To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)
Top

Eigenvector Calculator
Top
For a given $\lambda$, only solution is X = 0, except if det(A - $\lambda$ I) = 0. In that case, a nonzero X that satisfies (A - $\lambda$ I)X = 0 is guaranteed to exist, and that X is an eigen  vector.
Eigen vector Calculator is a online tool to calculate the eigen vector for any given matrix. You just have to enter the matrix either (2X2) or (3X3) in the space given and get the eigen vector instantly.
Default 3 x 3 matrix is given in the (3 x 3) eigenvector calculator below. By satisfying the characteristic equation of the matrix, you will get the eigen values. By substituting the eigen values in the characteristic equation, you will get the eigen vectors on clicking "Calculate Eigenvectors". Eigenvector for (2 x 2) matrix is also calculated as per the (3 x 3) matrix.
 

Steps for Eigenvector Calculator

Back to Top
Step 1 :  

Lets see how to solve Eigen vector:
Step 1 : Read the problem and note down the given matrix.
Step 2 : The characteristic equation is of the form |A - $\lambda$ I | = 0. Write it in that form to get the eigen values $\lambda_1$, $\lambda_2$....... $\lambda_n$.
Step 3 :
To get the eigen vectors substitute the eigen values in the characteristic equation and use cramers rule and get the answer.



Problems on Eigenvector Calculator

Back to Top
  1. Find the eigen vectors for the matrix:
    $\begin{vmatrix} -2 & -8 & -12\\ 1& 4 & 4\\ 0& 0& 1 \end{vmatrix}$


    Step 1 :  

    The given matrix is
    A = $\begin{vmatrix} -2 & -8 & -12\\ 1& 4 & 4\\ 0& 0& 1 \end{vmatrix}$



    Step 2 :  

    The characteristic equation is |A - $\lambda$ I | = 0
    i.e., $\begin{vmatrix} -2-\lambda & -8 & -12\\ 1& 4-\lambda & 4\\ 0& 0& 1-\lambda \end{vmatrix}$ = 0
    Solving the above determinant we get
    $\lambda^3$ - 3 $\lambda^2$ + 2 $\lambda$ = 0
    $\lambda$ ($\lambda^2$ -3 $\lambda$ + 2) = 0
    Hence $\lambda_1$ = 0, $\lambda_2$ = 1 and $\lambda_3$ = 2 are eigen values



    Step 3 :  

    To find eigen vectors take |A - $\lambda_1$ I | X1 = 0 Where $\lambda_1$ = 0
    $\begin{vmatrix} -2-0 & -8 & -12\\ 1& 4-0 & 4\\ 0& 0& 1-0 \end{vmatrix}$ $\begin{vmatrix}x_1\\ x_2\\ x_3\end{vmatrix}$ = 0
    Using cramer's rule we get
    X1 = $\begin{vmatrix}4\\ -1\\ 0\end{vmatrix}$
    To find eigen vectors take |A - $\lambda_2$ I | X1 = 0 where $\lambda_2$ = 1
    $\begin{vmatrix} -2-1 & -8 & -12\\ 1& 4-1 & 4\\ 0& 0& 1-1 \end{vmatrix}$ $\begin{vmatrix}x_1\\ x_2\\ x_3\end{vmatrix}$ = 0
    X2 = $\begin{vmatrix}4\\ 0\\ -1\end{vmatrix}$
    To find eigen vectors take |A - $\lambda_3$ I | X3 = 0 where $\lambda_3$ = 2
    $\begin{vmatrix} -2-2 & -8 & -12\\ 1& 4-2 & 4\\ 0& 0& 1-1 \end{vmatrix}$ $\begin{vmatrix}x_1\\ x_2\\ x_3\end{vmatrix}$ = 0
    Using cramer's rule we get
    X3 = $\begin{vmatrix}-2\\ 1\\ 0\end{vmatrix}$.



    Answer  :  

    Hence Eigen values are $\begin{vmatrix}X_1\\ X_2\\ X_3\end{vmatrix}$ = $\begin{pmatrix}
     4& -1 & 0\\
     -2& 1 & 0\\
     4&  0& -1
    \end{pmatrix}$



  2. Find the Eigen vectors for the matrix:
    A = $\begin{vmatrix} 8 & 3\\ 2 & 7 \end{vmatrix}$


    Step 1 :  

    The given matrix is
    A = $\begin{vmatrix} 8 & 3\\ 2 & 7 \end{vmatrix}$



    Step 2 :  

    The characteristic equation is |A - $\lambda$ I | = 0
    i.e., $\begin{vmatrix} 8-\lambda & 3\\ 2& 7- \lambda\end{vmatrix}$ = 0
    Solving the above determinant we get
    (8- $\lambda$)(7-$\lambda$) - 6 = 0
    Solving it we get $\lambda_1$ = 5 and $\lambda_2$ = 10 are eigen values



    Step 3 :  

    To find eigen vectors take |A - $\lambda_1$ I | X1 = 0 Where $\lambda_1$ = 5
    $\begin{vmatrix} 8 & 3\\ 2&7 \end{vmatrix}$ $\begin{vmatrix}x_1\\ x_2\end{vmatrix}$ = 0
    Using cramer's rule we get
    X1 = $\begin{vmatrix}1\\ -1\end{vmatrix}$
    To find eigen vectors take |A - $\lambda_2$ I | X2 = 0 where $\lambda_2$ = 10
    $\begin{vmatrix} -2 & 3\\ 2& -3 \end{vmatrix}$ $\begin{vmatrix}x_1\\ x_2\end{vmatrix}$ = 0
    X2 = $\begin{vmatrix} 3 \\2\end{vmatrix}$.



    Answer  :  

    The eigen vectors are

     

    X1 = $\begin{vmatrix}1\\ -1\end{vmatrix}$ and X2 = $\begin{vmatrix} 3 \\2\end{vmatrix}$.



*AP and SAT are registered trademarks of the College Board.