### Invariant subspaces of linear maps: Invariant subspaces

### Jordan normal form

To a linear map #L# of an #n#-dimensional vector space #V# to itself belongs a matrix #A=L_\alpha# as soon as we have chosen a basis #\alpha# for #V#. If #L_\beta# is the matrix with respect to a second basis #\beta# for #V#, then #L_\alpha# and #L_\beta# are conjugate, meaning: there exists an invertible #(n\times n)#-matrix #T# with the property that #L_\beta=T^{-1}\,L_\alpha\,T#. For example, the transition matrix #T = {}_\alpha I_\beta# has this property.

We keep in mind that a matrix #A# is diagonizable if it is conjugate to a diagonal matrix, and not all matrices are diagonizable. Here we will discuss a normal form that can be found for all matrices. We will start with matrices whose characteristic polynomial only has one root.

The Jordan form with one eigenvalue

Each #(n\times n)#-matrix #A# with eigenvalue #\lambda# and characteristic polynomial #(\lambda-x)^n# can be transformed to **Jordan normal form** (in short: **Jordan form**); meaning: there is an invertible matrix #S# such that

\[

S^{-1}AS=J=\left(\begin{array}{ccc} J_{i_1} & & \\ & \ddots & \\ & & J_{i_k}

\end{array}\right)

\] where the #(i\times i)#-matrix #J_{i}# is equal to

\[

J_{i}=\left(\begin{array}{cccc}\lambda & 1 & & \\ & \lambda & \ddots & \\

& & \ddots & 1\\ & & & \lambda

\end{array}\right)

\] and is called the **Jordan block** of size #i# corresponding eigenvalue #\lambda#.

The number #j_i# of Jordan blocks of size #i# satisfies

\[j_i =2\dim{\ker{(A-\lambda\,I_n)^{i}}}-\dim{\ker{(A-\lambda\,I_n)^{i+1}}}-\dim{\ker{(A-\lambda\,I_n)^{i-1}}}\] Assume that #B# is also an #(n\times n)#-matrix with eigenvalue #\lambda# whose characteristic polynomial is #(\lambda-x)^n#. Then #A# and #B# are conjugate if and only if the number of Jordan blocks of size #i# match for #i = 1,\ldots,n#.

For a diagonalizable matrix all Jordan blocks have size #1#.

Multiple Jordan blocks of the same dimension corresponding to eigenvalue #\lambda# are possible. The total number of Jordan blocks is equal to the dimension of the corresponding eigenspace #E_\lambda#, the null space #\ker{A-\lambda I}#.

We apply these results on the matrix of a linear map #L:V\to V# restricted to each of its generalized subspaces. The information on the Jordan blocks uniquely determines the conjugacy class of #L#.

Jordan normal form

Assume that #V# is a vector space of finite dimension #n#, that #L:V\to V# is a linear map, and that the characteristic polynomial #p_L(x)# of #L# is a product of linear factors:

\[p_L(x) = (\lambda_1-x)^{k_1}\cdot(\lambda_2-x)^{k_2}\cdots (\lambda_r-x)^{k_r}\]where #\lambda_1,\ldots,\lambda_r# are different from each other.

Then there are unique numbers \[ j_{\lambda_s,i}\phantom{xxx} \text{ with }\phantom{xxx} s=1,\ldots,r\phantom{xx}\text{ and }\phantom{xx} i = 1,\ldots,k_s\] such that the matrix of the restriction of #L# to the generalized eigenspace corresponding to #\lambda_s# with respect to a suitably chosen basis has a Jordan normal form with #j_{\lambda_s,i}# blocks of size #i#.

A linear map #M:V\to V# has the the same matrix as #L# relative to a suitably chosen basis for #V# if and only if it has the same eigenvalues as #L# and the same sizes #j_{\lambda_s,i}# (for #i=1,\ldots,k_s#) of Jordan blocks corresponding to each eigenvalue #\lambda_s# (for #s=1,\ldots,r#).

Which of the matrices below is a Jordan normal form of #A#?

Since the multiplicity of #{2}# in the characteristic polynomial is equal to #3#, the dimension of generalized eigenspace #E_{2}^*# equals #{3}#. Since the multiplicity of #{3}# in the characteristic polynomial is equal to #1#, the dimension of generalized eigenspace #E_{3}^*# equals #1#.

To determine the size of the Jordan blocks, we first calculate the dimensions #\ker{(A-\lambda\,I_4)^{i}}# for the eigenvalues #\lambda = 2#, # 3# and #i=1,2,\ldots#. If #i\ge 3#, this dimension is equal to #3# for #\lambda=2# and if #i\ge 1#, then this dimension is equal to #1# for #\lambda = 3#.

\[\begin{array}{l|r}\text{subspace}\phantom{i}&\text{dimension}\\

\hline

\ker{(A-2\,I_4)^{3}}& 3\\

\ker{(A-2\,I_4)^{2}}&2\\

\ker{A-2\,I_4}&1\\

\ker{A-3\,I_4}&1\\

\hline

\end{array}\]According to the theorem

*The Jordan form with one eigenvalue*, the numbers #j_1#, #j_2#, #j_3# of Jordan blocks of #A# corresponding to eigenvalue #2# of size #1#, #2#, #3#, respectively, can be determined as follows (not all steps are necessary because the dimension #3# has already reached the dimension of the generalized eigenspace after the first step):

\[\begin{array}{rcrcl}

j_3 &=&

2\dim{\ker{(A-2 I_4)^{3}}}-\dim{\ker{(A-2 I_4)^{4}}}-\dim{\ker{(A-2 I_4)^{2}}}&=&1\\

j_2 &=&

2\dim{\ker{(A-2 I_4)^{2}}}-\dim{\ker{(A-2 I_4)^{3}}}-\dim{\ker{(A-2I_4)}}&=&0\\

j_1 &=&

2\dim{\ker{(A-2 I_4)}}-\dim{\ker{(A-2 I_4)^{2}}}-\dim{\ker{I_4}}&=&0

\end{array}\]Similarly, we find that only the number of Jordan blocks of size #1# is nonzero for eigenvalue #3#. We conclude that #A# has

- exactly one Jordan block of size #3# on the generalized eigenspace corresponding to #2#
- exactly one Jordan block of size #1# on the generalized eigenspace corresponding to #3#.

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