# Jordan Normal Form problem with consistency

I'm trying to find the Jordan Normal Form of the matrix $A = \begin{bmatrix} 1 & 0 & 1 & -1 \\0 & 1 & 1 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1\end{bmatrix}$

(That is, a matrix J and a matrix S such that $A = S J S^{-1})$.

Finding the eigenvalues and eigenvectors is no problem. The only eigenvalue is $\lambda$=1 with multiplicity 4. The corresponding eigenvectors are $x = \begin{bmatrix}1\\0\\0\\0\end{bmatrix}$ and $y = \begin{bmatrix}0\\1\\0\\0\end{bmatrix}$.

However, I get stuck after that. In order to find the S and J matrix, I have to determine the consistency of the matrices $(A-I)w = x$ and $(A-I)w = y$. But both of these matrices are inconsistent:

$(A-I)w = x \rightarrow \begin{array}{cccc|c}0 & 0 & 1 & -1 &1\\0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\end{array} \rightarrow \begin{array}{cccc|c}0 & 0 & 1 & 0 &1\\0 & 0 & 1 & 0 &0\\0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\end{array}$

Which is clearly inconsistent since w3 = 0 from row 2 and w3 = 1 from row 1.

Also, $(A-I)w = x \rightarrow \begin{array}{cccc|c}0 & 0 & 1 & -1 &0\\0 & 0 & 1 & 0 & 1\\0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\end{array} \rightarrow \begin{array}{cccc|c}0 & 0 & 1 & 0 &0\\0 & 0 & 1 & 0 &1\\0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\end{array}$

Which is also clearly inconsistent.

So right now I do not know how to construct either the S or J matrix, since at least one of the systems $(A-I)w = x$ or $(A-I)w = y$ should have had a solution (and then also a solution for $(A-I)^2$ and possibly $(A-I)^3$ if there's one 1x1 and one 3x3 block).

Can someone help me out on how to construct the S and J matrices? Thanks!

• Note that $x+y$ is also an eigenvector, and $(A-I)w=x+y$ does have a solution. – Gerry Myerson Apr 10 '14 at 10:35

Since you have found a $2$-dimensional eigenspace, you know the image of $A-I$ only has dimension$~2$ (that is the rank of $A-I$). You also know $A-I$ is nilpotent, so $(A-I)^2$ must have smaller rank (and equivalently larger kernel), the question is whether it drops to $0$ or becomes$~1$. Simple calculation show that $(A-I)^2$ is nonzero (so has rank $1$) and in fact its image is spanned by $(1,1,0,0)$ (the image of the last standard basis vector). So you've got a Jordan block of size $3$, and another one with size $1$ is necessarily left.

Now an important point to notice is that the first basis vector for the Jordan block of size$~3$ is not just any eigenvector (although it must be one), but also a vector in the image of $(A-I)^2$. The latter means it must be $b_1=(1,1,0,0)$, up to a scalar. Now the last basis vector $b_3$ for this Jordan block can be any vector with $(A-I)^2\cdot b_3=b_1$, and we saw that $b_3=(0,0,0,1)$ has this property. Then $b_2=(A-I)\cdot b_3=(-1,0,1,0)$ is also clear. Finally $b_4$ is just any eigenvector linearly independent from $b_1$, so $b_4=(1,0,0,0)$ will do. This gives a Jordan basis.

• Hi Marc. Great answer, I'd like to get the general steps of your method...could you perhaps expand on that a little? In determining the matrix $S$ specifically. For example should the eigenvector which is the first basisvector for a larger size Jordan block always be in the image of $(A-\lambda I)^2$... – Christiaan Hattingh Apr 10 '14 at 15:10
• Hi Marc. Never mind, thanks I get it, $b_1$ must be in the image of $(A-I)^2$ so that it won't be 'killed' (be in the kernel), else the chain stops there...elegant answer - I +1ed – Christiaan Hattingh Apr 10 '14 at 16:25

Here is another way to determine $J$.

1. Find the characteristic polynomial. Since the matrix A is upper triangular it is easy to see that $c_A(x)=(x-1)^4$.
2. Find the minimum polynomial. Every linear factor of $c_A(x)$ must also be a factor of $m_A(x)$. So we must just find $n$ so that $(A-I)^n=0$. Turns out $n=3$.

Now the multiplicity of $\lambda=1$ in the minimum polynomial corresponds to the size of the largest simple Jordan block, and so we must have $$J=\begin{bmatrix} 1&0&0&0 \\ 1&1&0&0\\ 0&1&1&0 \\ 0&0&0&1 \end{bmatrix}.$$ So now you have $J$ but you still need $S$. We have $AS=SJ$; let $S=[S_1,S_2,S_3,S_4]$ where each $S_i$ is the $i$th column of $S$. Then we have $$AS=[AS_1, AS_2,AS_3,AS_4]=SJ=[S_1+S_2,S_2+S_3,S_3,S_4],$$so equating both sides (starting with the last column): $$AS_4=S_4 \Leftrightarrow (A-I)S_4=0,$$$$AS_3=S_3 \Leftrightarrow (A-I)S_3=0,$$$$AS_2=S_2+S_3 \Leftrightarrow (A-I)S_2=S_3 \Leftrightarrow (A-I)^2S_2=0,$$$$AS_1=S_1+S_2 \Leftrightarrow (A-I)S_1=S_2\Leftrightarrow (A-I)^2S_1=S_3 \Leftrightarrow (A-I)^3S_1=0.$$

So $S_3$ and $S_4$ are two linearly independent eigenvectors of $A$ associated with $\lambda=1$.

From this point I must say thanks to Marc, who's elegant answer switched the light on for me (please if you +1 me, do so for him too). In my answer I just want to expand the general method to find $S$: so here goes...

1. Since $S_1$ must be nonzero (The matrix $S$ must be invertible) and $(A-I)^2S_1=S_3$ we see that the eigenvector $S_3$ (which must also be nonzero) must be in the image of $(A-I)^2$. And we see that Ran$(A-I)^2=\{x(1,1,0,0)^T:x \in \mathbb{F}\}$. So let $S_3=(x_1,x_1,0,0)^T$ with $x_1 \neq 0$ (!).
2. Now we can calculate $S_1$ using the equation in 1 above. We get the general solution $S_1=(x_2,x_3,x_4,x_1)^T$ (so the only restriction is that the last coordinate of $S_1$ must equal $x_1$, other parameters are free).
3. Now we can calculate $S_2$ by $(A-I)S_1=S_2$. The general solution is $S_2=(x_4-x_1,x_4,x_1,0)^T$.
4. Now all we need is that $S_4$ must be linearly independent from $S_3$. So let $S_4=(x_5,0,0,0)^T$.

You can check that $$S=\begin{bmatrix} x_2 & x_4-x_1 & x_1 & x_5 \\ x_3 & x_4 & x_1 & 0 \\ x_4 & x_1 & 0 & 0 \\ x_1 & 0 & 0 & 0\end{bmatrix},$$ is such that $S^{-1}AS=J$