I am given, $P_{n\times n}$ column stochastic, I need to see whether a matrix of suitable order $B$ with an orthonormal column or rows can be constructed so that $BP=0$?
I started trialing like this:
$\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix} \begin{bmatrix}p_{11}&p_{12}\\ p_{21}&p_{22}\end{bmatrix}=\begin{bmatrix}0&0\\ 0&0\end{bmatrix}$
seems like I can get $B$ by solving the system, right?
And, then not sure about orthonormality etc.
Also, I don't need to construct in particular square $B$.
Thanks for helping.
This is not possible (if $B$ is square). Suppose $BP = O_n$ where $O_n$ is the $n\times n$ zero matrix. Let $\mathbf{p}_1$ be the first column of $P$. Then the first column of $BP$ is $B\mathbf{p}_1$, which must equal the zero vector $\mathbf{0}$ (first column of $O_n$). Since $P$ is column stochastic, $\mathbf{p}_1$ is a non-zero vector. Hence $B$ has a non-zero vector in its kernel, so $B$ is singular and thus cannot have orthonormal columns (equivalently, orthonormal rows).
EDIT: I missed the part where you said you did not need $B$ to be square. The above is assuming $B$ is square.
Some discussion for general (not necessarily square $B$): it is still impossible for $B$ to have orthonormal columns. This is because by same reasoning as above, $B$ must have non-trivial kernel, which is impossible for a $B$ with orthonormal columns (as orthonormal columns would necessarily be linearly independent).
It can be possible to make $B$ have orthonormal rows though. Basically, if $\mathbf{r}^T$ is a row of $B$ (so $\mathbf{r}\in \mathbb{R}^n$ is a column vector), having $BP = O$ is equivalent to requiring $\mathbf{r}^T \mathbf{p}_j = 0$ for every column $\mathbf{p}_j$ of $P$. In other words, the rows of $B$ should all be orthogonal to every column of $P$, or equivalently all come from $\mathrm{im}(P)^{\perp}\subseteq \mathbb{R}^n$ (orthogonal complement of the image of $P$).
So we can take $B$ to have as rows an orthonormal basis for $\mathrm{im}(P)^{\perp}$, provided such an orthonormal basis exists, which occurs iff $\mathrm{im}(P)^\perp$ has dimension at least $1$, or equivalently, $P$ is singular.
IN SUMMARY:
● We can never have a $B$ with orthonormal columns satisfy $BP = O$.
● If $P$ is singular, then it is possible to have $B$ with orthonormal rows with $BP = O$. This will happen iff the rows of $B$ are orthonormal elements of $\mathrm{im}(P)^\perp \subseteq \mathbb{R}^{n}$.
● If $P$ is non-singular (has full rank), then we cannot have a $B$ with orthonormal rows satisfying $BP = O$.
● Any possible $B$ must be non-square.
By the way, for any real matrix $P$, we have $\mathrm{im}(P)^\perp = \ker(P^T)$, which you can use to find orthogonal complements if you wish.