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In Wikipedia article, there seems to be an error saying that in $4 \times 5$ matrix $\dim(\text{rowspace}) = 4$, but I think it is $5$. Who is wrong?

Passage goes this:

Given a matrix J:

$$J = \begin{bmatrix} 2 & 4 & 1 & 3 & 2\\ -1 & -2 & 1 & 0 & 5\\ 1 & 6 & 2 & 2 & 2\\ 3 & 6 & 2 & 5 & 1\\ \end{bmatrix}$$

the rows are $r_1 = (2,4,1,3,2), r_2 = (−1,−2,1,0,5), r_3 = (1,6,2,2,2), r_4 = (3,6,2,5,1)$. Consequently, the row space of J is the subspace of $\mathbb{R}^5$ spanned by $\{r_1, r_2, r_3, r_4\}$. Since these four row vectors are linearly independent, the row space is 4-dimensional. Moreover, in this case it can be seen that they are all orthogonal to the vector $n = (6,−1,4,−4,0)$, so it can be deduced that the row space consists of all vectors in $\mathbb{R}^5$ that are orthogonal to $n$.

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    The row rank as well as the column rank cannot exceed the number of rows nor the the number of columns. Further, for any matrix, row rank = column rank. – 2017-01-30
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    I couldn't get why your argument implies the row rank of $J$ is 5. The argument in your question shows that the dimension of the row space is 4, and that's done. Could you give your argument why the row rank of $J$ is 5? – 2017-01-30
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    @HanulJeon I wrongly thought that the element numbers of row vectors (=5) is the dimension of the row space. It is 4 as it is spanned by the 4 linearly independent (row) vectors. The element number of the row, 5, shall be meaningful when I think about the dimension of the column space. Many thanks for asking me to clarify this. I realized this by replying to your comment. – 2017-01-30

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We have by definition that the rowspace of a matrix is the vector space spanned by the rows of the matrix, taken as vectors. Combine this with the fact that if a vector space is spanned by $m$ vectors, then a linearly independent set can have at most $m$ elements. This proves that the basis of the rowspace can have no more than the number of rows in the matrix.