-\lambda \right)}^2$(2 − λ)², and the eigenvalue 2 is repeated. For the quarter turn (0 −1; 1 0), it is ${\lambda }^2+1$λ² + 1, which is never 0 for a real $\lambda$λ, so the quarter turn has no real eigenvalues, as it keeps no direction.

For a 3 × 3 matrix the same determinant, $\det \left(A-\lambda I\right)$det(A − λI), is a cubic in $\lambda$λ, so a 3 × 3 matrix has up to three eigenvalues.

The usual mistakes

Taking $\lambda$λ off every entry. $\lambda I$λI has zeros off the diagonal, so −1 and 2 stay as they are.

Adding the second product. $\det \left(A-\lambda I\right)=\left(4-\lambda \right)\left(1-\lambda \right)-\left(-1\right)\left(2\right)$det(A − λI) = (4 − λ)(1 − λ) − (−1)(2); writing + (−1)(2) gives ${\lambda }^2-5\lambda +2$λ² − 5λ + 2, whose roots are not the eigenvalues.

Turning the signs of the roots over. $\left(\lambda -2\right)\left(\lambda -3\right)=0$(λ − 2)(λ − 3) = 0 gives $\lambda =2$λ = 2 and $\lambda =3$λ = 3, not −2 and −3.

Reading the eigenvalues off the diagonal of a matrix that is not triangular. The diagonal of A holds 4 and 1, and its eigenvalues are 2 and 3.

Bacteria and springs

In the applications below, a matrix describes how two strains of bacteria turn into each other each hour, and its characteristic polynomial gives two growth factors, of which only one has an eigenvector that can be a real culture. Then two trolleys joined by springs swing in two ways, and the eigenvalues of the spring matrix compare how fast.