Construct an example matrix such that $\mathbf A \mathbf A^T$ is not invertible

Question

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

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?

A proof of the theorem cited can be found on this link: https://inst.eecs.berkeley.edu/~ee127a/book/login/thm_full_rank_pd.html — Wei Zhong, Nov 22 '14 at 05:55
@John, forgot to mention a rank constrain for the example I would like to find. Already edited the post. — Wei Zhong, Nov 22 '14 at 06:01
@Nameless, John is right, I want to find a invertible matrix, but my fault not mention another constrain. Already edited the post. — Wei Zhong, Nov 22 '14 at 06:07
Neither inequality is possible; see http://math.stackexchange.com/questions/1025469/prove-that-for-a-real-matrix-a-kera-kerata/1025472#1025472 — Travis Willse, Nov 22 '14 at 06:40
@ Thanks @Travis, now I understand. Hopefully I can choose your comment as best answer because you are the first one to help me out. — Wei Zhong, Nov 23 '14 at 18:14
@WeiZhong I'm glad you found it helpful, Wei. I'd write up a separate answer, but there are already good answers below that explain that argument well. — Travis Willse, Nov 24 '14 at 00:50

score 3 · Accepted Answer · answered Nov 22 '14 at 09:16

3

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.

answered Nov 22 '14 at 09:16

Karthik Upadhya

764
7
15

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

LouisB · Answer 2 · 2014-11-22T07:54:53.573

2

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 $$rank(A^T A) = rank(A A^T) = rank(A) = rank(A^T)$$

edited Nov 22 '14 at 07:54

answered Nov 22 '14 at 06:21

LouisB

801

That's true, but I also want $\text{rank}(A A^T)<\text{rank}(A)$. – Wei Zhong Nov 22 '14 at 06:23
@WeiZhong You can't, as Louis said. – egreg Nov 22 '14 at 16:14

score 0 · Answer 3 · answered Nov 22 '14 at 15:53

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}}.

Construct an example matrix such that $\mathbf A \mathbf A^T$ is not invertible

3 Answers3