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I have a number of questions which seem linked to me, about basic (?) linear algebra:

Given a field (possibly skew) $K$, and an superfield $L$, one can do linear matrix algebra with coefficients in $L$ (any reference for basic facts about that ?)

So given a square matrix $M\in M_{n\times n}(L)$, it is left invertible if and only if it is right invertible. What are the known relations between the left kernel of $M$ in $L^n$ and its right kernel ?

By left kernel, I mean the space (left-vector space) of rows $\lambda\in L_{1\times n}$ (one row $n$ columns) such that $\lambda M=(0)$ and by right kernel, I mean the set of $\lambda \in L_{n\times 1}$ (one column n rows) such that $M\lambda=(0)$.

Question 1. For instance, if there is a non-trivial point of $K^n$ in the left-kernel of $M$, is there also a non-trivial point of $K^n$ in its right kernel?

If $K$ is commutative, I am trying to understand why the notion of dimension of a $K$-vector does not vary if one enlarges $K$. I suppose it is because the fact that a familly of vectors being linearly dependent is expressible by the determinant of a matrix being non-zero, which does not depend on larger $K$ (this must correspond to some kind of $\exists$-quantifier elimination I suppose).

Question 2. What happens if $K$ is skew ? In that case, there is no notion of determinant of a matrix (I mean, is there a notion of skew determinant ?). Does the notion of $K$-dimension of the $K$-vector $L$ space depends on possible enlargements of $K$ ?

Finally:

Question 3. Why does the Zariski dimension of an algebraic subset of $K^n$ ($K$ commutative, non necessarily algebraically closed) does not depend on possible enlargements of $K$ ?

I have a number of questions which seem linked to me, about basic (?) linear algebra:

Given a field (possibly skew) $K$, and an superfield $L$, one can do linear matrix algebra with coefficients in $L$ (any reference for basic facts about that ?)

So given a square matrix $M\in M_{n\times n}(L)$, it is left invertible if and only if it is right invertible. What are the known relations between the left kernel of $M$ in $L^n$ and its right kernel ?

By left kernel, I mean the space (left-vector space) of rows $\lambda\in L_{1\times n}$ (one row $n$ columns) such that $\lambda M=(0)$ and by right kernel, I mean the set of $\lambda \in L_{n\times 1}$ (one column n rows) such that $M\lambda=(0)$.

Question 1. For instance, given $M\in M_{n\times n}(L)$, if there is a non-trivial point of $K^n$ in the left-kernel of $M$, is there also a non-trivial point of $K^n$ in its right kernel?

If $K$ is commutative, I am trying to understand why the notion of dimension of a $K$-vector does not vary if one enlarges $K$. I suppose it is because the fact that a familly of vectors being linearly dependent is expressible by the determinant of a matrix being non-zero, which does not depend on larger $K$ (this must correspond to some kind of $\exists$-quantifier elimination I suppose).

Question 2. What happens if $K$ is skew ? In that case, there is no notion of determinant of a matrix (I mean, is there a notion of skew determinant ?). Does the notion of $K$-dimension of the $K$-vector $L$ space depends on possible enlargements of $K$ ?

Finally:

Question 3. Why does the Zariski dimension of an algebraic subset of $K^n$ ($K$ commutative, non necessarily algebraically closed) does not depend on possible enlargements of $K$ ?

I have a number of questions which seem linked to me, about basic (?) linear algebra:

Given a field (possibly skew) $K$, and a $K$-algebra $\mathcal A$ (again possibly non-commutative, but let us say an integral domain, unitary, acting on both sides of $\mathcal A$), one can do linear matrix algebra with coefficients in $\mathcal A$ (any reference for basic facts about that ?)

So given a square matrix $M\in M_{n\times n}(\mathcal A)$, it is left invertible if and only if it is right invertible. What are the known relations between the left kernel of $M$ in $\mathcal A^n$ and its right kernel ?

By left kernel, I mean the space (left-vector space) of rows $\lambda\in \mathcal A_{1\times n}$ (one row n columns) such that $\lambda M=(0)$ and by right kernel, I mean the set of $\lambda \in \mathcal A_{n\times 1}$ (one column n rows) such that $M\lambda=(0)$.

Question 1. For instance, if there is a non-trivial point of $K^n$ in the left-kernel of $M$, is there also a non-trivial point of $K^n$ in its right kernel?

If $K$ is commutative, I am trying to understand why the notion of dimension of a $K$-vector does not vary if one enlarges $K$. I suppose it is because the fact that a familly of vectors being linearly dependent is expressible by the determinant of a matrix being non-zero, which does not depend on larger $K$ (this must correspond to some kind of $\exists$-quantifier elimination I suppose).

Question 2. What happens if $K$ is skew ? In that case, there is no notion of determinant of a matrix (I mean, is there a notion of skew determinant ?). Does the notion of $K$-dimension of a $K$-vector space depends on possible enlargements of $K$ ?

Finally:

Question 3. Why does the Zariski dimension of an algebraic subset of $K^n$ ($K$ commutative, non necessarily algebraically closed) does not depend on possible enlargements of $K$ ?

I have a number of questions which seem linked to me, about basic (?) linear algebra:

Given a field (possibly skew) $K$, and an superfield $L$, one can do linear matrix algebra with coefficients in $L$ (any reference for basic facts about that ?)

So given a square matrix $M\in M_{n\times n}(L)$, it is left invertible if and only if it is right invertible. What are the known relations between the left kernel of $M$ in $L^n$ and its right kernel ?

By left kernel, I mean the space (left-vector space) of rows $\lambda\in L_{1\times n}$ (one row $n$ columns) such that $\lambda M=(0)$ and by right kernel, I mean the set of $\lambda \in L_{n\times 1}$ (one column n rows) such that $M\lambda=(0)$.

Question 1. For instance, if there is a non-trivial point of $K^n$ in the left-kernel of $M$, is there also a non-trivial point of $K^n$ in its right kernel?

If $K$ is commutative, I am trying to understand why the notion of dimension of a $K$-vector does not vary if one enlarges $K$. I suppose it is because the fact that a familly of vectors being linearly dependent is expressible by the determinant of a matrix being non-zero, which does not depend on larger $K$ (this must correspond to some kind of $\exists$-quantifier elimination I suppose).

Question 2. What happens if $K$ is skew ? In that case, there is no notion of determinant of a matrix (I mean, is there a notion of skew determinant ?). Does the notion of $K$-dimension of the $K$-vector $L$ space depends on possible enlargements of $K$ ?

Finally:

Question 3. Why does the Zariski dimension of an algebraic subset of $K^n$ ($K$ commutative, non necessarily algebraically closed) does not depend on possible enlargements of $K$ ?

I have a number of questions which seem linked to me, about basic (?) linear algebra:

Given a field (possibly skew) $K$, and a $K$-algebra $\mathcal A$ (again possibly non-commutative, but let us say an integral domain, unitary, acting on both sides of $\mathcal A$), one can do linear matrix algebra with coefficients in $\mathcal A$ (any reference for basic facts about that ?)

So given a square matrix $M\in M_{n\times n}(\mathcal A)$, it is left invertible if and only if it is right invertible. What are the known relations between the left kernel of $M$ in $\mathcal A^n$ and its right kernel ?

Question 1. For instance, if there is a non-trivial point of $K^n$ in the left-kernel of $M$, is there also a non-trivial point of $K^n$ in its right kernel?

If $K$ is commutative, I am trying to understand why the notion of dimension of a $K$-vector does not vary if one enlarges $K$. I suppose it is because the fact that a familly of vectors being linearly dependent is expressible by the determinant of a matrix being non-zero, which does not depend on larger $K$ (this must correspond to some kind of $\exists$-quantifier elimination I suppose).

Question 2. What happens if $K$ is skew ? In that case, there is no notion of determinant of a matrix (I mean, is there a notion of skew determinant ?). Does the notion of $K$-dimension of a $K$-vector space depends on possible enlargements of $K$ ?

Finally:

Question 3. Why does the Zariski dimension of an algebraic subset of $K^n$ ($K$ commutative, non necessarily algebraically closed) does not depend on possible enlargements of $K$ ?

I have a number of questions which seem linked to me, about basic (?) linear algebra:

Given a field (possibly skew) $K$, and a $K$-algebra $\mathcal A$ (again possibly non-commutative, but let us say an integral domain, unitary, acting on both sides of $\mathcal A$), one can do linear matrix algebra with coefficients in $\mathcal A$ (any reference for basic facts about that ?)

So given a square matrix $M\in M_{n\times n}(\mathcal A)$, it is left invertible if and only if it is right invertible. What are the known relations between the left kernel of $M$ in $\mathcal A^n$ and its right kernel ?

By left kernel, I mean the space (left-vector space) of rows $\lambda\in \mathcal A_{1\times n}$ (one row n columns) such that $\lambda M=(0)$ and by right kernel, I mean the set of $\lambda \in \mathcal A_{n\times 1}$ (one column n rows) such that $M\lambda=(0)$.

Question 1. For instance, if there is a non-trivial point of $K^n$ in the left-kernel of $M$, is there also a non-trivial point of $K^n$ in its right kernel?

If $K$ is commutative, I am trying to understand why the notion of dimension of a $K$-vector does not vary if one enlarges $K$. I suppose it is because the fact that a familly of vectors being linearly dependent is expressible by the determinant of a matrix being non-zero, which does not depend on larger $K$ (this must correspond to some kind of $\exists$-quantifier elimination I suppose).

Question 2. What happens if $K$ is skew ? In that case, there is no notion of determinant of a matrix (I mean, is there a notion of skew determinant ?). Does the notion of $K$-dimension of a $K$-vector space depends on possible enlargements of $K$ ?

Finally:

Question 3. Why does the Zariski dimension of an algebraic subset of $K^n$ ($K$ commutative, non necessarily algebraically closed) does not depend on possible enlargements of $K$ ?