2.2 - Eigenvalues and eigenvectors

2.2 - Eigenvalues and eigenvectors

Most vectors change direction under a matrix transformation, but some special directions are preserved. You will identify these eigenvectors, find their scale factors from a characteristic equation, and solve for every corresponding eigendirection of a two-by-two matrix.

Unchanged directions

Which vectors can a matrix stretch without turning them away from their original line? For A=(4123)\mathbf A=\begin{pmatrix}4&1\\2&3\end{pmatrix}, the vector (1,1)T(1,1)^{\mathsf T} becomes (5,5)T(5,5)^{\mathsf T}. Its length changes but its direction line stays the same.

Eigenvalue and eigenvector

For a square matrix A\mathbf A, a nonzero vector e\mathbf e is an eigenvector with eigenvalue λ\lambda when

Ae=λe.\mathbf A\mathbf e=\lambda\mathbf e.

The scalar λ\lambda describes the action along that direction. Every nonzero multiple of e\mathbf e is another eigenvector with the same eigenvalue.

If λ>0\lambda>0, the sense is preserved; if λ<0\lambda<0, it reverses. If λ=0\lambda=0, the vector is sent to zero. That is still an eigenvector, although its line collapses instead of being mapped onto itself. The zero vector itself is excluded: it would satisfy the equation for every λ\lambda and identify no direction.

[DIAGRAM: asset_name: Two special directions of a matrix; asset_slug: cam9231-eigendirections; caption: Along an eigendirection the image stays on the same line; description: Exact coordinate panels for the matrix with rows 4 1 and 2 3. A vector (1,1) and its image (5,5) lie on y equals x, while (1,minus 2) and its image (2,minus 4) lie on y equals minus 2x. A second panel compares (1,0) with its image (4,2), showing a direction that is not an eigendirection. Inputs and images are labelled with pale blue and sage accents.]
Along an eigendirection the image stays on the same line

Check the whole vector equation, not just one coordinate ratio; a coordinate may be zero.

Worked example 1: verifying an eigenvector

Question. For A=(4123)\mathbf A=\begin{pmatrix}4&1\\2&3\end{pmatrix}, determine whether e=(1,1)T\mathbf e=(1,1)^{\mathsf T} and v=(1,0)T\mathbf v=(1,0)^{\mathsf T} are eigenvectors.

Plan. Multiply each vector and test whether the result is one scalar multiple of the original.

Ae=(55)=5e,Av=(42).\mathbf A\mathbf e=\begin{pmatrix}5\\5\end{pmatrix}=5\mathbf e, \qquad \mathbf A\mathbf v=\begin{pmatrix}4\\2\end{pmatrix}.

Every multiple of v\mathbf v has second coordinate zero; its image does not.

Answer. e\mathbf e is an eigenvector with eigenvalue 55. The vector v\mathbf v is not an eigenvector.

Check. The coordinate rules give X−Y=(4x+y)−(2x+3y)=2(x−y)X-Y=(4x+y)-(2x+3y)=2(x-y). Thus y=xy=x is an invariant line. On it, X=5xX=5x, confirming the multiplier 55. The horizontal direction of v\mathbf v has slope zero, whereas its image has slope 2/4=1/22/4=1/2, so that direction changes.

Now subtract 4I4\mathbf I from the matrix. This changes a scale factor and can reverse its sign, without changing the direction vectors.

Worked example 2: a negative eigenvalue

Question. For B=(012−1)\mathbf B=\begin{pmatrix}0&1\\2&-1\end{pmatrix}, verify that f=(1,−2)T\mathbf f=(1,-2)^{\mathsf T} is an eigenvector and find its eigenvalue.

Plan. Multiply first, then compare both coordinates with the original vector.

Bf=(−24)=−2(1−2).\mathbf B\mathbf f=\begin{pmatrix}-2\\4\end{pmatrix} =-2\begin{pmatrix}1\\-2\end{pmatrix}.

The multiplier is negative, so the vector reverses sense and doubles in length.

Answer. f\mathbf f is an eigenvector with eigenvalue −2-2.

Check. Since B=A−4I\mathbf B=\mathbf A-4\mathbf I and Af=(2,−4)T=2f\mathbf A\mathbf f=(2,-4)^{\mathsf T}=2\mathbf f, subtracting 4f4\mathbf f gives −2f-2\mathbf f, independently confirming the multiplier.

An eigenvector with eigenvalue 11 is fixed. Other nonzero eigenvalues preserve its line while moving nonzero points on it. The methods in this sequence concern matrices whose eigenvalues are real and distinct.

Watch out

λ\lambda is a scalar and e\mathbf e is a nonzero column vector. An answer such as “the eigenvector is 55” confuses the multiplier with the direction.

Verify the multiplier directly.

The characteristic equation

To discover the possible scale factors, rearrange the defining equation as (A−λI)e=0(\mathbf A-\lambda\mathbf I)\mathbf e=\mathbf0. A nonsingular coefficient matrix would force e=0\mathbf e=\mathbf0, which is forbidden. Therefore this matrix must be singular.

Characteristic equation

The characteristic equation of A\mathbf A is

det⁡(A−λI)=0.\det(\mathbf A-\lambda\mathbf I)=0.

For A=(abcd)\mathbf A=\begin{pmatrix}a&b\\c&d\end{pmatrix}, it becomes

(a−λ)(d−λ)−bc=0.(a-\lambda)(d-\lambda)-bc=0.

Subtract λ\lambda from diagonal entries only. The roots are the eigenvalues. Not in the formula booklet: learn it.

Using det⁡(λI−A)=0\det(\lambda\mathbf I-\mathbf A)=0 instead gives the same roots. Keep one convention throughout an expansion.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account