Eigenspace and Eigenspectrum Values in a Matrix


For a given matrix A the set of all eigenvectors of A associated with an eigenvalue \lambda spans a subspace, which is called the Eigenspace of A with respect to \lambda and is denoted by E_\lambda. The set of all eigenvalues of A is called Eigenspectrum, or just spectrum, of A.
If \lambda is an eigenvalue of A, then the corresponding eigenspace E_\lambda is the solution space of the homogeneous system of linear equations (A-\lambda I)x = 0. Geometrically, the eigenvector corresponding to a non – zero eigenvalue points in a direction that is stretched by the linear mapping. The eigenvalue is the factor by which it is stretched. If the eigenvalue is negative, then the direction of the stretching is flipped.

Below are some useful properties of eigenvalues and eigenvectors in addition to the properties which are already listed in the article Mathematics | Eigen Values and Eigen Vectors.

  • A matrix A and its transpose A^{T} possess the same eigenvalues but not necessarily the same eigenvectors.
  • The eigenspace E_\lambda is the null space of A - \lambda I since
    Ax = \lambda x \Longleftrightarrow Ax - \lambda x = 0 \Longleftrightarrow (A - \lambda I)x = 0 \Longleftrightarrow x\in ker(A - \lambda I)
  • Note: ker stands for Kernel which is another name for null space.

    Computing Eigenvalues, Eigenvectors, and Eigenspaces:

    Consider given 2 X 2 matrix:
     A =  \begin{bmatrix}  4 & 2 \\ 1 & 3  \end{bmatrix} 
    Step 1: Characteristic polynomial and Eigenvalues.
    The characteristic polynomial is given by 
    det(A - \lambda I) 
     = det(\begin{bmatrix}  4 & 2 \\ 1 & 3  \end{bmatrix} - \begin{bmatrix}  \lambda & 0 \\ 0 & \lambda  \end{bmatrix}) = \begin{vmatrix}  4-\lambda & 2 \\ 1 & 3-\lambda  \end{vmatrix} 
    = (4-\lambda)(3-\lambda) - 2.1
    After we factorize the characteristic polynomial, we will get
    which gives eigenvalues as \lambda_1 = 2 and \lambda_2 = 5
    Step 2: Eigenvectors and Eigenspaces
    We find the eigenvectors that correspond to these eigenvalues by looking 
    at vectors x such that 
      \begin{bmatrix}  4-\lambda & 2 \\ 1 & 3-\lambda  \end{bmatrix}  % x =  0  
    For \lambda = 5 we obtain
      \begin{bmatrix}  4-5 & 2 \\ 1 & 3-5  \end{bmatrix} % \begin{bmatrix}  x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix}  -1 & 2 \\ 1 & -2  \end{bmatrix} % \begin{bmatrix}  x_1 \\ x_2 \end{bmatrix} = 0 
    After solving the above homogeneous system of equations,
    we will obtain a solution space
    E_5 = span(\begin{bmatrix}  2 \\ 1 \end{bmatrix})
    This eigenspace is one dimensional as it possesses a single basis vector.
    Similarly, we find eigenvector for \lambda = 2 by solving
    the homogeneous system of equations
      \begin{bmatrix}  4-2 & 2 \\ 1 & 3-2  \end{bmatrix} % \begin{bmatrix}  x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix}  2 & 2 \\ 1 & 1  \end{bmatrix} % \begin{bmatrix}  x_1 \\ x_2 \end{bmatrix} = 0 
    This means any vector x = \begin{bmatrix}  x_1 \\ x_2 \end{bmatrix} , where x_2=-x_1  
    such as \begin{bmatrix}  1 \\ -1 \end{bmatrix}  is an eigenvector with 
    eigenvalue 2. This means eigenspace is given as 
    E_2 = span(\begin{bmatrix}  1 \\ -1 \end{bmatrix})

    The two eigenspaces E_5 and E_2 in the above example are one dimensional as they are each spanned by a single vector. However, in other cases, we may have multiple identical eigenvectors and the eigenspaces may have more than one dimension.

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