The matrices are both of size m x n over some field F, obviously. The first direction of this proposition is clear enough, however the opposite direction (same nullspace -> row-equivalence) is challenging to me. Would be thankful for a hint, at the very least.
2
-
$\begingroup$ Let $A,B$ be the matrices that defines linear maps from $E$ to $F$. $(e_1, \dots, e_m, e_{m+1}, \dots ,e_n)$ be a basis of $E$ such that $(e_1, \dots, e_m)$ is a basis of the shared null space. How can you map $A.e_j$ to $B.e_j$ for $m+1 \le j \le n$ in order to conclude? $\endgroup$– mathcounterexamples.netCommented May 24 at 19:06
-
$\begingroup$ Each row operation is a multiplication on the left by an invertible matrix. Such multiplication obviously doesn't affect the null space. $\endgroup$– KakashiCommented May 24 at 19:18
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Let $A, B$ be the two matrices, that define linear maps from the space $E$ and $F$.
Let $(e_1, \dots e_l, e_{l+1}, \dots, e_n)$ be a basis of $E$ where $(e_1, \dots e_l)$ is a basis of the shared null space. $A.e_{l+1}, \dots, A.e_n$ are linearly independent in $F$. If necessary, complete this family of vectors into a basis $(A.e_{l+1}, \dots, A.e_n, a_{n+1}, \dots, a_m)$ of $F$. In a similar way build a second basis $(B.e_{l+1}, \dots, B.e_n, b_{n+1}, \dots, b_m)$. The linear map $P$ that maps those two basis vector to vector is invertible and satisfies the equality $B=PA$. You’re done.
answered May 24 at 20:19