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Resultados del cálculo
Calculadora de eigenvectores con pasos
Calcula eigenvectores y eigenvalores fácilmente con explicaciones paso a paso.
1. Definition & Formula
An eigenvector of a square matrix $A$ is a nonzero vector $\mathbf{v}$
such that
$$A\mathbf{v}=\lambda\mathbf{v}$$
Here $\lambda$ is the eigenvalue associated with $\mathbf{v}$. To find
eigenvalues solve the characteristic equation:
$$\det(A-\lambda I)=0$$
For each eigenvalue $\lambda$, find eigenvectors by solving:
$$(A-\lambda I)\mathbf{v}=\mathbf{0}$$
The set of all eigenvectors (plus the zero vector) for a given $\lambda$ is
the eigenspace. Eigenvectors can be scaled by any nonzero scalar — they represent
directions.
Since there is only one independent eigenvector for a double root, $A$ is
defective (not diagonalizable).
3. Errores comunes
Usar el vector cero: $\mathbf{0}$ is not an eigenvector.
Detenerse tras hallar los autovalores: Finding $\lambda$ is only half the job
— you must solve $(A-\lambda I)\mathbf{v}=0$.
Errores aritméticos en los determinantes: 3×3 determinants and sign
errors are common.
Confundir las multiplicidades: Check geometric multiplicity (number of
independent eigenvectors) when eigenvalues repeat.
No simplificar los subespacios propios: Present eigenvectors in simplest
integer or rational form when possible.
4. Problemas de práctica
Intenta estos ejercicios. Haz clic para ver autovectores modelo.
Ejercicio 1
$$\begin{bmatrix}2 & 1\\[4pt]0 & 3\end{bmatrix}$$
Mostrar respuesta
Eigenvalues $2,3$. Eigenvectors: for $2$, $[1,0]^T$; for $3$,
$[1,0]^T$ (note: check rows). Actually compute: for $\lambda=2$, $(A-2I)=[0\ 1;0\
1]$ gives $y=0\Rightarrow v=[1,0]^T$. For $\lambda=3$, $(A-3I)=[-1\ 1;0\ 0]$ gives
$-x+y=0\Rightarrow v=[1,1]^T$.
Ejercicio 2
$$\begin{bmatrix}4 & 0\\[4pt]0 & 1\end{bmatrix}$$
Mostrar respuesta
Eigenvalues 4 and 1. Eigenvectors: $\lambda=4\Rightarrow [1,0]^T$;
$\lambda=1\Rightarrow [0,1]^T$.
Ejercicio 3
$$\begin{bmatrix}1 & 2\\[4pt]2 & 1\end{bmatrix}$$
Mostrar respuesta
Characteristic polynomial: $(1-\lambda)^2-4 =
\lambda^2-2\lambda-3=(\lambda-3)(\lambda+1)$. Eigenvalues 3 and -1. Eigenvectors:
for 3, $[1,1]^T$; for -1, $[1,-1]^T$.
Ejercicio 4
$$\begin{bmatrix}5 & 4\\[4pt]2 & 1\end{bmatrix}$$
Mostrar respuesta
Characteristic polynomial:
$(5-\lambda)(1-\lambda)-8=\lambda^2-6\lambda-3$. Eigenvalues
$3\pm\sqrt{12}=3\pm2\sqrt3$. For $\lambda=3+2\sqrt3$, solve to get an eigenvector;
for $\lambda=3-2\sqrt3$ similarly. Numerical eigenvector approximations are
acceptable on calculators.
Preguntas Frecuentes (FAQ)
¿Qué son los eigenvectores?
Son vectores que solo cambian de escala cuando se multiplican por una matriz.
¿La calculadora muestra los pasos?
Sí, muestra el proceso completo para calcular eigenvalores y eigenvectores.
