Úloha 3.2.14 - charakteristicke polynomy a podobnosť matic
Posted: Tue May 13, 2025 9:59 pm
$A = \begin{pmatrix}
1 & 4 & -2 \\
4 & 1 & -2 \\
-2 & -2 & -2
\end{pmatrix},$
$B = \begin{pmatrix}
1 & 4 & -2 \\
1 & 1 & -8 \\
-2 & -2 & -2
\end{pmatrix}$
\[
ch_A(x), \ ch_B(x) = ?, \quad A \sim B \ \text{?}
\]
\[
ch_A(x) = \det(A - xI) = \det \begin{pmatrix}
1 - x & 4 & -2 \\
4 & 1 - x & -2 \\
-2 & -2 & -2 - x
\end{pmatrix}
\]
\[
ch_A(x) = (1 - x) \det \begin{pmatrix}
1 - x & -2 \\
-2 & -2 - x
\end{pmatrix}
- 4 \det \begin{pmatrix}
4 & -2 \\
-2 & -2 - x
\end{pmatrix}
+ (-2) \det \begin{pmatrix}
4 & 1 - x \\
-2 & -2
\end{pmatrix}
\]
Prvý minor:
\[
\det \begin{pmatrix}
1 - x & -2 \\
-2 & -2 - x
\end{pmatrix} = (1 - x)(-2 - x) - (-2)(-2) = (1 - x)(-2 - x) - 4
= x^2 + x - 2 - 4 = x^2 + x - 6
\]
\[
a_{11} \cdot M_{11} = (1 - x)(x^2 + x - 6)
\]
Druhý minor:
\[
\det \begin{pmatrix}
4 & -2 \\
-2 & -2 - x
\end{pmatrix} = 4(-2 - x) - (-2)(-2) = -8 - 4x -4 = -4(x +3);
\]
\[
a_{12} \cdot M_{12} = 16(x + 3)
\]
Tretí minor:
\[
\det \begin{pmatrix}
4 & 1 - x \\
-2 & -2
\end{pmatrix} = 4(-2) - (1 - x)(-2) = -8 - (-2 + 2x) = -8 + 2 - 2x = -6 - 2x = -2(x + 3)
\]
\[
a_{13} \cdot M_{13} = 4(x + 3)
\]
\[
\text{Ch}_A(x) = (1 - x)(x^2 + x - 6) + 16(x + 3) + 4(x + 3)
\]
\[
= (1 - x)(x^2 + x - 6) + 20(x + 3) = -x^3 +7x - 6 + 20x + 60
= -x^3 + 27x + 54 = x^3 - 27x - 54
\]
\[
\text{Ch}_B(x) = \det(B - xI) = \det \begin{pmatrix}
1 - x & 4 & -2 \\
1 & 1 - x & -8 \\
-2 & -2 & -2 - x
\end{pmatrix}
\]
\[
\text{Ch}_B(x) = (1 - x) \det \begin{pmatrix}
1 - x & -8 \\
-2 & -2 - x
\end{pmatrix} - 4 \det \begin{pmatrix}
1 & -8 \\
-2 & -2 - x
\end{pmatrix} + (-2) \det \begin{pmatrix}
1 & 1 - x \\
-2 & -2
\end{pmatrix}
\]
Prvý minor:
\[
\det \begin{pmatrix}
1 - x & -8 \\
-2 & -2 - x
\end{pmatrix} = (1 - x)(-2 - x) - (-8)(-2) = x^2 + x - 2 - 16 = x^2 + x - 18
\]
\[
a_{11} \cdot M_{11} = (1 - x)(x^2 + x - 18)
\]
Druhý minor:
\[
\det \begin{pmatrix}
1 & -8 \\
-2 & -2 - x
\end{pmatrix} = 1(-2 - x) - (-8)(-2) = -2 - x - 16 = -x - 18
\]
\[
a_{12} \cdot M_{12} = 4(x + 18)
\]
Tretí minor:
\[
\det \begin{pmatrix}
1 & 1 - x \\
-2 & -2
\end{pmatrix} = 1(-2) - (1 - x)(-2) = -2 + 2(1 - x) = -2 + 2 - 2x = -2x
\]
\[
a_{13} \cdot M_{13} = 4x
\]
\[
\text{Ch}_B(x) = (1 - x)(x^2 + x - 18) + 4(x + 18) + 4x
\]
\[
= -x^3 + 19x - 18 + 4x + 72 + 4x = -x^3 + 27x + 54 = x^3 - 27x - 54
\]
Matice majú rovnaké charakteristické polynómy.
Ešte obe majú rovnakú stopu: $\text{Tr}(A) = 1 + 1 + (-2) = 0, \text{Tr}(B) = 0$; aj majú rovnaké determinanty:
\[
\det(A) = 1 \cdot \begin{vmatrix}
1 & -2 \\
-2 & -2
\end{vmatrix}
- 4 \cdot \begin{vmatrix}
4 & -2 \\
-2 & -2
\end{vmatrix}
+ (-2) \cdot \begin{vmatrix}
4 & 1 \\
-2 & -2
\end{vmatrix}
\]
\[
= 1(-6) - 4(-12) + (-2)(-6) = -6 + 48 + 12 = 54
\]
\[
\det(B) = 1 \cdot \begin{vmatrix}
1 & -8 \\
-2 & -2
\end{vmatrix}
- 4 \cdot \begin{vmatrix}
1 & -8 \\
-2 & -2
\end{vmatrix}
+ (-2) \cdot \begin{vmatrix}
1 & 1 \\
-2 & -2
\end{vmatrix}
\]
\[
= 1(-18) - 4(-18) + (-2)(0) = -18 + 72 = 54
\]
\[
x^3 - 27x - 54 = 0 \quad \Rightarrow \quad (x - 6)(x^2 + 6x + 9) = (x - 6)(x + 3)^2
\]
$ \lambda_1 = 6 $ (má algebraickú násobnosť - 1)
$ \lambda_2 = -3 $ (má algebraickú násobnosť - 2)
Najdeme geometrickú násobnosť pre každé vl. číslo:
$ \text{Pre } \lambda_{1} = 6$:
\[
B - 6I = \begin{pmatrix}
-5 & 4 & -2 \\
1 & -5 & -8 \\
-2 & -2 & -8
\end{pmatrix}
\text{Redukujem maticu do stupň. tvaru:}
\]
\[
B' =
\begin{pmatrix}
-5 & 4 & -2 \\
1 & -5 & -8 \\
-2 & -2 & -8
\end{pmatrix}
\begin{matrix}
<-\\
<-\\
\phantom{3}
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
-5 & 4 & -2 \\
-2 & -2 & -8
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
+5I\\
\phantom{3}
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
0 & -21 & -42 \\
-2 & -2 & -8
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
\phantom{2}\\
+2I
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
0 & -21 & -42 \\
0 & -12 & -24
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
⊙\frac{1}{21}\\
⊙ ( - \frac{1}{12} )
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
0 & 1 & 2 \\
0 & 1 & 2
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
\phantom{2}\\
- II
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
0 & 1 & 2 \\
0 & 0 & 0
\end{pmatrix}
h(B') = 2; geom. násobnosť: 3 - 2 = 1
\]
$ \text{Pre } \lambda_{2} = -3$:
\[
B'' =
B + 3I = \begin{pmatrix}
4 & 4 & -2 \\
1 & 4 & -8 \\
-2 & -2 & 1
\end{pmatrix}
\text{Redukujem maticu do stupň. tvaru:}
\]
\[
B'' =
\begin{pmatrix}
4 & 4 & -2 \\
1 & 4 & -8 \\
-2 & -2 & 1
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
- \frac{1}{4}\\
\phantom{3}
\end{matrix}
\sim
\begin{pmatrix}
4 & 4 & -2 \\
0 & 3 & - \frac{15}{2} \\
-2 & -2 & 1
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
\phantom{2}\\
+ \frac{1}{2} I
\end{matrix}
\sim
\begin{pmatrix}
4 & 4 & -2 \\
0 & 3 & - \frac{15}{2} \\
0 & 0 & 0
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
⊙ \frac{1}{3}\\
\phantom{3}
\end{matrix}
\sim
\begin{pmatrix}
4 & 4 & -2 \\
0 & 1 & - \frac{15}{6} \\
0 & 0 & 0
\end{pmatrix}
h(B'') = 2; geom. násobnosť: 3 - 2 = 1
\]
\[
\text{Pre } \lambda_{2} = -3 \Rightarrow \text{geom. násobnosť je menšia ako algebraická} \Rightarrow B \ \text{nie je diagonalizovateľná, pričom matica A je symetrická:}
\]
\[
(a_{12} = a_{21} = 4, \quad a_{13} = a_{31} = -2, \quad a_{23} = a_{32} = -2)
\]
\[
\Rightarrow \text{je diagonalizovateľná} \Rightarrow \text{matice A, B nie sú podobné.}
\]
1 & 4 & -2 \\
4 & 1 & -2 \\
-2 & -2 & -2
\end{pmatrix},$
$B = \begin{pmatrix}
1 & 4 & -2 \\
1 & 1 & -8 \\
-2 & -2 & -2
\end{pmatrix}$
\[
ch_A(x), \ ch_B(x) = ?, \quad A \sim B \ \text{?}
\]
\[
ch_A(x) = \det(A - xI) = \det \begin{pmatrix}
1 - x & 4 & -2 \\
4 & 1 - x & -2 \\
-2 & -2 & -2 - x
\end{pmatrix}
\]
\[
ch_A(x) = (1 - x) \det \begin{pmatrix}
1 - x & -2 \\
-2 & -2 - x
\end{pmatrix}
- 4 \det \begin{pmatrix}
4 & -2 \\
-2 & -2 - x
\end{pmatrix}
+ (-2) \det \begin{pmatrix}
4 & 1 - x \\
-2 & -2
\end{pmatrix}
\]
Prvý minor:
\[
\det \begin{pmatrix}
1 - x & -2 \\
-2 & -2 - x
\end{pmatrix} = (1 - x)(-2 - x) - (-2)(-2) = (1 - x)(-2 - x) - 4
= x^2 + x - 2 - 4 = x^2 + x - 6
\]
\[
a_{11} \cdot M_{11} = (1 - x)(x^2 + x - 6)
\]
Druhý minor:
\[
\det \begin{pmatrix}
4 & -2 \\
-2 & -2 - x
\end{pmatrix} = 4(-2 - x) - (-2)(-2) = -8 - 4x -4 = -4(x +3);
\]
\[
a_{12} \cdot M_{12} = 16(x + 3)
\]
Tretí minor:
\[
\det \begin{pmatrix}
4 & 1 - x \\
-2 & -2
\end{pmatrix} = 4(-2) - (1 - x)(-2) = -8 - (-2 + 2x) = -8 + 2 - 2x = -6 - 2x = -2(x + 3)
\]
\[
a_{13} \cdot M_{13} = 4(x + 3)
\]
\[
\text{Ch}_A(x) = (1 - x)(x^2 + x - 6) + 16(x + 3) + 4(x + 3)
\]
\[
= (1 - x)(x^2 + x - 6) + 20(x + 3) = -x^3 +7x - 6 + 20x + 60
= -x^3 + 27x + 54 = x^3 - 27x - 54
\]
\[
\text{Ch}_B(x) = \det(B - xI) = \det \begin{pmatrix}
1 - x & 4 & -2 \\
1 & 1 - x & -8 \\
-2 & -2 & -2 - x
\end{pmatrix}
\]
\[
\text{Ch}_B(x) = (1 - x) \det \begin{pmatrix}
1 - x & -8 \\
-2 & -2 - x
\end{pmatrix} - 4 \det \begin{pmatrix}
1 & -8 \\
-2 & -2 - x
\end{pmatrix} + (-2) \det \begin{pmatrix}
1 & 1 - x \\
-2 & -2
\end{pmatrix}
\]
Prvý minor:
\[
\det \begin{pmatrix}
1 - x & -8 \\
-2 & -2 - x
\end{pmatrix} = (1 - x)(-2 - x) - (-8)(-2) = x^2 + x - 2 - 16 = x^2 + x - 18
\]
\[
a_{11} \cdot M_{11} = (1 - x)(x^2 + x - 18)
\]
Druhý minor:
\[
\det \begin{pmatrix}
1 & -8 \\
-2 & -2 - x
\end{pmatrix} = 1(-2 - x) - (-8)(-2) = -2 - x - 16 = -x - 18
\]
\[
a_{12} \cdot M_{12} = 4(x + 18)
\]
Tretí minor:
\[
\det \begin{pmatrix}
1 & 1 - x \\
-2 & -2
\end{pmatrix} = 1(-2) - (1 - x)(-2) = -2 + 2(1 - x) = -2 + 2 - 2x = -2x
\]
\[
a_{13} \cdot M_{13} = 4x
\]
\[
\text{Ch}_B(x) = (1 - x)(x^2 + x - 18) + 4(x + 18) + 4x
\]
\[
= -x^3 + 19x - 18 + 4x + 72 + 4x = -x^3 + 27x + 54 = x^3 - 27x - 54
\]
Matice majú rovnaké charakteristické polynómy.
Ešte obe majú rovnakú stopu: $\text{Tr}(A) = 1 + 1 + (-2) = 0, \text{Tr}(B) = 0$; aj majú rovnaké determinanty:
\[
\det(A) = 1 \cdot \begin{vmatrix}
1 & -2 \\
-2 & -2
\end{vmatrix}
- 4 \cdot \begin{vmatrix}
4 & -2 \\
-2 & -2
\end{vmatrix}
+ (-2) \cdot \begin{vmatrix}
4 & 1 \\
-2 & -2
\end{vmatrix}
\]
\[
= 1(-6) - 4(-12) + (-2)(-6) = -6 + 48 + 12 = 54
\]
\[
\det(B) = 1 \cdot \begin{vmatrix}
1 & -8 \\
-2 & -2
\end{vmatrix}
- 4 \cdot \begin{vmatrix}
1 & -8 \\
-2 & -2
\end{vmatrix}
+ (-2) \cdot \begin{vmatrix}
1 & 1 \\
-2 & -2
\end{vmatrix}
\]
\[
= 1(-18) - 4(-18) + (-2)(0) = -18 + 72 = 54
\]
\[
x^3 - 27x - 54 = 0 \quad \Rightarrow \quad (x - 6)(x^2 + 6x + 9) = (x - 6)(x + 3)^2
\]
$ \lambda_1 = 6 $ (má algebraickú násobnosť - 1)
$ \lambda_2 = -3 $ (má algebraickú násobnosť - 2)
Najdeme geometrickú násobnosť pre každé vl. číslo:
$ \text{Pre } \lambda_{1} = 6$:
\[
B - 6I = \begin{pmatrix}
-5 & 4 & -2 \\
1 & -5 & -8 \\
-2 & -2 & -8
\end{pmatrix}
\text{Redukujem maticu do stupň. tvaru:}
\]
\[
B' =
\begin{pmatrix}
-5 & 4 & -2 \\
1 & -5 & -8 \\
-2 & -2 & -8
\end{pmatrix}
\begin{matrix}
<-\\
<-\\
\phantom{3}
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
-5 & 4 & -2 \\
-2 & -2 & -8
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
+5I\\
\phantom{3}
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
0 & -21 & -42 \\
-2 & -2 & -8
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
\phantom{2}\\
+2I
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
0 & -21 & -42 \\
0 & -12 & -24
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
⊙\frac{1}{21}\\
⊙ ( - \frac{1}{12} )
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
0 & 1 & 2 \\
0 & 1 & 2
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
\phantom{2}\\
- II
\end{matrix}
\sim
\begin{pmatrix}
1 & -5 & -8 \\
0 & 1 & 2 \\
0 & 0 & 0
\end{pmatrix}
h(B') = 2; geom. násobnosť: 3 - 2 = 1
\]
$ \text{Pre } \lambda_{2} = -3$:
\[
B'' =
B + 3I = \begin{pmatrix}
4 & 4 & -2 \\
1 & 4 & -8 \\
-2 & -2 & 1
\end{pmatrix}
\text{Redukujem maticu do stupň. tvaru:}
\]
\[
B'' =
\begin{pmatrix}
4 & 4 & -2 \\
1 & 4 & -8 \\
-2 & -2 & 1
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
- \frac{1}{4}\\
\phantom{3}
\end{matrix}
\sim
\begin{pmatrix}
4 & 4 & -2 \\
0 & 3 & - \frac{15}{2} \\
-2 & -2 & 1
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
\phantom{2}\\
+ \frac{1}{2} I
\end{matrix}
\sim
\begin{pmatrix}
4 & 4 & -2 \\
0 & 3 & - \frac{15}{2} \\
0 & 0 & 0
\end{pmatrix}
\begin{matrix}
\phantom{1}\\
⊙ \frac{1}{3}\\
\phantom{3}
\end{matrix}
\sim
\begin{pmatrix}
4 & 4 & -2 \\
0 & 1 & - \frac{15}{6} \\
0 & 0 & 0
\end{pmatrix}
h(B'') = 2; geom. násobnosť: 3 - 2 = 1
\]
\[
\text{Pre } \lambda_{2} = -3 \Rightarrow \text{geom. násobnosť je menšia ako algebraická} \Rightarrow B \ \text{nie je diagonalizovateľná, pričom matica A je symetrická:}
\]
\[
(a_{12} = a_{21} = 4, \quad a_{13} = a_{31} = -2, \quad a_{23} = a_{32} = -2)
\]
\[
\Rightarrow \text{je diagonalizovateľná} \Rightarrow \text{matice A, B nie sú podobné.}
\]