Let's say I have a matrix whose rank is $\rho(A) =1$.

How does it affect its Jordan Normal Form?

I know that a matrix A is similar to a Jordan Matrix and that means it has the same rank. So if a matrix has a rank of one, doesn't it mean that the Jordan Normal form is:

\begin{pmatrix} \lambda & 1 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{pmatrix}

A matrix whose rank is 1?



1 Answer 1


No, because the matrix you're showing is not a Jordan normal form at all (unless $\lambda=0$). Every off-diagonal $1$ need to appear inside a single Jordan block, that is, connecting two diagonal elements with the same value.

The possible Jordan normal forms of rank 1 are

$$\left(\begin{array}{c|cc|c} \mathbf{0}_{a\times a} \\ \hline & 0 & 1 \\ & 0 & 0 \\ \hline &&& \mathbf{0}_{b\times b} \end{array}\right)$$ and $$\operatorname{Diag}(0,\ldots,0,\lambda,0,\ldots,0) = \left(\begin{array}{c|c|c} \mathbf{0}_{a\times a} \\ \hline & \lambda \\ \hline && \mathbf{0}_{b\times b} \end{array}\right) \text{ with }\lambda\ne 0$$

  • $\begingroup$ Thank you. But we can say for sure that A Jordan Normal Form has only one row (and column) with no zero values, right? That is, the definition of the rank? $\endgroup$
    – Alan
    Commented Oct 31, 2015 at 10:11
  • 1
    $\begingroup$ @Alan: The rank of a block diagonal matrix is the sum of the ranks of the diagonal blocks, so for a rank-1 JNF matrix there must be exactly one block of rank 1. And the only way for a Jordan block to have rank 1 is if it has one of the two forms above -- which implies that there must be exactly one nonzero element in the matrix, because that happens to be true for both of the possible forms. $\endgroup$ Commented Oct 31, 2015 at 10:14

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