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enter image description here First, I used row elementary operation on the matrix, $$\begin{pmatrix} 1 & 2 & 3 & 4\\ 1 & -1 & 2 & 3\\ 1 & -3 & 3 & 5\\ 1 & 4 & 2 & 2\\ \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 2 & 3 & 4\\ 0 & 3 & 1 & 1\\ 0 & 0 & 5 & 8\\ 0 & 0 & 0 & 0\\ \end{pmatrix}$$

I had referred to Difference between dimension and rank of matrix before asking this and I am aware that $\dim(V) = \operatorname{rank}(A) + \operatorname{null}(A)$. So, I believed that the dimension of the vector space $V$ is $3 + 1 = 4$. However, I checked my answers and found out the dimension is only $3$. Could someone please explain to me why?

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    $\begingroup$ ``dimension" of a matrix doesn't exist, the theorem should state as $\dim(V) = \text{rank}(A) + \text{null}(A)$. $\endgroup$ – Zhanxiong Oct 22 '15 at 2:10
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    $\begingroup$ Just so that you're clear, it's not the dimension of the matrix, but rather the dimension of the vector space that the matrix acts on. The arithmetic $3+1=4$ describes a linear map from a $4$-dimensional space to itself, whose range consists of a $3$-dimensional subspace, and the kernel (null space) is $1$-dimensional. $\endgroup$ – Sammy Black Oct 22 '15 at 2:13
  • $\begingroup$ thanks, I did an edit @Zhanxiong $\endgroup$ – Lily L Oct 22 '15 at 2:13
  • $\begingroup$ oh alright, thank you for the clarification! @SammyBlack $\endgroup$ – Lily L Oct 22 '15 at 2:18
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    $\begingroup$ @LilyL Thank you, it's clear now. If you know that the dim of the range space is the rank of the matrix (actually it is one of the various ways to define the rank of a matrix), the problem has been solved by yourself. $\endgroup$ – Zhanxiong Oct 22 '15 at 3:03

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