A question about (0,0,0) vector

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SUMMARY

The only vector in the kernel (ker) of a matrix is (0,0,0), which is indeed a vector. According to the rank-nullity theorem, the dimension of the kernel, dim(ker A), is calculated as the number of columns of matrix A minus its rank. Since (0,0,0) is the only element in the kernel, it forms a subspace with dimension 0, confirming that a basis for this kernel is empty. Therefore, the dimension of the kernel is 0.

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when i find that the only vector of ker is (0,0,0)

what is the dim(ker)

is it 1
or is it 0??

does (0,0,0)
counts as a vector

or in case if we find two vectors and one of them is (0,0,0)

what is the dimension of ker now ??
 
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Let A be a matrix. By the rank-nullity theorem, rank A + nullity A = rank A + dim(ker A) = number of columns of A, say n(A). So dim(ker A) = n(A) - rank A.

What is the null space of a full-rank matrix? What is its rank?
 
Yes, (0,0,0) is a vector. In fact, since adding (0,0,0) to itself gives, again, (0,0,0) and multiplying (0,0,0) by any number gives, again, (0,0,0), (0,0,0) is the only vector in which the set containing only it is a subspace! Since any vector in a basis must be non-zero, a "basis" for {(0,0,0)} must be empty: its dimension is 0.
 

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