2.2 - Eigenvectors in three dimensions

2.2 - Eigenvectors in three dimensions

A three-by-three matrix has a characteristic cubic, but its eigenvectors still come from linear equations. You will use determinant structure, find vectors for real distinct eigenvalues, and prove consequences of the defining equation for powers and inverses.

The characteristic cubic

Can a changed matrix keep the same eigenvalues? Compare

B=(310130005),A=(310130105).\mathbf B=\begin{pmatrix}3&1&0\\1&3&0\\0&0&5\end{pmatrix},\qquad \mathbf A=\begin{pmatrix}3&1&0\\1&3&0\\1&0&5\end{pmatrix}.

Only the bottom-left entry changes. An expansion down the third column of A−λI\mathbf A-\lambda\mathbf I never uses that entry, so the characteristic equations will agree. The eigenvectors need not agree.

For three dimensions, det⁡(A−λI)=0\det(\mathbf A-\lambda\mathbf I)=0 is cubic. Equivalently, use det⁡(λI−A)=0\det(\lambda\mathbf I-\mathbf A)=0, which changes the determinant's sign but not its zeros. The latter produces a monic cubic; below we state the final equation in that form.

Worked example 1: exploiting a sparse column

Question. Find the characteristic equation and eigenvalues of B=(310130005)\mathbf B=\begin{pmatrix}3&1&0\\1&3&0\\0&0&5\end{pmatrix}.

Plan. Expand down column 3, then factor the remaining quadratic.

det⁡(B−λI)=(5−λ)((3−λ)2−1)=(5−λ)(2−λ)(4−λ).\begin{aligned} \det(\mathbf B-\lambda\mathbf I) &=(5-\lambda)\bigl((3-\lambda)^2-1\bigr)\\ &=(5-\lambda)(2-\lambda)(4-\lambda). \end{aligned}

Equating this to zero and multiplying the equation by −1-1 gives a monic form.

Answer. The characteristic equation is

(λ−2)(λ−4)(λ−5)=0,orλ3−11λ2+38λ−40=0.(\lambda-2)(\lambda-4)(\lambda-5)=0, \quad\text{or}\quad\lambda^3-11\lambda^2+38\lambda-40=0.

The eigenvalues are 2,4,52,4,5.

Check. The roots add to the diagonal sum 3+3+5=113+3+5=11, while their product is 40=5(9−1)=det⁡B40=5(9-1)=\det\mathbf B. Their pair products are 8+10+20=388+10+20=38, checking the expanded cubic.

Change only the bottom-left entry to 11. This shows why a nontriangular matrix can still have an efficient determinant expansion.

Worked example 2: a coupling that leaves the roots unchanged

Question. Find the characteristic equation and eigenvalues of A=(310130105)\mathbf A=\begin{pmatrix}3&1&0\\1&3&0\\1&0&5\end{pmatrix}.

Plan. Expand down the same column; its two zeros still remove two terms.

det⁡(A−λI)=(5−λ)∣3−λ113−λ∣=(5−λ)(2−λ)(4−λ).\det(\mathbf A-\lambda\mathbf I) =(5-\lambda)\begin{vmatrix}3-\lambda&1\\1&3-\lambda\end{vmatrix} =(5-\lambda)(2-\lambda)(4-\lambda).

The bottom-left entry is outside this minor.

Answer. The characteristic equation is again λ3−11λ2+38λ−40=0\lambda^3-11\lambda^2+38\lambda-40=0, with eigenvalues 2,4,52,4,5.

Check. Independently expand det⁡(λI−A)\det(\lambda\mathbf I-\mathbf A) along row 1:

(λ−3)2(λ−5)−(λ−5)=(λ−5)((λ−3)2−1).(\lambda-3)^2(\lambda-5)-(\lambda-5) =(\lambda-5)\bigl((\lambda-3)^2-1\bigr).

This has exactly the stated roots.

For a triangular matrix the characteristic determinant is the product of its diagonal factors. For a general matrix, inspect sparse rows and columns first. If a cubic does not factor immediately, test plausible rational roots, divide by the factor found, and solve the remaining quadratic.

Watch out

In three dimensions, det⁡(A−λI)\det(\mathbf A-\lambda\mathbf I) has leading term −λ3-\lambda^3. Multiplying the whole characteristic equation by −1-1 is valid; changing only some signs is not.

Use the same sparse-column strategy with different numbers.

Real distinct eigenvalues

Once a root is known, a three-dimensional eigenvector comes from (A−λI)e=0(\mathbf A-\lambda\mathbf I)\mathbf e=\mathbf0, just as in two dimensions. Use all three equations: the first two alone may leave a coordinate unchecked.

The eigenvalue calculations here concern real distinct eigenvalues. For a three-by-three matrix this gives three independent eigendirections. Choose convenient integer components, avoiding the zero vector. Repeated eigenvalues and non-real eigendirections require further theory outside this sequence.

Worked example 3: using the third coordinate equation

Question. Find eigenvectors of A=(310130105)\mathbf A=\begin{pmatrix}3&1&0\\1&3&0\\1&0&5\end{pmatrix} for λ=2,4,5\lambda=2,4,5.

Plan. Solve the three homogeneous equations for each eigenvalue, clearing fractions where useful.

For λ=2\lambda=2, the equations reduce to x+y=0x+y=0 and x+3z=0x+3z=0. Take x=3x=3, giving y=−3y=-3, z=−1z=-1.

For λ=4\lambda=4, they reduce to −x+y=0-x+y=0 and x+z=0x+z=0. Take x=1x=1, giving y=1y=1, z=−1z=-1.

For λ=5\lambda=5, −2x+y=0-2x+y=0 and x−2y=0x-2y=0 force x=y=0x=y=0; zz is free and must be nonzero.

Answer. Corresponding eigenvectors are

λ=2: (3−3−1),λ=4: (11−1),λ=5: (001).\lambda=2:\ \begin{pmatrix}3\\-3\\-1\end{pmatrix},\qquad \lambda=4:\ \begin{pmatrix}1\\1\\-1\end{pmatrix},\qquad \lambda=5:\ \begin{pmatrix}0\\0\\1\end{pmatrix}.

Every nonzero multiple of each is valid.

Check. Multiplication by the original matrix gives (6,−6,−2)T(6,-6,-2)^{\mathsf T}, (4,4,−4)T(4,4,-4)^{\mathsf T}, and (0,0,5)T(0,0,5)^{\mathsf T}, respectively, verifying all coordinates and pairings.

Change only the bottom-right entry from 55 to 66. The first two coordinate relations stay the same; the final coordinate changes for the first two directions.

Worked example 4: adapting the eigenvectors

Question. Find eigenvalues and corresponding eigenvectors of E=(310130106)\mathbf E=\begin{pmatrix}3&1&0\\1&3&0\\1&0&6\end{pmatrix}.

Plan. Use the characteristic factorisation, then recalculate each third-coordinate relation.

The determinant factors as (6−λ)(2−λ)(4−λ)(6-\lambda)(2-\lambda)(4-\lambda), so the eigenvalues are 2,4,62,4,6.

At λ=2\lambda=2, y=−xy=-x and x+4z=0x+4z=0, giving (4,−4,−1)T(4,-4,-1)^{\mathsf T}. At λ=4\lambda=4, y=xy=x and x+2z=0x+2z=0, giving (2,2,−1)T(2,2,-1)^{\mathsf T}. At λ=6\lambda=6, the first two equations force x=y=0x=y=0, so take (0,0,1)T(0,0,1)^{\mathsf T}.

Answer. The three displayed vectors correspond to 2,4,62,4,6, respectively, with arbitrary nonzero scalar multiples allowed.

Check. Their original-matrix images are (8,−8,−2)T(8,-8,-2)^{\mathsf T}, (8,8,−4)T(8,8,-4)^{\mathsf T}, and (0,0,6)T(0,0,6)^{\mathsf T}, the required multiples.

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