If two linear transformations between finite-dimensional vector spaces, we say
$$L,T : V \longrightarrow W$$
has the same kernel, and kernel is not the space zero. Is $T$ a multiple of $L$?
Asked
Active
Viewed 933 times
2
-
Of course not. For example are bijections $\mathbb R^n \to \mathbb R^n$ multiples? – Andres Mejia Feb 01 '19 at 05:10
-
What would happen if $W = \mathbb{C}$ and $Ker(T)\neq 0$? – Pedro Zapata Feb 01 '19 at 05:14
2 Answers
0
No, for example take the linear map $f\in\operatorname{End} K^2$ associated (using the standard basis) to the matrix $$ \begin{bmatrix} 0 & 1 \\ 1&0 \end{bmatrix} $$ This matrix is invertible and hence $f$ has trivial kernel. The identity map on $K^2$ is also invertible (of course!), but $f$ and $\operatorname{id}$ are not multiples of each other.
For a general counterexample to your edited question. Let $K$ be a field and $n>1$. Consider the maps $f, g\in \operatorname{End} K^n$ defined by $f(x)=(x_1,0,\ldots,0)$ and $g(x)=(x_1,\ldots,x_1)$.
ffffforall
- 400
-
-
It would still fail. For example, row operations (multiplication by elementary matrices) preserves the kernel, but will almost always yield matrices that are not multiples of each other. Thus you can start with your favorite matrix that is not of full rank, perform some row operations and get a counterexample. – ffffforall Feb 01 '19 at 05:22
-
Thanks for the answers, I think I should have asked what happens in the case of linear functions. – Pedro Zapata Feb 01 '19 at 05:29
-
-
-
Yes, it is true then, see https://math.stackexchange.com/questions/890101/if-two-functionals-have-the-same-kernel-then-one-is-a-multiple-of-the-other and the two linked questions at the top. – ffffforall Feb 01 '19 at 05:58
-
0
Hint: Take square matrices of order $2$ like $A=[a_{ij}]$ where $a_{11}=1, a_{12}=a_{21}=a_{22}=0$ and $B=[b_{ij}]$ where $b_{21}=1, b_{11}=b_{12}=b_{22}=0$
Nitin Uniyal
- 7,946