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I am trying to understand if there are any relation between a square matrix and its transpose when it comes to compare their range and rank and then their kernel and nullity. In my specific case the only think that they have in common is the fact that the nullity of both are the same ( 2 ) . I try to find anything that could help me to understand but I only found properties about the rank and nullity for non square matrices. In my case I am not sure if it is a coincide or if in fact there is any property that can connect the nullity of a n x n matrix and its transpose.

Can anyone help on this or suggest a website where I can read more about this?

Thanks

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    Are you saying you have a concrete example of a matrix that doesn't have the same rank as its transpose? You must be doing something wrong. – 2017-02-20
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    yes I did something wrong. Both ranks are the same thank you for your help. – 2017-02-20
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    $\operatorname{range}(A^T)=\operatorname{null}(A)^\perp$ and $\operatorname{null}(A^T)=\operatorname{range}(A)^\perp$. – 2017-02-20
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    Where can I find these properties? – 2017-02-21

1 Answers 1

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The rank of a matrix is the same as the rank of its transpose.

To see this, consider this characterization of rank, which is clearly invariant under transposition:

A rank-1 matrix is a nonzero matrix that can be written as the product of a column (to the left) and a row (to the right).

The rank of a matrix $A$ is the smallest $k\in\mathbb N$ such that $A$ is the sum of $k$ rank-1 matrices.

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    I did something wrong and the two ranks are the same. About the nullity? is there anything we can say about it? Or being the same is just a coincidence? Thank you for your help – 2017-02-20
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    @user290335: In the case of a square matrix, $A$ and $A^T$ will have the same nullity as a _consequence_ of having equal rank, through the rank-nullity theorem. – 2017-02-20
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    Thank you very much for your help. – 2017-02-20