Let $K$ be a field, $K\subseteq \Bbb C$. $V$ is a linear space over $K$, $\dim(V)=n(n\geq2)$. Choose ordered basis $\epsilon_1,\epsilon_2,\dotsc,\epsilon_n$ for $V$.
$\bf A,B$ are two linear transformations from $V$ to $V$, with transformation matrices $A, A^*$ relative to the ordered basis , respectively. ($A^*$ is the adjugate matrix of $A$). then
1). $\bf AB=\bf BA$
2). Let $M=\{\alpha\in V|\bf B\alpha =0\}$. If $0$ is a eigenvalue of $\bf A$, Can one find a basis of $M$?
For 1), Since $AA^*=A^*A$ , so $\bf AB=\bf BA$
For 2), $\dim(M)=n $ or $n-1$. But what is a basis of $M$?
Any help will be appreciated!
If $0$ is an eigenvalue of $A$, then $A^*A=\det(A)I=0$, so that the columns of $A$ is in $M$, and $M$ is not trivial - so yes, then we can find a basis for $M$. How? Well , the same as for finding the kernel of any linear transformation...just solve the homogenous system with coefficient matrix, the matrix representation of the linear transformation, in this case $A^*$.
For further investigation: does the columns of $A$ span $M$? If so, then the basis can just be derived by finding a linearly independent subset from the columns of $A$.
Edit. After thinking about it I realized, if $0$ is an eigenvalue of $A$, and $A$ is rank $n-1$, then the columns of $A$ do indeed span $M$, as there is at least one nonzero cofactor of some entry in $A$ so that $A^*$ is not the zero matrix.
And if $A$ is rank $n-2$ or less, then all the cofactor matrices of the entries of $A$ are rank deficient, so that $A^*=0$ and so then $V=M$.