Question

Determine the Jordan form of the matrix \(\begin{bmatrix}1&1&0\\0&1&0\\0&1&1\end{bmatrix}\)

Answer 1

I found that the characteristic polynomial is \((\lambda-1)^3\), and that there is an eigenvector \((1,0,1)^t\) with eigenvalue 1. I'm just not sure what I do after that.

Answer 2

It seems to me that the answer is
\[
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{array}
\right)
\]

Answer 3

You have to work it out and check your answer against mine.

Answer 4

That looks right, I'm just having trouble understanding how I know that it is that and how I can check it. I'm learning this all from the book, without any lessons, and the chapter focuses more on proofs than on methods. Is it that because the exponent of the generalized eigenvector is 2, so we have a 2x2 Jordan block, and then the eigenvalue was 1 for both the original matrix and for the \(T-\lambda\) matrix?