# How to generate a rectangular matrix that has no left/right inverse?

Is it possible to generate non-invertible rectangular matrices? If so, how we can prove it? And is there any way to generate randomly such matrices in MATLAB (with preferably uniform distribution)?

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When you say singular in the context of rectangular matrices, in what sense are you referring to? Singular often means determinant $0$, but it doesn't make sense for a general rectangular matrix to have a determinant. – Cameron Williams Jun 26 '13 at 4:35
You r right, it was better to say non-invertible rectangular matrix. I want to know how we can generate non-invertible rectangular matrix that cannot be inverted even by pseudoinverse. – user1468089 Jun 26 '13 at 4:44
Generating non-invertible matrices is hard because any given matrix is almost surely going to have an inverse. Well this is true for square matrices, anyway. Almost singular matrices occur much more frequently though. – Cameron Williams Jun 26 '13 at 4:52
I read somewhere that we can generate non-invertible rectangular matrices by multiplying a singular, e.g. A, matix and a rectangular, e.g. B, then the product C=A*B would be non-invertible. Is it true? If so, can we prove it? – user1468089 Jun 26 '13 at 4:59
This would be true, yes. You could probably argue this easily because $A$ has at least two rows that are not independent and thus neither would $C$ and so your resulting rank would be less than the row space. I would worry that you're only picking out a subset of all of the possible singular matrices when doing this but it's at least a start in the right direction. What you could do is this: start with a known singular (non-sparce!) matrix $A$ and then let $B$ be a random matrix (choose the entries uniformly I suppose). Odds are B will be invertible but even if it's not, that's even better. – Cameron Williams Jun 26 '13 at 5:02

A wide (resp. tall) matrix $A$ has no right (resp. left) inverses if and only if it has deficient row (resp. column) rank, i.e. if and only if the rows (resp. columns) of $A$ are linearly dependent. Owing to the dependency between rows/columns, what you mean by "uniform distribution" is unclear.
At any rate, here are some methods that you may consider. Without loss of generality, suppose $A$ is an $m\times n$ wide matrix (so that $m<n$). Presumably, the entries of $A$ are real numbers.
1. Generate a random integer $k$ between $0$ and $m-1$. If $k=0$, set $A=0$. Otherwise, generate $k$ random row vectors $\mathbf{r}_1,\ldots,\mathbf{r}_k$, each of length $n$. For $i=1,2,\ldots,m$, generate a probability vector $(p_1,p_2,\ldots,p_k)$ (i.e. $0\le p_i\le 1$ and $\sum_ip_i=1$). Set the $i$-th row of $A$ to $\sum_i p_i\mathbf{r}_i$.
2. Generate a random integer $k$ between $0$ and $m-1$. If $k=0$, set $A=0$. Otherwise, generate $k$ random column vectors $\mathbf{c}_1,\ldots,\mathbf{c}_k$, each of length $m$. For $j=1,2,\ldots,n$, generate a probability vector $(p_1,p_2,\ldots,p_k)$. Set the $j$-th column of $A$ to $\sum_j p_j\mathbf{c}_j$.
3. Generate a random integer $k$ between $0$ and $m-1$. Generate $k$ random nonnegative real numbers $\sigma_1,\ldots,\sigma_k$ within some range. Generate an $m\times m$ real orthogonal matrix $U$ and an $n\times n$ real orthogonal matrix $V$ (see Wikipedia for an algorithm). Then set $$A=U\left[\operatorname{diag}(\sigma_1,\ldots,\sigma_k,0,\ldots,0),\ 0_{m\times(n-m)}\right]V.$$