Skip to main content
Post Closed as "Not suitable for this site" by Franz Lemmermeyer, YCor, Jan-Christoph Schlage-Puchta, Hugh Thomas, user21574
edited body
Source Link

Given any $n \times k$ real matrix $M$, where $n<k$ and $rank(M)=k$$rank(M)=n$, I consider the following equation (where $M'$ is the transpose of $M$):

$$ MM' = MAM' $$

Then clearly, $A = \mathbb{1}_k $, the $k\times k$ identity matrix, is a possible solution.

Is that unique, or there are other possible solutions?

Given any $n \times k$ real matrix $M$, where $n<k$ and $rank(M)=k$, I consider the following equation (where $M'$ is the transpose of $M$):

$$ MM' = MAM' $$

Then clearly, $A = \mathbb{1}_k $, the $k\times k$ identity matrix, is a possible solution.

Is that unique, or there are other possible solutions?

Given any $n \times k$ real matrix $M$, where $n<k$ and $rank(M)=n$, I consider the following equation (where $M'$ is the transpose of $M$):

$$ MM' = MAM' $$

Then clearly, $A = \mathbb{1}_k $, the $k\times k$ identity matrix, is a possible solution.

Is that unique, or there are other possible solutions?

Source Link

Unique solution to a matrix equations

Given any $n \times k$ real matrix $M$, where $n<k$ and $rank(M)=k$, I consider the following equation (where $M'$ is the transpose of $M$):

$$ MM' = MAM' $$

Then clearly, $A = \mathbb{1}_k $, the $k\times k$ identity matrix, is a possible solution.

Is that unique, or there are other possible solutions?