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From linear algebra we know that the rank of a matrix is the maximal number of linearly independent columns or rows in a matrix. So, for a matrix, the rank can be determined by simple row reduction, determinant, etc. However, I am wondering how the concept of a rank applies to a single vector, i.e., $\mathbf{v} = [a, \ b, \ c]^{\top}$. My intuition suggests that the rank must be equal to 1, but I'm not even sure if it is defined for a vector. Can anyone help shed some light on this issue?

Thanks in advance.

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    $\begingroup$ Warning: not all vectors have the form of a column of numbers. This is not how we define things in mathematics (any more). A vector is simply an element in a vector space. A vector space is defined axiomatically. A vector in it has no a-priori shape or form. $\endgroup$ – Ittay Weiss Sep 3 '20 at 9:23
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    $\begingroup$ You may consider $1$ for a non-zero vector, $0$ for the zero vector. $\endgroup$ – Yves Daoust Sep 3 '20 at 9:25
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    $\begingroup$ See thread below why that would not be a good idea. $\endgroup$ – Ittay Weiss Sep 3 '20 at 9:50
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A vector is an element in a vector space. As such, it has no rank. A matrix, in the context of linear algebra, is interesting because it represents a linear transformation between vector spaces. It's the linear transformation we care about, not the rectangle of numbers we call a matrix. A linear transformation has a rank and that rank is the dimension of the image of the linear transformation. It's an interesting concept since it's a measurement of how large the linear transformation is. It just so happens that when the linear transformation is represented as a matrix the rank can be computed in various convenient ways.

With this in mind, the superficial resemblance of a vector to a matrix is misleading. A vector is not a representation of a linear transformation between vector spaces. For that reason, it simply makes no sense to ask what the rank of a vector is.

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    $\begingroup$ the rank of a $1 \times p$ matrix is defined just like the rank of any matrix. It's $0$ if all entries are $0$ and $1$ otherwise. It simply has nothing to do with the rank of a vector, nor should the superficial resemblance of this matrix to a vector lead one to any fantasies about rank of a vector making any kind of sense. $\endgroup$ – Ittay Weiss Sep 3 '20 at 10:10
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    $\begingroup$ @BenGrossmann the rank of that transformation is at most $1$. Regardless, this only works if you choose the standard basis in $\mathbb R ^p$ and the basis $1$ in $\mathbb R$ to represent the linear transformation. If the vector you start with is in an infinite dimensional vector space, then this is no longer possible. So, this sort of observation (without being explicit about the choice of bases) reinforces the confusion between a vector/transformation and its representation as a column/matrix. There is no such thing as "the matrix representing a transformation". You need to specify bases. $\endgroup$ – Ittay Weiss Sep 3 '20 at 11:13
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    $\begingroup$ @BenGrossmann my main aim is to emphasise the difference between a transformation and its representations. I know that this is usually avoided in many approaches to elementary algebra but that does not mean it's a good thing. The apparent simplicity of saying "a vector is just a column of numbers and a matrix is just a rectangle of numbers" is a double-edged sword. Regardless, your claim that $\mathbb R^p$ has a canonical basis is incorrect. There is nothing canonical about the standard basis. It's merely a very useful one, hence naming it standard. But it is not canonical. $\endgroup$ – Ittay Weiss Sep 3 '20 at 11:28
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    $\begingroup$ very nice answer emphasizing the fact that the concept of "rank" should be reserved specifically to linear transformations, and that a vector by itself is not a linear transformation (unless of course one unnecessarily assumes finite-dimensionality to identify $V\cong V^{**}$... but most likely if a student is capable of understanding this isomorphism more than likely they wouldn't have this question). Also, while matrices are extremely useful for computational purposes, they're a conceptual nightmare $\endgroup$ – user580918 Sep 3 '20 at 11:41
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    $\begingroup$ @BenGrossmann even if you want to concentrate on vector spaces and linear transformations of a particular form (say only consider $\mathbb R^n$) it is still not the case that there is a canonical basis. The whole point of diagonalisation is to represent a given matrix in terms of (usually) the non-standard basis. So, the standard basis is not only not canonical but also mainly useful due to a particular syntactic representation of vector spaces. If you call that a question of elegance and choose to portray that negatively, well, ok, I guess. $\endgroup$ – Ittay Weiss Sep 3 '20 at 11:56
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Another definition of rank of a matrix is the dimension of the vector space spanned by its column, so any non-zero vector spans a space of dimension 1.

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  • $\begingroup$ It is not true to claim that a vector is a matrix with just one column. In the vector space of all continuous functions on the reals (just as an example) a vector is a continuous function. Not a matrix. $\endgroup$ – Ittay Weiss Sep 3 '20 at 9:16
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    $\begingroup$ But you are still implying that a vector is a matrix of a special form. It's irrelevant that a vector spans a subspace of dimension at most 1. The vector is not a representation of a transformation, so it's got no rank. $\endgroup$ – Ittay Weiss Sep 3 '20 at 9:20
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    $\begingroup$ @YvesDaoust I disagree. In what sense does a single non-zero vector in a high dimensional vector space have full rank? If anything, all this is saying is that a single vector is linearly independent unless it is the zero vector. That is not a new concept and it has nothing to do with rank. What is being suggested is superficially a natural extension but it ignores the fact that rank is a concept defined for a linear transformation, not to vectors at all. Rows, columns, and rectangles of numbers are merely representations of vectors and linear transformations. And not even all of them. $\endgroup$ – Ittay Weiss Sep 3 '20 at 9:32
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    $\begingroup$ Moreover, saying "an extension of rank from matrices to vectors" already is problematic. An extension suggests that someone vector is more general than matrix. That is manifestly incorrect. The two are completely different concepts. It's just so happens that under a particular representations it appears that a vector is a special form of matrix. $\endgroup$ – Ittay Weiss Sep 3 '20 at 9:35
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    $\begingroup$ @YvesDaoust yes, there are several reasons to refuse this generalisation. Firstly, it's not a generalisation (the vector concept does not extend the matrix notion). Second, it does not capture anything other than "a single vector is linearly independent unless it is zero" so it gives nothing new. Thirdly, it adds to the confusion where some people think a vector is a matrix of a special form where in fact the two notions are completely distinct. $\endgroup$ – Ittay Weiss Sep 3 '20 at 9:37

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