Can we solve $\mathit{a}\mathit{P}=\mathit{b}$ where $\mathit{a}$ and $\mathit{b}$ are each $1\times n$ row vectors and are known, and $\mathit{P}$ is an $n\times n$ permutation matrix that is unknown?

Ivan Buckley
2022-09-25
Answered

Can we solve $\mathit{a}\mathit{P}=\mathit{b}$ where $\mathit{a}$ and $\mathit{b}$ are each $1\times n$ row vectors and are known, and $\mathit{P}$ is an $n\times n$ permutation matrix that is unknown?

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ticotaku86

Answered 2022-09-26
Author has **12** answers

No, It is not possible in general.

Consider $P=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$a=(1,1)$ and $b=(2,2)$

Now solving

$aP=b$

amounts to solving the following equation

$a+c=2$

$b+d=2$

That means $a+b+c+d=4$

This is a contradiction as by definition a permutation matrix is a square binary matrix that has exactly one entry of 1 in each row and each column and 0s elsewhere.

So sum of all entries of a $n\times n$ permutation matrix is n.

So for a $2\times 2$ Permutation matrix sum of all entries is equal to 2.

Consider $P=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$a=(1,1)$ and $b=(2,2)$

Now solving

$aP=b$

amounts to solving the following equation

$a+c=2$

$b+d=2$

That means $a+b+c+d=4$

This is a contradiction as by definition a permutation matrix is a square binary matrix that has exactly one entry of 1 in each row and each column and 0s elsewhere.

So sum of all entries of a $n\times n$ permutation matrix is n.

So for a $2\times 2$ Permutation matrix sum of all entries is equal to 2.

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For a function of $n+1$ variables $f({x}_{0},{x}_{1},{x}_{2},...{x}_{n})$ can a gradient exist?

When I asked my professor this during class he said, "no, at most a gradient will exist for a function of three variables f(x,y,z) because there are only at most three standard basis vectors with which to represent a vector."

This is a calculus 3 class so perhaps this answer was given to keep the concept of the gradient within the scope of the class, but I suspect this isn't the whole story and there is more to this than my professor is telling.

Edit:

The definition of the gradient for a function of two variables given during class was: Let z=f(x,y) be a function, then the gradient of f is defined as $\mathrm{\nabla}f={f}_{x}\overrightarrow{i}+{f}_{y}\overrightarrow{j}$