Jordan normal form In linear algebra, the Jordan normal form, also called the Jordan canonical form, named in honor of the 19th and early 20th-century French mathematician Camille Jordan, answers the question, for a given square matrix M over a field K, to what extent M can be simplified into a standard shape by changing basis. Motivation Consider the situation of matrix diagonalization. A square matrix is diagonalizable if the sum of the dimensions of the eigenspaces is the number of rows or columns of the matrix. Let us examine the following matrix We have eigenvalues of A being only λ = 5, 5, 5, 5. Now, the dimension of the kernel of A-5I is 1, so A is not diagonalizable. However, we can construct the Jordan form of this matrix. Since the kernel of A-5I is 1, we know that the Jordan form is comprised of only one Jordan block, that is, the Jordan form of A is Observe that J can be written as 5I+N, where N is a nilpotent matrix. Since we have now A similar to such a simple matrix, we can perform calculations involving A by using the Jordan form, which can ease the calculations in many cases. For example calculating powers of matrices is significantly easier by using the Jordan form. General case It is not possible to make all such matrices M diagonal, even when K is algebraically closed: what the Jordan normal form does is to quantify the failure. In abstract terms, any M is written as a sum D + N where D is diagonalizable, N is nilpotent, and D commutes with N. The way the normal form is usually written is explicitly as the direct sum of block square matrices, known as Jordan blocks. Jordan blocks are of the form λI + N=Jn(λ), where λ is one of the eigenvalues of M, n the number of rows or columns of the Jordan block, and N is a special nilpotent matrix defined as Nij=δi,j+1 (where δ is the Kronecker delta). This form is valid over an algebraically closed field. That is, one typical Jordan block looks like $\backslash begin$ \lambda & 1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & 0 & \cdots & 0 \\ 0 & 0 & \lambda & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & 0 & 0 & \lambda \\\end If one knows the dimensions of the kernels (M-λI)k for 1 ≤ k ≤ m, where m is the algebraic multiplicity of the eigenvalue λ, one can determine the Jordan form that exists for M. Calculating the invertible transition matrix P such that P-1MP=J can be done by considering eigenvectors. The proof of the Jordan normal form is usually carried out as an application to the ring K[X] of the structure theorem for finitely-generated modules over principal ideal domains, of which it is a corollary. Algorithms and methods Let us examine the methods of determining the transition matrix by example. Example 1 Consider the calculation of the transition matrix for the matrix above. Recall We concern ourselves with obtaining generalized eigenvectors, that is, solutions to $(A-\backslash lambda\; I)^k\backslash mathbf\; =\; \backslash mathbf$ which will allow us to calculate "chains" of vectors, whose elements form the columns of the transition matrix. For A above, we know there is only one Jordan block (see above), so we firstly obtain one generalized eigenvector - since (A-5I)^4 is the zero matrix, ker (A-5I)^4 is the entire space, so we can pick one of the standard basis vectors for the space, v=(1,0,0,0)T, since none of the standard basis vectors are an eigenvector of (A-5I)^3, (A-5I)^2, or A-5I. Then, forming the chain $\backslash left\backslash ,\; (A-5I)^2\backslash mathbf,\; (A-5I)\backslash mathbf,\; \backslash mathbf\backslash right\backslash \}$ $=\backslash left\backslash \; 6048\; \backslash \backslash \; 4320\; \backslash \backslash \; -3648\; \backslash \backslash \; -5280\; \backslash end,$ \begin 2832 \\ 2352 \\ -1952 \\ -2368 \end, \begin 322 \\ 236 \\-260 \\ -242 \end, \begin 1 \\ 0 \\0 \\ 0 \end\right\} so, we can form the transition matrix as \end Example 2 Say we have $B\; =$ \begin 5 & 4 & 2 & 1 \\ 0 & 1 & -1 & -1 \\ -1 & -1 & 3 & 0 \\ 1 & 1 & -1 & 2 \\ \end The eigenvalues of B are 4, 4, 2 and 1. Now, we have $\backslash mathrm\backslash \; \backslash ker\; =\; 1,\; \backslash mathrm\backslash \; \backslash ker\; =\; 1,\; \backslash mathrm\backslash \; \backslash ker\; =\; 1,\; \backslash mathrm\backslash \; \backslash ker^2\; =\; 2$ so we can say that the Jordan form of the matrix is $J=J\_1(1)\backslash oplus\; J\_1(2)\backslash oplus\; J\_2(4)$ since vectors in ker B-4I are also in ker (B-4I)2. We have that $\backslash ker^4\; =\; \backslash mathrm\backslash \; \backslash left\backslash \; 0\; \backslash \backslash \; 0\; \backslash \backslash \; -1\; \backslash \backslash \; 1\backslash end,\; \backslash begin\; 1\; \backslash \backslash \; 0\; \backslash \backslash \; -1\; \backslash \backslash \; 1\backslash end\backslash right\backslash \}$ but we pick a vector in the span that is not in any of the kernels of (B-4I)3, (B-4I)2, or B-4I, so choose v=(0,0,-1,1)T since (1,0,-1,1) is in the kernel of B-4I. Now, there are three chains, , , and , where w=(1,-1,0,1)T is the basis vector of the 1-dimensional kernel of B-2I and likewise x=(-1,1,0,0) is the basis vector of the 1-dimensional kernel of B-I. Form the transition matrix from these chain vectors as follows: and If we had interchanged the order of which the chain vectors appeared, that is, changing the order of w, 'x, and together, the Jordan blocks would be interchanged, giving equivalent Jordan forms, however. 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