Solution.
1.
A = ( 1 2 − 8 4 ) A=\begin{pmatrix}
1 & 2\\
-8& 4
\end{pmatrix} A = ( 1 − 8 2 4 ) Solve equation for finding eigenvalues:
∣ A − λ I ∣ = 0 |A-\lambda I|=0 ∣ A − λ I ∣ = 0
∣ 1 − λ 2 − 8 4 − λ ∣ = λ 2 − 5 λ + 20. \begin{vmatrix}
1- \lambda& 2 \\
-8 & 4-\lambda
\end{vmatrix}=\lambda^2-5\lambda+20. ∣ ∣ 1 − λ − 8 2 4 − λ ∣ ∣ = λ 2 − 5 λ + 20. From here we have eigenvaluesλ 1 = 5 − 55 i 2 , λ 2 = 5 + 55 i 2 . \lambda_1=\frac{5-\sqrt{55}i}{2}, \lambda_2=\frac{5+\sqrt{55}i}{2}. λ 1 = 2 5 − 55 i , λ 2 = 2 5 + 55 i .
Solve equation for finding the first eigenvector
( A − λ 1 I ) v 1 = 0. (A-\lambda_1 I)v_1=0. ( A − λ 1 I ) v 1 = 0.
Hence
( − 3 + 55 i 2 2 − 8 3 + 55 i 2 ) → ( 1 − 3 − 55 i 16 0 0 ) \begin{pmatrix}
\frac{-3+\sqrt{55}i}{2} & 2 \\
-8 & \frac{3+\sqrt{55}i}{2}
\end{pmatrix}\to
\begin{pmatrix}
1 & \frac{-3-\sqrt{55}i}{16}\\
0& 0
\end{pmatrix} ( 2 − 3 + 55 i − 8 2 2 3 + 55 i ) → ( 1 0 16 − 3 − 55 i 0 ) then x 2 = x 2 , x 1 = 3 + 55 i 16 . x_2=x_2, x_1=\frac{3+\sqrt{55}i}{16}. x 2 = x 2 , x 1 = 16 3 + 55 i .
So, v 1 = ( 3 + 55 i 16 1 ) . v_1=\begin{pmatrix}
\frac{3+\sqrt{55}i}{16} \\
1
\end{pmatrix}. v 1 = ( 16 3 + 55 i 1 ) .
Solve equation for finding the second eigenvector
( A − λ 2 I ) v 2 = 0. (A-\lambda_2 I)v_2=0. ( A − λ 2 I ) v 2 = 0.
Hence
( − 3 − 55 i 2 2 − 8 3 − 55 i 2 ) → ( 1 − 3 + 55 i 16 0 0 ) \begin{pmatrix}
\frac{-3-\sqrt{55}i}{2} & 2 \\
-8 & \frac{3-\sqrt{55}i}{2}
\end{pmatrix}\to
\begin{pmatrix}
1 & \frac{-3+\sqrt{55}i}{16}\\
0& 0
\end{pmatrix} ( 2 − 3 − 55 i − 8 2 2 3 − 55 i ) → ( 1 0 16 − 3 + 55 i 0 ) then x 2 = x 2 , x 1 = 3 − 55 i 16 . x_2=x_2, x_1=\frac{3-\sqrt{55}i}{16}. x 2 = x 2 , x 1 = 16 3 − 55 i .
So, v 2 = ( 3 − 55 i 16 1 ) . v_2=\begin{pmatrix}
\frac{3-\sqrt{55}i}{16} \\
1
\end{pmatrix}. v 2 = ( 16 3 − 55 i 1 ) .
2.
B = ( a b − b a ) B=\begin{pmatrix}
a& b\\
-b& a
\end{pmatrix} B = ( a − b b a ) Solve equation for finding eigenvalues:
∣ B − λ I ∣ = 0 |B-\lambda I|=0 ∣ B − λ I ∣ = 0
∣ a − λ b − b a − λ ∣ = a 2 − λ 2 + b 2 . \begin{vmatrix}
a- \lambda& b \\
-b & a-\lambda
\end{vmatrix}=a^2-\lambda^2+b^2. ∣ ∣ a − λ − b b a − λ ∣ ∣ = a 2 − λ 2 + b 2 . From here we have eigenvaluesλ 1 = a 2 + b 2 , λ 2 = − a 2 + b 2 . \lambda_1=\sqrt{a^2+b^2}, \lambda_2=-\sqrt{a^2+b^2}. λ 1 = a 2 + b 2 , λ 2 = − a 2 + b 2 .
Solve equation for finding the first eigenvector
( B − λ 1 I ) v 1 = 0. (B-\lambda_1 I)v_1=0. ( B − λ 1 I ) v 1 = 0.
Hence
( a − a 2 + b 2 b − b a − a 2 + b 2 ) → ( b − a − a 2 + b 2 0 2 b a 2 + b 2 ) \begin{pmatrix}
a- \sqrt{a^2+b^2}& b\\
-b & a-\sqrt{a^2+b^2}
\end{pmatrix}\to
\begin{pmatrix}
b & -a-\sqrt{a^2+b^2}\\
0& 2b\sqrt{a^2+b^2}
\end{pmatrix} ( a − a 2 + b 2 − b b a − a 2 + b 2 ) → ( b 0 − a − a 2 + b 2 2 b a 2 + b 2 ) x 2 = 2 b a 2 + b 2 , x 1 = 2 a a 2 + b 2 + 2 ( a 2 + b 2 ) . x_2=2b\sqrt{a^2+b^2},
x_1=2a\sqrt{a^2+b^2}+2(a^2+b^2). x 2 = 2 b a 2 + b 2 , x 1 = 2 a a 2 + b 2 + 2 ( a 2 + b 2 ) .
So, v 1 = ( 2 a a 2 + b 2 + 2 ( a 2 + b 2 ) 2 b a 2 + b 2 ) . v_1=\begin{pmatrix}
2a\sqrt{a^2+b^2}+2(a^2+b^2)\\
2b\sqrt{a^2+b^2}
\end{pmatrix}. v 1 = ( 2 a a 2 + b 2 + 2 ( a 2 + b 2 ) 2 b a 2 + b 2 ) .
Solve equation for finding the second eigenvector
( B − λ 2 I ) v 2 = 0. (B-\lambda_2 I)v_2=0. ( B − λ 2 I ) v 2 = 0.
Hence
( a + a 2 + b 2 b − b a + a 2 + b 2 ) → ( − b a + a 2 + b 2 0 − 2 b a 2 + b 2 ) \begin{pmatrix}
a+\sqrt{a^2+b^2}& b\\
-b & a+\sqrt{a^2+b^2}
\end{pmatrix}\to
\begin{pmatrix}
-b & a+\sqrt{a^2+b^2}\\
0& -2b\sqrt{a^2+b^2}
\end{pmatrix} ( a + a 2 + b 2 − b b a + a 2 + b 2 ) → ( − b 0 a + a 2 + b 2 − 2 b a 2 + b 2 ) x 2 = − 2 b a 2 + b 2 , x 1 = 2 a a 2 + b 2 − 2 ( a 2 + b 2 ) . x_2=-2b\sqrt{a^2+b^2},
x_1=2a\sqrt{a^2+b^2}-2(a^2+b^2). x 2 = − 2 b a 2 + b 2 , x 1 = 2 a a 2 + b 2 − 2 ( a 2 + b 2 ) .
So, v 2 = ( 2 a a 2 + b 2 − 2 ( a 2 + b 2 ) − 2 b a 2 + b 2 ) . v_2=\begin{pmatrix}
2a\sqrt{a^2+b^2}-2(a^2+b^2)\\
-2b\sqrt{a^2+b^2}
\end{pmatrix}. v 2 = ( 2 a a 2 + b 2 − 2 ( a 2 + b 2 ) − 2 b a 2 + b 2 ) .
#339914 on Dec 2023
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