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If the kernel of a matrix is the $0$ vector, why is the basis of the kernel non existent?

If I have the matrix $\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}$

The kernel is the $0$ vector

Why is the basis of the kernel non-existent? Shouldn't the basis of the kernel be $0$?

Brad
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    It's not that the basis doesn't exit so much as the basis is the empty set, if that makes sense. – Jürgen Sukumaran Feb 24 '16 at 03:43
  • @Prospect I think made a typo. I know that the basis exists. I do not understand why the basis of the kernel in the matrix above does not exist. – Brad Feb 24 '16 at 03:43
  • Sorry, I misread your question. See Tony's comment. – Future Feb 24 '16 at 03:44
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    A basis is a set of linearly independent vectors. A set of vectors is linearly independent if no nontrivial linear combination of them is zero. Therefore, ${0}$ is not a linearly independent set. –  Feb 24 '16 at 03:50

1 Answers1

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This is by definition. The dimension is defined to be the minimum cardinality of the sets of vectors that span $V$. In the case $V=\{0\}$, we have that any set of vectors in this space will be linearly dependent and thus cannot be a basis.

Jürgen Sukumaran
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