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I found a strange question

give me the basis of kernel?

i go to find the set of kernel by slove the eqation f(u) = O

i do a matrix of the systhem linear (f(u)=0) that i found but

the matrix is note invertible that mean the

set of solution is empty set

i mean that kernel is a Empty set

the problem how i find basis of a Empty set ?

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1 Answer 1

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The kernel of a linear transformation is never empty, since for any vector spaces $V$ and $W$, and any linear transformation $f:V\to W$, it must be true that $$f(0_V)=0_W$$ and therefore $0_V\in\ker(f)$. However, it is true that $$f\text{ is injective}\iff \ker(f)=\{0_V\}$$ in which case you have the question of how to find a basis of the subspace $\{0_V\}$ of $V$ (again, this is not empty). But even though the subspace $\{0_V\}$ is not empty, the empty set $\varnothing$ is a basis of that subspace.

Note that, in particular, it is wrong to say that $\{0_V\}$ is a basis of $\{0_V\}$, since for example, $$17\cdot 0_V = 0_V$$ shows how a nontrivial linear combination of the set $\{0_V\}$ can produce the zero vector.

Any linear combination of the vectors in the set $\varnothing$ (of course, there actually aren't any vectors in $\varnothing$) that produces the zero vector must be a trivial linear combination (after all, which of the coefficients is a non-zero scalar?)

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  • $\begingroup$ ya the kernel never empty thx u that mean i have problem in calculation $\endgroup$
    – zerocool
    Aug 31, 2016 at 1:11
  • $\begingroup$ but how u find the basis of set has just the zero vector $\endgroup$
    – zerocool
    Aug 31, 2016 at 1:12
  • $\begingroup$ The empty set $\varnothing$ is a basis of $\{0_V\}$ - I've added more to my answer. $\endgroup$ Aug 31, 2016 at 1:15
  • $\begingroup$ ok thanks u bro $\endgroup$
    – zerocool
    Aug 31, 2016 at 1:17

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