### Invariant subspaces of linear maps: Invariant subspaces

### Generalized eigenspace

We have seen that a linear map #L# of a finite-dimensional real or complex vector space #V# to itself, is not always diagonizable, even if #V# is complex. The problem is that the dimension of the eigenspace of #L# with respect to a root of the characteristic polynomial can be smaller than the multiplicity of that root in #p_L(x)#. Consider for example #V = \mathbb{R}^2# and #L=L_A# for #A=\matrix{0&1\\ 0&0}# with characteristic polynomial #p_L(x) = x^2#. The root #0# has multiplicity #2# and the dimension of the eigenspace of #L# at eigenvalue #0# is #1#.

We will treat this second problem in two steps. First we point out an invariant subspace of #V# that can be bigger than the eigenspace at a given root #\lambda# of the characteristic polynomial, but on which the restriction of #L# to that subspace has a characteristic polynomial that is a power of #x-\lambda#. *Later* we will indicate a basis for that subspace where the restriction of #L# approximates a diagonal form.

Generalized eigenspaceLet #V# be a vector space, #L:V\to V# a linear map, and #\lambda# an eigenvalue of #L#. The **generalized eigenspace** of #L# with respect to #\lambda# is the subspace \(E_\lambda^*\) consisting of all vectors #\vec{v}# of #V# for which there exists a natural number #k# that satisfies #(L-\lambda\, I_V)^k(\vec{v}) = \vec{0}#. This subspace is invariant under #L#.

In the finite-dimensional case, the dimension of the generalized eigenspace of #\lambda# is equal to the multiplicity of #\lambda# in the characteristic polynomial of #L#:

Assume that #V# is a finite-dimensional vector space and that #L:V\to V# is a linear map such that #\lambda# is a root of the characteristic polynomial #p_L(x)#.

- If #\lambda# has multiplicity #\ell# in the minimal polynomial #m_L(x)#, then \(E_\lambda^* =\ker{(L-\lambda\, I_V)^{\ell}}\).
- If #\lambda# has multiplicity #k# in #p_L(x)#, then \(E_\lambda^* =\ker{(L-\lambda\, I_V)^k}\) and we have \(\dim{E_\lambda^*}= k\).
- Write #e_i=\dim{\ker{L-\lambda\,I_V)^i}}# so that #e_\ell=k#. The sequence #e_1,e_2,e_3,\ldots,e_{\ell}# is strictly increasing.

Determine a basis for the generalized eigenspace of #A# corresponding to the eigenvalue #2#.

Since the multiplicity of # 2 # in the characteristic polynomial is equal to #3#, the generalized eigenspace #E_{2}^*# coincides with #\ker{(A-2 \,I_4)^{3}}#.

The obvious method of determining #\ker{(A-2 \,I_4)^{3}}# is as follows: By squaring \[(A-2 \,I_4)^2 =\matrix{2 & -2 & -2 & 0 \\ -2 & 2 & 3 & 1 \\ 3 & -3 & -4 & -1 \\ -2 & 2 & 3 & 1 \\ }\] and subsequently multiplying this matrix by \(A-2 \,I_4\) we find

\[ (A-2 \,I_4)^{3} = \matrix{2 & -2 & -2 & 0 \\ -1 & 1 & 1 & 0 \\ 2 & -2 & -2 & 0 \\ -1 & 1 & 1 & 0 \\ }\] Next, we compute the kernel of this linear map by solving the system of equations \((A-2 \,I_4)^{3} (\vec{x}) = \vec{0}\). This leads to the following basis for \(\ker{(A-2 \,I_4)^{3}}\): \[\basis{ \matrix{0 \\ -2 \\ 2 \\ 0 \\ } , \matrix{0 \\ 0 \\ 0 \\ -2 \\ } , \matrix{2 \\ 2 \\ 0 \\ 0 \\ } } \] Alternatively, we may use theorem

*Invariant direct sum*according to which the requested basis is also a basis of #\im{A-3 \,I_4 }#. This subspace is spanned by the columns of the matrix \[A-3 \,I_4 = \matrix{2 & -3 & -4 & -1 \\ -1 & -1 & 1 & 1 \\ 3 & -2 & -5 & -2 \\ -2 & 1 & 3 & 1 \\ }\] By thinning we find that the following columns are a basis for \(\ker{(A-2 \,I_4)^{3}}\): \[\basis{\cv{ 2 \\ -1 \\ 3 \\ -2 } , \cv{ -3 \\ -1 \\ -2 \\ 1 } , \cv{ -4 \\ 1 \\ -5 \\ 3 } } \]

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