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Q&A

If a matrix has lots of values in a column but, not in row than, what that actually called?

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$\begin{bmatrix}a & b & c & d\end{bmatrix}$

$\begin{bmatrix}a \\ b \\ c \\ d \end{bmatrix}$

Which one is $1$ dimensional Matrix? I think that first matrix is $1$ dimensional. But, when we put $2$ values in row and two values in column. Then, we call that $2$ dimensional Matrix. So, what's the dimension of second matrix? It has lots of values in column and, only $1$ in every row.

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Answer (1 comment)
Wikipedia (1 comment)

1 answer

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in short

Both matrix, in the example, can be considered 1 dimensional matrix, or 4 dimensional vectors.

some explanation

for convenience, let's consider v1, v2 such as:


       a
v1 = [ b ]
       c

v2 = [a, b, c]

where a,b,c are elements of R

  • v1 is a vector, a column vector of R^3, a one dimensional matrix,

    or a 3x1 matrix

  • v2 is a vector, a row vector of R^3, a one dimensional matrix,

    or a 1x3 matrix


but how it comes that both v1 and v2 has the same dimension?

ok, first what is a dimension (matrix dimension)?

What is a dimension?

def 1: dimension of a matrix

The dimensions of a matrix are the number of rows by the number of columns. If a matrix has a rows and b columns, it is an a×b matrix.

_ High school definition

if we add to this definition the statement that either of v1 or v2 is a matrix of dimension 1 (which is technically correct, but not with this definition), things will become understandably confusing.

But, if you accept that:

  • each algebraic vector (just a set of scalar elements) can be considered a matrix of dimension 1 ( the same way that each real number x, can be seen as a complex number x + i.0 ),
  • and that v1 (or v2) is an algebraic vector by def.

(which you already do, since your question imply that one matrix at list is of dimension 1)

we can check why the other is of the same dimension.

in our examples, you can see that:

  • the row vector v2 is the same as the column vector v1, with the difference that the one is row while the other is a column.
  • and that we can transform v1 to v2 by writing v1 rows as columns for v2,

This transformation is called a transposition; and, in this case, v2 is called the transpose of v1 (and vice-versa)


With this definitions and the rules of matrix multiplication/addition, you can demonstrate that:

(1) : for each vector v of M_3x1 (the group of matrix of dimension 3x1),

exists a unique element u of M_1x3 (the group of matrix of dimension 1x3),

such as

u = T(v)

and

(2) :

T(v + w) = T(v) + T(w) and T(k.v) = k.T(v)

where T represents transform operation, and k an element from R

  • (1): means that transpose is a bijection (or correspondence) from 3x1 to 1x3
  • (2): means that transpose is a linear transformation: it preserves structure

a transformation that has this tow proprieties is called an isomorphism, and if there is one between 2 vectorial spaces, they are considered equivalent.

You can think of it as the relation between a group of students and their IDs, (assuming that each one of them has a unique ID) while they are not the same, counting the students or their IDs will give the same result..

And in case of spaces of finite dimension the existing of such relationship between them, also imply that they have the same dimension.

def 2: dimension of vector space

In mathematics, the dimension of a vector space V is the cardinality (i.e. the number of vectors) of a basis of V over its base field.

_ wikipedia

I think that this definition* and the examples associated with it may make it easier to understand why both vectors can be considered of dimension one, by writing each of them in a base of dimension 1.

*after some simplification: cardinality? basis?


https://en.wikipedia.org/wiki/Dimension_(vector_space) https://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces https://en.wikipedia.org/wiki/Rank_(linear_algebra)#Proofs_that_column_rank_=_row_rank http://www.math.lsa.umich.edu/~speyer/417/Transpose.pdf

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1 comment thread

Technically (almost) correct but inconsistent and misleading (1 comment)

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