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Consider the linear transformation $T:\mathbb{R}^2\rightarrow \mathbb{R}^3$ such that $T(x_1,x_2)=(x_1-x_2, x_2-x_1, -x_1)$. Find dimension of null space.

Now in this question, I find that the only member of the null space is the zero vector. So the dimension should be $1$. But the answer is $0$. Why is it so ?

Mathematician 42
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Ka Sikh
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  • The dimension of the zero space is zero, not one. – Mathematician 42 Apr 25 '17 at 07:30
  • How did you find that out – Ka Sikh Apr 25 '17 at 07:31
  • Dear Ka Sikh, since you are rather new to SE I would like to remind you that mathematical expressions should be written with the help of the MathJax software to ensure readability of your questions :) – Jakob Elias Apr 25 '17 at 07:32
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    By definition of dimension. The dimension of a vector space is defined as the number of elements in a basis. By definition, the set $\left{0\right}$ is not linearly independent (Indeed, $5\cdot 0=0$ is a non-trivial linear combination of the vectors in $\left{0\right}$ that is zero), hence it's not a basis of the zero space. In fact the zero space is generated by a basis with no elements. – Mathematician 42 Apr 25 '17 at 07:35

1 Answers1

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The subspace consisting of only the zero vector, has dimension $0$.

Take a look at "Why $\mathbf{0}$ vector has dimension zero?" for more details, but note that in this question/title, it should be the "subspace $\left\{ {\bf 0} \right\}$" rather than only the "element ${\bf 0}$".

Now in this question, I find that the only member of the null space is the zero vector. So the dimension should be $1$. But the answer is $0$. Why is it so ?

If the dimension would be $1$, any basis for this subspace would consist of exactly one (non-zero) vector (by the definition of dimension), since a basis has to be linearly independent. But then the subspace spanned by this basis necessarily has an infinite number of elements, since all scalar multiples of the basis vector are in the subspace.

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