### Invariant subspaces of linear maps: Invariant subspaces

### Direct sum decomposition into invariant subspaces

We will now show that finding a more simple form of the matrix of a linear map #V\to V# relative to the basis for a finite-dimensional vector space #V# can be reduced to the case in which the minimal polynomial is either the power of a single linear polynomial (in the case #V# is real) or the power of a quadratic polynomial with negative discriminant.

We keep in mind that the notation #V = U\oplus W# means that #V# is the *direct sum* of the subspaces #U# and #W#; meaning: #V = U+W# and #U\cap W =\{\vec{0}\}#.

Invariant direct sumLet #V# be a vector space with finite dimension #n# and let #L:V\to V# be a linear map which is a zero of the polynomial #f(x)#. If #g(x)# and #h(x)# are polynomials with \[\gcd(g(x),h(x))=1\phantom{xx}\text{ and }\phantom{xx}f(x) = g(x)\cdot h(x)\] then we have

\[\begin{array}{rcl} \im{g(L)} &=& \ker{h(L)}\\ \im{h(L)} &=& \ker{g(L)}\\ V &=& \ker{g(L)}\oplus \ker{h(L)}\end{array}\]

The involved subspaces #\ker{h(L)}# and # \ker{g(L)}# are both invariant under #L#.

Hence, if we choose the bases #\alpha# for #\ker{g(L)}# and #\beta# for #\ker{h(L)}#, then the composition #\gamma# is a basis for #V# and the matrix of #L# relative to #\gamma# has the form \[L_\gamma = \matrix{L_\alpha&0\\ 0 & L_\beta}\]where #L_\alpha# is the matrix of #\left.L\right|_{\ker{g(L)}}# relative to #\alpha# and #L_\beta# the matrix of #\left.L\right|_{\ker{h(L)}}# relative to #\beta#.

In particular, the characteristic polynomial of #L# on #V# is the product of the characteristic polynomials of the restrictions of #L# to #\ker{g(L)}# and to #\ker{h(L)}#.

Application to a complete factorization of the characteristic polynomial of #L# gives:

If #V# is a finite-dimensional vector space and #L:V\to V# a linear map with characteristic polynomial equal to \[(x-\lambda_1)^{k_1} \cdots (x-\lambda_r)^{k_r}\]for certain differing numbers #\lambda_1,\ldots,\lambda_r# and natural numbers #k_1,\ldots,k_r#, then we have \[ V = \ker{(L-\lambda_1\,I_V)^{k_1}}\oplus \cdots \oplus \ker{(L-\lambda_r\,I_V)^{k_r}}\]

In particular, the restriction of #L# to \(\ker{(L-\lambda_i\,I_V)^{k_i}}\) has the characteristic polynomial \((x-\lambda_i)^{k_i}\) and we have \(\dim{\ker{(L-\lambda_i\,I_V)^{k_i}}}=k_i\) for #i=1,\ldots,r#.

Determine a #(3\times3)#-matrix #T# whose first two columns are a basis for #\ker{( A-2 I_3)^2}# and whose third column is a basis for #\ker{ A-3I_3}#.

According to the theorem

*Invariant direct sum*, the requested basis is also a basis of #\im{A-3I_3 }# supplemented with a basis of #\im{( A-2 I_3)^2 }#. These subspaces are spanned by the columns of the matrices \[ A-3I_3 = \matrix{-4 & 4 & 5 \\ 2 & -3 & -3 \\ -4 & 4 & 5 \\ }\phantom{xx}\text{ and }\phantom{xx}( A-2 I_3)^2= \matrix{-3 & 0 & 3 \\ 2 & 0 & -2 \\ -4 & 0 & 4 \\ } \] After thinning out, we find that the following columns form a basis:

\[\begin{array}{rcl}\basis{\cv{ -4 \\ 2 \\ -4 } , \cv{ 4 \\ -3 \\ 4 } } \phantom{xx}&\text{ for } &\ker{( A-2 I_3)^2 }\phantom{xxxx}\\ &\text{ and }&\\ \basis{\cv{ -3 \\ 2 \\ -4 } } \phantom{xx}&\text{ for } &\ker{A-3I_3 }\end{array}\] This leads to the answer

\[T = \matrix{-4 & 4 & -3 \\ 2 & -3 & 2 \\ -4 & 4 & -4 \\ }\]

The matrix #L_A# relative to the basis consisting of the columns of #T# is

\[ T^{-1}\, A\, T = \matrix{2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ }\]

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