2×2 Matrix Eigenvector Formulas
Introduction
A=[abcd]
If matrix A is a 2×2 matrix and x is a nonzero vector and both are in R2 or C2, then x is an eigenvector of A if Ax is a scalar multiple of x and expressed as Ax=λx
Eigenvalues of A are λ1=a+d+√(a−d)2+4bc2
For matrix A, we present two sets of general eigenvector formulas as well as a pair of special case formulas. The pair of eigenvector formulas associated with λ1 are
x1=(a−d+√(a−d)2+4bc2c)
The pair of eigenvector formulas associated with λ2 are
x2=(a−d−√(a−d)2+4bc2c)
When b or c is 0, the pair of special case formulas are
xλ=a=(a−dc) and xλ=d=(bd−a).
Eigenvalues
The characteristic equation for eigenvalues of A is det(A−λI)=0 and written out is
|a−λbcd−λ|=0
and the equation is λ2−(a+d)λ+(ad−bc)=0. The characteristic polynomial p(λ) is λ2−(a+d)λ+(ad−bc).
Eigenvalues of A are determined by solving the characteristic equation and they are λ1=a+d+√(a−d)2+4bc2
If b=0 or c=0, then λ={a,d}.
When b=0 or c=0 the characteristic polynomial is λ2−(a+d)λ+ad with roots λ={a,d}.
Eigenvector formulas
General formulas for λ1
λ1 is an eigenvalue of A associated with eigenvectors x1=(2bd−a+√(d−a)2+4bc)
Let x=(x1,x2) and x is in R2 or C2 and
D=(a−d)2+4bc=(d−a)2+4bc=a2−2ad+d2+4bc.
The equation for determining eigenvectors corresponding to the eigenvalue λ1 is (A−λ1I)x=0.
(A−λ1I)x=(12(a−d−√D)x1+bx2cx1+12(d−a−√D)x2)=0.
Solving the first component in (1), 12(a−d−√D)x1+bx2=0, we have x1=2bd−a+√Dx2
[abcd](2bd−a+√D)=(b(a+d+√D)2bc+d(d−a+√D)).
To verify Ax1=λ1x1, we compute the scalar product of λ1 and the eigenvector x1.
a+d+√D2(2bd−a+√D)=(b(a+d+√D)12(a+d+√D)(d−a+√D))=(b(a+d+√D)12(2d2+2d√D−2ad+4bc))=(b(a+d+√D)2bc+d(d−a+√D)).
Solving the second component in (1), cx1+12(d−a−√D)x2=0, we find
x1=a−d+√D2cx2
and the eigenvector is
x′1=(a−d+√D2c).
Next, we compute the dot product of the eigenvector x′1 and A.
[abcd](a−d+√D2c)=(a(a−d+√D)+2bcc(a+d+√D)).
To verify Ax′1=λ1x′1, we compute the scalar product of λ1 and the eigenvector x′1.
a+d+√D2(a−d+√D2c)=(12(a+d+√D)(a−d+√D)c(a+d+√D))=(12(2a2+2a√D−2ad+4bc)c(a+d+√D))=(a(a−d+√D)+2bcc(a+d+√D)).
General formulas for λ2
λ2 is an eigenvalue of A associated with x2 which are given by x2=(2bd−a−√(d−a)2+4bc)
Let x=(x1,x2) and x in R2 or C2 and
D=(a−d)2+4bc=(d−a)2+4bc=a2−2ad+d2+4bc.
The equation for determining eigenvectors corresponding to the eigenvalue λ2 is (A−λ2I)x=0.
(A−λ2I)x=(12(a−d+√D)x1+bx2cx1+12(d−a+√D)x2)=0.
Solving the first component in (2), 12(a−d+√D)x1+bx2=0, we find
x1=2bd−a−√Dx2
and the eigenvector is
x2=(2bd−a−√D).
Next, we compute the dot product of the eigenvector x2 and A.
[abcd](2bd−a−√D)=(b(a+d−√D)2bc+d(d−a−√D)).
To verify Ax2=λ2x2, we compute the scalar product of λ2 and the eigenvector x2.
(a+d−√D2)(2bd−a−√D)=(b(a+d−√D)12(a+d−√D)(d−a−√D))=(b(a+d−√D)12(2d2−2d√D−2ad+4bc))=(b(a+d−√D)2bc+d(d−a−√D)).
Solving the second component in (2), cx1+12(d−a+√D)x2=0, we find
x1=a−d−√D2cx2
and the eigenvector is
x′2=(a−d−√D2c).
Next, we compute the dot product of the eigenvector x′2 and A.
[abcd](a−d−√D2c)=(a(a−d−√D)+2bcc(a+d−√D)).
To verify Ax′2=λ2x′2, we compute the scalar product of λ2 and the eigenvector x′2.
(a+d−√D2)(a−d−√D2c)=(12(a+d−√D)(a−d−√D)c(a+d−√D))=(12(2a2−2ad−2a√D+4bc)c(a+d−√D))=(a(a−d−√D)+2bcc(a+d−√D)).
Special case formulas
A set of special case eigenvector formulas exist when b or c is 0.
When b=0 then λ=a belongs to the eigenvector xλ=a=(a−dc).
When b=0 and λ=a, the equation to solve for the eigenvector is (A−aI)x=0 and written out is
(0cx1+(d−a)x2)=0.
Solving the second component we find
x1=a−dcx2
and the eigenvector is
xλ=a=(a−dc).
We confirm xλ=a is an eigenvector by showing Ax=λx and we have
[a0cd](a−dc)=(a(a−d)ac)=a(a−dc).
If c=0 then λ=d belongs to the eigenvector xλ=d=(bd−a).
When c=0 and λ=d
(A−dI)x=((a−d)x1+bx20)=0
Solving the first component we find (d−a)x1=bx2 and the eigenvector is
xλ=d=(bd−a).
We confirm xλ=d is an eigenvector by showing Ax=λx and we have
[ab0d](bd−a)=(bdd(d−a))=d(bd−a).
Examples
Example 1
Let A=[1345]. The characteristic polynomial of A is p(λ)=λ2−6λ−7 and the eigenvalues are λ1=7,λ2=−1. The eigenvectors of A are
x1=((a−d)+√(a−d)2+4bc2c)=((1−5)+√16+4(3)(4)2(4))=4(12)x1=(2b(d−a)+√(d−a)2+4bc)=(2(3)(5−1)+√16+4(3)(4))=6(12)x2=((a−d)−√(a−d)2+4bc2c)=((1−5)−√16+4(3)(4)2(4))=4(−32)x2=(2b(d−a)−√(d−a)2+4bc)=(2(3)(5−1)−√16+4(3)(4))=−2(−32)
Example 2
Let A=[1i−i−1]. The characteristic polynomial of A is p(λ)=λ2−2 and the eigenvalues are λ1=√2,λ2=−√2. The eigenvectors of A are
x1=((a−d)+√(a−d)2+4bc2c)=((1−(−1))+√(1−(−1))2+4(i)(−i)2(−i))=2((1+√2)i1)x1=(2b(d−a)+√(d−a)2+4bc)=(2(i)(−1−1)+√(−1−1)2+4(i)(−i))=2((1+√2)i1)x2=((a−d)−√(a−d)2+4bc2c)=((1−(−1))−√(1−(−1))2+4(i)(−i)2(−i))=2((1−√2)i1)x2=(2b(d−a)−√(d−a)2+4bc)=(2(i)(−1−1)−√(−1−1)2+4(i)(−i))=2((1−√2)i1)
Example 3
Let A=[1305]. The characteristic polynomial of A is p(λ)=λ2−6λ+5 and the eigenvalues are λ1=5,λ2=1. The eigenvectors of A are
x1=((a−d)+√(a−d)2+4bc2c)=((1−5)+√(1−5)2+4(3)(0)2(0))=(00)xλ=d=(bd−a)=(34)x1=(2b(d−a)+√(d−a)2+4bc)=(2(3)(5−1)+√16)=2(34)x2=((a−d)−√(a−d)2+4bc2c)=((1−5)−√(1−5)2+4(3)(0)2(0))=−8(10)x2=(2b(d−a)−√(d−a)2+4bc)=(2(3)(5−1)−√16)=6(10)
Example 4
Let A=[1043]. The characteristic polynomial of A is p(λ)=λ2−4λ+3 and the eigenvalues are λ1=3,λ2=1. The eigenvectors of A are
x1=((a−d)+√(a−d)2+4bc2c)=((1−3)+√(1−3)22(4))=8(01)x1=(2b(d−a)+√(d−a)2+4bc)=(2(0)(3−1)+√(3−1)2)=4(01)x2=((a−d)−√(a−d)2+4bc2c)=((1−3)−√(1−3)22(4))=4(−12)x2=(2b(d−a)−√(d−a)2+4bc)=(2(0)(3−1)−√(3−1)2)=(00)xλ=a=(a−dc)=2(−12)
Example 5
Let A=(104i). The characteristic polynomial of A is p(λ)=λ2−(1+i)λ+i and the eigenvalues are λ1=1,λ2=i. The eigenvectors of A are
x1=((a−d)+√(a−d)2+4bc2c)=((1−i)+√(1−i)2+4(0)(4)2(4))=(1−i+1−i8)=2(1−i4)x1=(2b(d−a)+√(d−a)2+4bc)=(2(0)(i−1)+√(i−1)2+4(0)(4))=(0i−1+1−i)=(00)xλ=a=(a−dc)=(1−i4)x2=((a−d)−√(a−d)2+4bc2c)=((1−i)−√(1−i)2+4(0)(4)2(4))=(1−i−1+i8)=8(01)x2=(2b(d−a)−√(d−a)2+4bc)=(2(0)(i−1)−√(i−1)2+4(0)(4))=(0i−1−1+i)=2(i−1)(01)√(i−1)2=√−2i=1−i
Example 6
Let A=(0−110). The characteristic polynomial of A is p(λ)=λ2+1 and the eigenvalues are λ1=i,λ2=−i. The eigenvectors are A are
x1=((a−d)+√(a−d)2+4bc2c)=((0−0)+√(0−0)2+4(−1)(1)2(1))=2(i1)x1=(2b(d−a)+√(d−a)2+4bc)=(2(−1)(0−0)+√(0−0)+4(−1)(1))=2i(i1)x2=((a−d)−√(a−d)2+4bc2c)=((0−0)−√(0−0)2+4(−1)(1)2(1))=2(−i1)x2=(2b(d−a)−√(d−a)2+4bc)=(2(−1)(0−0)−√(0−0)+4(−1)(1))=−2i(−i1)
Example 7
Let A=(4−52−2). The characteristic polynomial of A is p(λ)=λ2−2λ+2 and the eigenvalues are λ1=1+i,λ2=1−i. The eigenvectors are A are
x1=((a−d)+√(a−d)2+4bc2c)=((4−(−2))+√(4−(−2))2+4(−5)(2)2(2))=2(3+i2)x1=(2b(d−a)+√(d−a)2+4bc)=(2(−5)(−2−4)+√(−2−4)2+4(−5)(2))=−10(3+i2)x2=((a−d)−√(a−d)2+4bc2c)=((4−(−2))−√(4−(−2))2+4(−5)(2)2(2))=2(3−i2)x2=(2b(d−a)−√(d−a)2+4bc)=(2(−5)(−2−4)−√(−2−4)2+4(−5)(2))=−10(3−i2)
Eigenvector tool
The eigenvector tool only utilizes the previously defined formulas to calculate eigenvectors. Additionally a link to WolframAlpha that solves the same matrix is provided to confirm the results.