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I have been under the impression that the dimension of a matrix is simply whatever dimension it lives in. More precisely, if a vector space contained the vectors $(v_1, v_2,...,v_n)$, where each vector contained $3$ components $(a,b,c)$ (for some $a$, $b$ and $c$), then its dimension would be $\Bbb R^3$.

Recently I was told this is not true, and the dimension of this vector space would be $\Bbb R^n$.

In other words, I was under the belief that the dimension is the number of elements that compose the vectors in our vector space, but the dimension is how many vectors the vector space contains?!

Where am I going wrong?

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  • $\begingroup$ It is not true that the dimension is the number of vectors it contains. If this were the case, then $\mathbb{R}$ would have dimension infinity... $\endgroup$ – copper.hat Mar 4 '13 at 19:19
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    $\begingroup$ my APOLOGIES. i was actually told the number of vectors in any BASIS of V is the dim[v] $\endgroup$ – Johnathon Svenkat Mar 4 '13 at 19:20
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    $\begingroup$ Your vectors have $3$ coordinates/components. The vector space $\mathbb{R}^3$ has dimension $3$, ie every basis consists of $3$ vectors. $\endgroup$ – Julien Mar 4 '13 at 19:21
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    $\begingroup$ @JohnathonSvenkat: That is the definition of dimension, so is necessarily true. If you have a collection of vectors, and each has three components as in your example above, then the dimension is at most three. $\endgroup$ – copper.hat Mar 4 '13 at 19:24
  • $\begingroup$ so what would dim[M_2(R)] be? 4? $\endgroup$ – Johnathon Svenkat Mar 4 '13 at 19:25
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The dimension of a vector space is the number of coordinates you need to describe a point in it. Thus, a plane in $\mathbb{R}^3$, is of dimension $2$, since each point in the plane can be described by two parameters, even though the actual point will be of the form $(x,y,z)$.

If you take the rows of a matrix as the basis of a vector space, the dimension of that vector space will give you the number of independent rows. For a vector space whose basis elements are themselves matrices, the dimension will be less or equal to the number of elements in the matrix, this $\dim[M_2(\mathbb{R})]=4$

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  • $\begingroup$ then why is the dim[M_2(r)] = 4? by the first line of your definition wouldn't it just be 2? $\endgroup$ – Johnathon Svenkat Mar 4 '13 at 19:28
  • $\begingroup$ @JohnathonSvenkat - no. The dimension of a single matrix is indeed what I wrote. The dimension of a vector space who's basis is composed of $2\times2$ matrices is indeed four, because you need 4 numbers to describe the vector space. $\endgroup$ – nbubis Mar 4 '13 at 19:32
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    $\begingroup$ I would argue that a matrix does not have a dimension, only vector spaces do. We can ask for the number of rows and the number of columns of a matrix, which determine the dimension of the image and codomain of the linear mapping that the matrix represents. $\endgroup$ – Chris Godsil Mar 4 '13 at 19:34
  • $\begingroup$ @ChrisGodsil - good point. I'll clarify my answer. $\endgroup$ – nbubis Mar 4 '13 at 19:36
  • $\begingroup$ great. thanks guys $\endgroup$ – Johnathon Svenkat Mar 4 '13 at 19:39
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The following literature, from Friedberg's "Linear Algebra," may be of use here:


Definitions. A vector space is called finite-dimensional if it has a basis consisting of a finite number of vectors. The unique number of vectors in each basis for $V$ is called the dimension of $V$ and is denoted by $\dim(V)$.


A basis, if you didn't already know, is a set of linearly independent vectors that span some vector space, say $W$, that is a subset of $V$.

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The dimension is the number of bases in the COLUMN SPACE of the matrix representing a linear function between two spaces. i.e. if you have a linear function mapping R3 --> R2 then the column space of the matrix representing this function will have dimension 2 and the nullity will be 1. This is a result of the rank + nullity theorem --> e.g. dimension of R3 = rank(col(A)) + null(A), or 3 = 2 + 1. Essentially, one of the basis vectors in R3 collapses (or is mapped) into the 0 vector (the kernel) in R2.

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  • $\begingroup$ This answer would be improved if you used mathJax formatting (LaTeX syntax). Cheers $\endgroup$ – 6005 Sep 16 '16 at 19:25

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