# Jordan Normal Form Issues

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Let's take the simple example where the matrix, A, is 3 by 3 and has the single eigenvalue 3. Then the Characteristic equation is $(x- 3)^3= 0$. Since every matrix satisfies its own characteristice equiation, it must be true that for every vector, v, $(A- 3)^3= 0$. It might be the case that (A- 3I)v= 0 for every vector v. In that there are three independent vectors such that Av= 3v and we can use those three vectors as a basis for the vector space. Written in that basis, A would be diagonal: $$\begin{bmatrix}3 & 0 & 0 \\ 0 & 3 & 1\\ 0 & 0 & 3\end{bmatrix}$$. Or, it might be that (A- 3I)v= 0 only for multiples of a single vector or linear combinations of two vectors. In the first case, it must still be true that $(A- 3I)^3u= 0$ for all vectors so we must have, for some vector, u, $(A- 3I)v= w\ne 0$ but that $(A- 3I)^3v= (A- 3I)^2w= 0$. Of course that is the same as saying that $(A- 3I)^2w$ is a multiple of v, the eigenvector, and so, letting x be (A- 3I)w, (A- 3I)x= v. Those two vectors, x such that (A- 3I)x= v, and w such that (A- I3)w= x, are called "generalized eigenvectors". Of course, to find the matrix representation of a linear transformation in a given basis, we apply the matrix to each basis vector in turn, writing the result as a linear combination of the basis vectors, so that the coefficients are the columns of the matrix. (If you are not aware of that [very important!] fact, suppose A is a linear transformation from a three dimensional vector space to itself. Further, suppose $\{v_1, v_2, v_3\}$ is a basis for that vector space. Then $Av_1$ is a vector in the space and so can be written as a linear combination of the basis vectors, say, $Av_1= av_1+ bv_2+ cv_3$. In that basis, $v_1$ itself is written as the column $$\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$$ since $v_1= (1)v_1+ (0)v_2+ (0)v_3[/tex] Since [itex]Av_1= av_1+ bv_2+ cv_3$, we must have $$Av_1= A\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ b \\ c\end{bmatrix}$$ so obviously, the first column of a must be $$\begin{bmatrix}a \\ b \\ c\end{bmatrix}$$.) Here, our basis vectors are v, x, and w such that (A- 3I)v= 0, (A- 3I)x= v, and (A- 3I)w= u. From (A- 3I)v= 0, which is the same as Av= 3v+ 0x+ 0w, we see that the first column of the matrix is $$\begin{bmatrix}3 \\ 0 \\ 0\end{bmatrix}$$ From (A- 3I)x= v, which is the same as Ax= v+ 3x+ 0w, we see that the second column of the matrix is $$\begin{bmatrix}1 \\ 3 \\ 0\end{bmatrix}$$ Finally, from (A- 3I)w= x, which is the same as Aw= 0v+ x+ 3w, we see that the third column of the matrix is $$\begin{bmatrix}0 \\ 1 \\ 3\end{bmatrix}$$ so that, in this ordered basis, the linear transformation is represented by the matrix $$\begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3\end{bmatrix}$$