El pasado día vimos que para calcular los valores propios o autovalores necesitamos el polinomio característico.
Ejercicio: Determinar los autovalores de la matriz \[\begin{bmatrix}1 & 1 & 0\\
Solución:
(%i3) A : matrix ([1 ,1 ,0 ],[2 ,0 ,2 ],[– 3 ,– 2 ,– 1 ])$ ecu_caract : expand (determinant (A – x * ident (3 )))= 0 ; solve (ecu_caract ,[x ]);
(ecu_caract)−x3−x=0(%o3) [x=−%i,x=%i,x=0]
Ejercicio: Determinar los autovalores de la matriz \[\begin{bmatrix}4 & 3 & 0\\
Solución:
(%i3) A : matrix ([4 ,3 ,0 ],[– 6 ,– 5 ,0 ],[3 ,3 ,1 ])$ ecu_caract : expand (determinant (A – x * ident (3 )))= 0 ; solve (ecu_caract ,[x ]);
(ecu_caract)−x3+3x−2=0(%o3) [x=−2,x=1]
Recodad que definíamos los autovectores, o vectores propios, como
Recordemos que dada una matriz, \(\mathbf{A}\in\mathcal{M}_n(\mathbb{K})\), decimos que \(\vec{v} \in\mathbb{K}^n\) es un autovector, o vector propio de la matriz, sí \[\mathbf{A}\vec{v}=\lambda\vec{v},\] siendo \(\lambda\in\mathbb{K}\). Al escalar \(\lambda\) se le denomina autovalor o valor propio de la matriz.
Si determinamos los valores propios, o autovalores, saber los vectores propios, o autovalores, será equivalente a resolver el sistema \[(\mathbf{A}-\lambda\, I)\vec{x}=\vec{0},\] para cada valor propio \(\lambda\) raíz de la ecuación característica(el resultado de igualar el polinomio característico a cero). Así, para cada valor propio habrá un vector propio, al menos.
Cada valor propio tiene asociado un conjunto \(C_\lambda=\{\vec{v}\in\mathbb{K}^n\}\), que se determina resolviendo el sistema homogeneo \((\mathbf{A}-\lambda\, I)\vec{x}=\vec{0}\), que se denomina subespacio propio de \(\lambda\). Las bases de estos subespacios son los vectores propios de la matriz.
Ejercicio: Determinar los subespacios propios de la matriz \[\begin{bmatrix}2&1\\1&2\end{bmatrix}\]
Solución:
(%i6) A : matrix ( [ 2 , 1 ] , [ 1 , 2 ] ) $ determinant ( A − x · ident ( 2 ) ) ; s : solve ( % , x ) ; aut : [ ] $ for i : 1 thru length ( s ) do ( aut : append ( aut , [ ev ( x , s [ i ] ) ] ) ) $ aut ;
\[{{\left( 2-x\right) }^{2}}-1\]
\[\left[ x=3,x=1\right] \]
\[\left[ 3,1\right] \]
(%i10) X : matrix ( [ x ] , [ y ] ) $ ( A − aut [ 1 ] · ident ( 2 ) ) . X ; s : linsolve ( [ % [ 1 , 1 ] , % [ 2 , 1 ] ] , [ x , y ] ) ; v : [ ev ( [ x , y ] , [ ev ( s , %rnum_list [ 1 ] = 1 ) ] ) ] ;
\[\begin{bmatrix}y-x\\x-y\end{bmatrix}\]\[solve: dependent equations eliminated: (2)\]
\[\left[ x={\mathrm{\% r1}},y={\mathrm{\% r1}}\right] \]
\[\left[ \left[ 1,1\right] \right] \]
(%i13) ( A − aut [ 2 ] · ident ( 2 ) ) . X ; s : linsolve ( [ % [ 1 , 1 ] , % [ 2 , 1 ] ] , [ x , y ] ) ; v : append ( v , [ ev ( [ x , y ] , [ ev ( s , %rnum_list [ 1 ] = 1 ) ] ) ] ) ;
\[\begin{bmatrix}y+x\\y+x\end{bmatrix}\]\[solve: dependent equations eliminated: (2)\]
\[\left[ x=-{\mathrm{\% r2}},y={\mathrm{\% r2}}\right] \]
\[\left[ \left[ 1,1\right] ,\left[ -1,1\right] \right] \]
Ejercicio: Determinar los subespacios propios de la matriz \[\begin{bmatrix}3&2&{-1}\\2&3&1\\0&0&5\end{bmatrix}\]
Solución:
(%i6) A : matrix ( [ 3 , 2 , − 1 ] , [ 2 , 3 , 1 ] , [ 0 , 0 , 5 ] ) $ rat ( determinant ( A − x · ident ( 3 ) ) ) = 0 ; s : solve ( % , x ) ; aut : [ ] $ for i : 1 thru length ( s ) do ( aut : append ( aut , [ ev ( x , s [ i ] ) ] ) ) $ aut ;
\[-{{x}^{3}}+11 {{x}^{2}}-35 x+25=0\]
\[\left[ x=1,x=5\right] \]
\[\left[ 1,5\right] \]
(%i10) X : matrix ( [ x ] , [ y ] , [ z ] ) $ ( A − aut [ 1 ] · ident ( 3 ) ) . X ; s : linsolve ( [ % [ 1 , 1 ] , % [ 2 , 1 ] , % [ 3 , 1 ] ] , [ x , y , z ] ) ; v : [ ev ( [ x , y , z ] , [ ev ( s , %rnum_list [ 1 ] = 1 ) ] ) ] ;
\[\begin{bmatrix}-z+2 y+2 x\\z+2 y+2 x\\4 z\end{bmatrix}\]\[solve: dependent equations eliminated: (1)\]
\[\left[ x={\mathrm{\% r1}},y=-{\mathrm{\% r1}},z=0\right] \]
\[\left[ \left[ 1,-1,0\right] \right] \]
(%i13) X : matrix ( [ x ] , [ y ] , [ z ] ) $ ( A − aut [ 2 ] · ident ( 3 ) ) . X ; s : linsolve ( [ % [ 1 , 1 ] , % [ 2 , 1 ] , % [ 3 , 1 ] ] , [ x , y , z ] ) ;
\[\begin{bmatrix}-z+2 y-2 x\\z-2 y+2 x\\0\end{bmatrix}\]\[solve: dependent equations eliminated: (3 2)\]
\[\left[ x=-\frac{{\mathrm{\% r2}}-2 {\mathrm{\% r3}}}{2},y={\mathrm{\% r3}},z={\mathrm{\% r2}}\right] \]
(%i15) v : append ( v , [ ev ( [ x , y , z ] , [ ev ( s , [ %rnum_list [ 1 ] = 1 , %rnum_list [ 2 ] = 0 ] ) ] ) ] ) $ v : append ( v , [ ev ( [ x , y ] , z , [ ev ( s , [ %rnum_list [ 1 ] = 0 , %rnum_list [ 2 ] = 1 ] ) ] ) ] ) ;
\[\left[ \left[ 1,-1,0\right] ,\left[ -\frac{1}{2},0,1\right] ,\left[ 1,1\right] \right] \]
Ejercicio: Cuántos vectores propio tiene la matriz \[\begin{bmatrix}4 & -1 & 6 & 0\\
Solución:
(%i6) A : matrix ( [ 4 , − 1 , 6 , 0 ] , [ 2 , 1 , 6 , 0 ] , [ 2 , − 1 , 8 , 0 ] , [ 0 , 2 , 0 , 1 ] ) $ rat ( determinant ( A − x · ident ( 4 ) ) ) = 0 ; s : solve ( % , x ) ; aut : [ ] $ for i : 1 thru length ( s ) do ( aut : append ( aut , [ ev ( x , s [ i ] ) ] ) ) $ aut ;
\[{{x}^{4}}-14 {{x}^{3}}+53 {{x}^{2}}-76 x+36=0\]
\[\left[ x=9,x=1,x=2\right] \]
\[\left[ 9,1,2\right] \]
(%i10) X : matrix ( [ x ] , [ y ] , [ z ] , [ t ] ) $ ( A − aut [ 1 ] · ident ( 4 ) ) . X ; s : linsolve ( [ % [ 1 , 1 ] , % [ 2 , 1 ] , % [ 3 , 1 ] , % [ 4 , 1 ] ] , [ x , y , z , t ] ) ; v : [ ev ( [ x , y , z , t ] , [ ev ( s , %rnum_list [ 1 ] = 1 ) ] ) ] ;
\[\begin{bmatrix}6 z-y-5 x\\6 z-8 y+2 x\\-z-y+2 x\\2 y-8 t\end{bmatrix}\]\[solve: dependent equations eliminated: (1)\]
\[\left[ x={\mathrm{\% r1}},y={\mathrm{\% r1}},z={\mathrm{\% r1}},t=\frac{{\mathrm{\% r1}}}{4}\right] \]
\[\left[ \left[ 1,1,1,\frac{1}{4}\right] \right] \]
(%i12) ( A − aut [ 2 ] · ident ( 4 ) ) . X ; s : linsolve ( [ % [ 1 , 1 ] , % [ 2 , 1 ] , % [ 3 , 1 ] , % [ 4 , 1 ] ] , [ x , y , z , t ] ) ;
\[\begin{bmatrix}6 z-y+3 x\\6 z+2 x\\7 z-y+2 x\\2 y\end{bmatrix}\]\[solve: dependent equations eliminated: (1)\]
\[\left[ x=0,y=0,z=0,t={\mathrm{\% r2}}\right] \]
(%i13) v : append ( v , [ ev ( [ x , y , z , t ] , [ ev ( s , %rnum_list [ 1 ] = 1 ) ] ) ] ) ;
\[\left[ \left[ 1,1,1,\frac{1}{4}\right] ,\left[ 0,0,0,1\right] \right] \]
(%i15) ( A − aut [ 3 ] · ident ( 4 ) ) . X ; s : linsolve ( [ % [ 1 , 1 ] , % [ 2 , 1 ] , % [ 3 , 1 ] , % [ 4 , 1 ] ] , [ x , y , z , t ] ) ;
\[\begin{bmatrix}6 z-y+2 x\\6 z-y+2 x\\6 z-y+2 x\\2 y-t\end{bmatrix}\]\[solve: dependent equations eliminated: (3 1)\]
\[\left[ x={\mathrm{\% r3}},y=6 {\mathrm{\% r4}}+2 {\mathrm{\% r3}},z={\mathrm{\% r4}},t=12 {\mathrm{\% r4}}+4 {\mathrm{\% r3}}\right] \]
(%i17) v : append ( v , [ ev ( [ x , y , z , t ] , [ ev ( s , [ %rnum_list [ 1 ] = 1 , %rnum_list [ 2 ] = 0 ] ) ] ) ] ) $ v : append ( v , [ ev ( [ x , y , z , t ] , [ ev ( s , [ %rnum_list [ 1 ] = 0 , %rnum_list [ 2 ] = 1 ] ) ] ) ] ) ;
\[\left[ \left[ 1,1,1,\frac{1}{4}\right] ,\left[ 0,0,0,1\right] ,\left[ 1,2,0,4\right] ,\left[ 0,6,1,12\right] \right] \]
Ejercicio: Cuál es el producto escalar de [1,-2,3,-4] por el autovector unitario del mayor valor propio de la matriz \[\begin{bmatrix}4 & 2 & 0 & 0 \\
Solución:
(%i6) A : matrix ( [ 4 , 2 , 0 , 0 ] , [ 3 , 3 , 0 , 0 ] , [ 0 , 0 , 2 , 5 ] , [ 0 , 0 , 0 , 2 ] ) $ rat ( determinant ( A − x · ident ( 4 ) ) ) = 0 ; s : solve ( % , x ) ; aut : [ ] $ for i : 1 thru length ( s ) do ( aut : append ( aut , [ ev ( x , s [ i ] ) ] ) ) $ aut ;
\[{{x}^{4}}-11 {{x}^{3}}+38 {{x}^{2}}-52 x+24=0\]
\[\left[ x=6,x=1,x=2\right] \]
\[\left[ 6,1,2\right] \]
(%i10) X : matrix ( [ x ] , [ y ] , [ z ] , [ t ] ) $ ( A − 6 · ident ( 4 ) ) . X ; s : linsolve ( [ % [ 1 , 1 ] , % [ 2 , 1 ] , % [ 3 , 1 ] , % [ 4 , 1 ] ] , [ x , y , z , t ] ) ; v : ev ( [ x , y , z , t ] , [ ev ( s , %rnum_list [ 1 ] = 1 ) ] ) ;
\[\begin{bmatrix}2 y-2 x\\3 x-3 y\\5 t-4 z\\-4 t\end{bmatrix}\]\[solve: dependent equations eliminated: (1)\]
\[\left[ x={\mathrm{\% r1}},y={\mathrm{\% r1}},z=0,t=0\right] \]
\[\left[ 1,1,0,0\right] \]
(%i12) [ 1 , − 2 , 3 , − 4 ] . v / sqrt ( v . v ) ; % , numer ;
\[-\frac{1}{\sqrt{2}}\]
\[-0.7071067811865475\]
Ejercicio: Cuál es el producto escalar de [1,-2,3,-4] por el autovector unitario del mayor valor propio de la matriz \[\begin{bmatrix}4 & 1 & 0 & 1\\
Solución:
(%i6) A : matrix ( [ 4 , 1 , 0 , 1 ] , [ 2 , 3 , 0 , 1 ] , [ − 2 , 1 , 2 , − 3 ] , [ 2 , − 1 , 0 , 5 ] ) $ rat ( determinant ( A − x · ident ( 4 ) ) ) = 0 ; s : solve ( % , x ) ; aut : [ ] $ for i : 1 thru length ( s ) do ( aut : append ( aut , [ ev ( x , s [ i ] ) ] ) ) $ aut ;
\[{{x}^{4}}-14 {{x}^{3}}+68 {{x}^{2}}-136 x+96=0\]
\[\left[ x=6,x=4,x=2\right] \]
\[\left[ 6,4,2\right] \]
(%i10) X : matrix ( [ x ] , [ y ] , [ z ] , [ t ] ) $ ( A − 6 · ident ( 4 ) ) . X ; s : linsolve ( [ % [ 1 , 1 ] , % [ 2 , 1 ] , % [ 3 , 1 ] , % [ 4 , 1 ] ] , [ x , y , z , t ] ) ; v : ev ( [ x , y , z , t ] , [ ev ( s , %rnum_list [ 1 ] = 1 ) ] ) ;
\[\begin{bmatrix}y-2 x+t\\-3 y+2 x+t\\-4 z+y-2 x-3 t\\-y+2 x-t\end{bmatrix}\]\[solve: dependent equations eliminated: (4)\]
\[\left[ x={\mathrm{\% r1}},y={\mathrm{\% r1}},z=-{\mathrm{\% r1}},t={\mathrm{\% r1}}\right] \]
\[\left[ 1,1,-1,1\right] \]
(%i11) [ 1 , − 2 , 3 , − 4 ] . v / sqrt ( v . v ) ;
\[-4\]
Ejercicio: Cuántos suman los coeficientes del polinomio característico de la matriz \[\left[\begin{smallmatrix}0 & -1 & -1 & 0\\ -2 & 1 & -1 & 0 \\ -2 & 2 & 2 & 0\\ 0 & 0 & 0 & -1\end{smallmatrix}\right].\]
Solución:
C.)
(%i3) A : matrix ([0 ,– 1 ,– 1 ,0 ],[– 2 ,1 ,– 1 ,0 ],[– 2 ,2 ,2 ,0 ],[0 ,0 ,0 ,– 1 ])$ define (f (x ),expand (determinant (A – x * ident (4 )))); f (1 );
(%o2) f ( x ) := x 4 − 2 x 3 − 3 x 2 + 4 x + 4 (%o3) 4
Para el primer autovalor