Eigenvalues and Eigenvectors of 22× Matrix
A matrix is a rectangular array of numbers arranged in rows and columns.
A 22× Matrix is a rectangular array of numbers arranged in 2 rows and 2 columns. A 22× Matrix is represented in the following:
ab A = cd
10 We call special matrix identity matrix. In general, we use I to represent identity matrix, i.e, 01
10 I = 01
ab11 ab22 x1 Operations of 22× Matrix: Let A = , B = , x = , α is a scalar number. cd11 cd22 x2 We have
ααab11 a. α A = ααcd11
a11 b a 22 b aa 1212++ bb b. AB+= + = cd1 1 c 2 d 2 cc 12++ dd 1 2
a1 b 1 x 1 ax 11+ bx 12 c. Ax = = c1 d 1 x 2 cx 11+ dx 12
a1 b 1 a 2 b 2 aa 1 2++ bc 12 ab 12 bd 1 2 d. AB = = cd1 1 c 2 d 2 ca 1 2++ dc 12 cbdd 12 1 2
14 79 Example 1: Given A = , B = . Find 23 58
a. 2AB− b. AB
Answer: −−51 a. 2AB−= −−12
27 41 b. AB = 29 49
ab The Determinant of 22× matrix A = is cd
ab A= = ad − bc cd
26 Example 2: Find the determinant of −17
26 Answer: =14 −− ( 6) = 20 −17
Characteristic Equation of 22× matrix A :
ab− λ AI−=λ =0 cd− λ
The above equation is called the Characteristic Equation of matrix A .
12− Example 3: Find the characteristic equation of 34−
Answer: The characteristic equation is
12−−λ = 0 34−−λ (1−λλ )(4 − − ) − 3 ×− (2)0 = −+44λλλ − +2 + 60 = λλ2 +3 += 20
Eigenvalues of matrix A . The solutions to the characteristic equation of matrix A are eigenvalues of matrix A .
Review the solution formula to quadric equations.
For quadric equation:
ax2 + bx += c 0,
−±b b2 −4 ac the solutions are x = 1,2 2a
12− Example 4: Find the eigenvalues of 34−
Answer: From example 3, we know the characteristic equation is
λλ2 +3 += 20 (1)
Solving characteristic equation (1), we have λ1,2 =−−2, 1
12− Thus, the eigenvalues of are λ1,2 =−−2, 1 34−
Eigenvectors of of matrix A .
ab x1 Given λ is a eigenvalue of 22× matrix A = , then any nonzero vector x = is called cd x2 eigenvector of matrix A if
Ax= λ Ix
or (A−=λ Ix )0
ab− λ x1 or = 0 cd− λ x2
x1 cx1 Remark: If x = is an eigenvector of matrix A , then cx = is an eigenvector of matrix x2 cx2 A for any nonzero scalar number c . 12− Example 5: Find the eigenvectors of 34−
12− From Example 4, we have λ1,2 =−−2, 1are the eigenvalues of . 34−
For eigenvalue λ = −2 , (A−=λ Ix )0 becomes (A+= 2) Ix 0 or
12+− 2 x1 = 0 3−+ 42x2 ⇓
32xx12−= 0 32xx12−= 0 ⇓ 2 xt= 1 3 xt2 =
2 Let t =1, we have one eigenvector x = 3 . 1
For eigenvalue λ = −1, (A−=λ Ix )0 becomes (A+= Ix )0 or
11+− 2 x1 = 0 3−+ 41x2 ⇓
22xx12−= 0 33xx12−= 0 ⇓
xt1 = xt2 = 1 Let t =1, we have one eigenvector x = . 1
1 −−1 Example 6: Find the eigenvectors of 2 23−
First, we get the characteristic equation:
1 −−1 λ − 2 = 0 23−−λ ⇓ (1−−λλ )(3 −− ) + 1 = 0 ⇓ λλ2 +4 += 40
Solving it, we have the eigenvalues are λ1,2 = −2 .
For eigenvalue λ = −2 , we have (A−=λ Ix )0 becomes (A+= 2) Ix 0 or
1 −+12 − x 1 = 0 2 x2 2−+ 32 ⇓ 1 xx−=0 122 20xx12−= ⇓ 1 xt= 1 2 xt2 =
1 Let t =1, we have one eigenvector x = . 2 1
32− Example 7: Find the eigenvectors of 41−
First, we get the characteristic equation:
32−−λ = 0 41−−λ ⇓ (3−λλ )( −− 1 ) + 8 = 0 ⇓ λλ2 −2 += 50 ⇓ 2±− 4 20 λ = =12 ± i 1,2 2
Solving it, we have the eigenvalues are λ1,2 =12 ± i .
For eigenvalue λ =12 + i , we have (A−=λ Ix )0 becomes (A−+ (1 2 iIx ) ) = 0 or
3−+ (1 2i ) − 2 x1 = 0 4−− 1 (1 + 2i ) x2 ⇓
(2− 2ix )12 −= 2 x 0 4x12−+ (2 2 ix ) = 0 ⇓ 2 1 (1+i ) (1 ++ ii ) 1 x1 = tt = = t= tt = 2− 2i 1 − i (1 −+ ii )(1 ) 122 − i 2 xt2 =
1 (1+ i ) Let t =1, we have one eigenvector x = . 2 1
For eigenvalue λ =12 − i , we have (A−=λ Ix )0 becomes (A−− (1 2 iIx ) ) = 0 or
3−− (1 2i ) − 2 x1 = 0 4−− 1 (1 − 2i ) x2 ⇓
(2+ 2ix )12 −= 2 x 0 4x12−− (2 2 ix ) = 0 ⇓ 2 1 (1−i ) (1 −− ii ) 1 x1 = tt = = t= tt = 2+ 2i 1 + i (1 −+ ii )(1 ) 122 − i 2 xt2 =
1 (1− i ) Let t =1, we have one eigenvector x = . 2 1