One theorem says: A matrix $\mathbf A \in \mathbb R^{m \times n}$ is:

  • full column rank iif $\mathbf A^T \mathbf A$ is invertible
  • full row rank iif $\mathbf A \mathbf A^T$ is invertible

(proof link)

Now, I want to find just one example matrix $\mathbf A$, that is a singular matrix (i.e. degenerate or not full rank), such that $\mathbf A^T \mathbf A$ or $\mathbf A \mathbf A^T$ is not invertible and with a constrain that:

$\text{rank}(\mathbf A^T \mathbf A) < \text{rank}(\mathbf A)$ or $\text{rank}(\mathbf A \mathbf A^T) < \text{rank}(\mathbf A)$

I tried many random numbers to construct a small size of matrix (e.g. $4 \times 5$) using R language , but found no such matrix. Is this kind of degenerate matrix that makes $\mathbf A^T \mathbf A$ or $\mathbf A \mathbf A^T$ not invertible very unlikey to find (or it is impossilbe)? Any idea or method to construct such a matrix?


What you are asking for is impossible.

From the relation $$rank(AA^T) = rank(A^T) - dim(N(A) \cap R(A^T)) $$.

The second term is always zero and the equality is always satisfied.

  • $\begingroup$ Thanks, but could you give me a link to explain the background theory behind this equation? Or a proof for this equation? $\endgroup$ – Wei Zhong Nov 23 '14 at 18:17
  • $\begingroup$ You can find it in page 210 of this book goo.gl/F6n07c $\endgroup$ – Karthik Upadhya Nov 23 '14 at 18:56
  • $\begingroup$ Exactly what I want, thanks a lot! $\endgroup$ – Wei Zhong Nov 23 '14 at 19:22

To generate matrices that are not full rank, start with a square matrix that is full rank (r > 1), and replace one row (or column) with a linear combinations of the other rows (or columns). That will reduce the rank by 1, in the case of row operations.

According to this Wikipedia entry, the rank of a real matrix is equal to the rank of its corresponding Gram matrix, so $$rank(A^T A) = rank(A A^T) = rank(A) = rank(A^T)$$

  • $\begingroup$ That's true, but I also want $\text{rank}(A A^T)<\text{rank}(A)$. $\endgroup$ – Wei Zhong Nov 22 '14 at 6:23
  • $\begingroup$ @WeiZhong You can't, as Louis said. $\endgroup$ – egreg Nov 22 '14 at 16:14

Let $A$ be a 3x2 matrix (with 3rd row being proportional to the 2nd row):

  A={{1, 2 }, 
     {2, 10}, 
     {3, 15}}.

Then $A.A^T$ is a 3x3 singular matrix:

  A.A^T={{5,  22,  33 }, 
         {22, 104, 156}, 
         {33, 156, 234}}

And $A^T.A$ is a 2x2 nonsigular matrix:

  A^T.A={{14, 67 }, 
         {67, 329}}.

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