Are a row vector and a column vector the same thing?

Question

Suppose I have $$ A = \begin{bmatrix} a\\b\\c\\d\\ \end{bmatrix}$$

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

Now, I know $A = B^T$. But in what sense are these different mathematical objects?

$$ \begin{bmatrix} 1\\ 2\\ 3\\ 4\\ \end{bmatrix} + \begin{bmatrix} 1\\ 2\\ 3\\ 4\\ \end{bmatrix}= \begin{bmatrix} 2\\ 4\\ 6\\ 8\\ \end{bmatrix} $$

$$ \begin{bmatrix} 1&2&3&4\\ \end{bmatrix} + \begin{bmatrix} 1&2&3&4\\ \end{bmatrix}= \begin{bmatrix} 2&4&6&8\\ \end{bmatrix} $$

To me, these seem to behave the same way. Is there any difference between a row vector and a column vector? How are $A$ and $B$ different?

They are different when it comes to multiplication, if you consider them matrices. i.e. using your example if M is a 4x4 matrix, then you can write MB or AM, but you can't write BM or MA — Bamboo, May 21 '15 at 03:08
When you make operations you just need to be careful with regard to the implementation of functions and appropriately organize your vectors and matrices - as you observe from your example. — Moti, May 21 '15 at 03:09
It's worth noting that, in an abstract sense, the relationship between the space of row vectors and the space of column vectors is that one is the dual space $V^$ of the other, which, for finite dimensional spaces, is isomorphic. It's also worth noting that, if you have a map (like transposition) taking $V$ to analogous elements of $V^$, you are now working in an inner product space. (This is a case where abstracting might help you understand what structures you're implicitly using when you work in a "concrete" example like $\mathbb R^n$.) — Milo Brandt, May 21 '15 at 03:14

score 3 · Accepted Answer · answered May 21 '15 at 03:08

3

These two representations do work in the same way. The set of all row vectors over a giving ring, $R$, is isomorphic to the set of all column vectors over the same ring, $C$, via the linear transformation $\phi: R \rightarrow C$ defined by $\phi (v)=v^T$. You can check that this is a bijection and preserves scalar multiplication since you already proved that they behave the same under addition.

But I will add that often vectors of a given space are represented as column vectors as convention.

answered May 21 '15 at 03:08

TheBluegrassMathematician

826

How does the existence of this linear transformation relate to the row space / column space distinction? Is the column space simply a linear transformation of all the vectors in the row space? – Stan Shunpike May 21 '15 at 03:13
Transpose is a linear map? – Stan Shunpike May 21 '15 at 03:13
In this case, yes the transpose is a linear map... We're only dealing with finite dimensional row and column vectors, not matrices. – TheBluegrassMathematician May 21 '15 at 03:30
Are you saying the transpose doesn't act as a linear map for matrices? – Stan Shunpike May 21 '15 at 03:30
It does since $(kA+kB)^T=kA^T +k B^T$. – TheBluegrassMathematician May 21 '15 at 03:41

score 2 · Answer 2 · answered May 21 '15 at 03:11

Taken by themselves, they are isomorphic to each other. However, in matrices, the rows and columns specify two different vector spaces. There are special relationships between row spaces and column spaces of a matrix but they are different ways of specifying a space or linear transformation. See here.

score 2 · Answer 3 · answered May 21 '15 at 03:24

2

Once you declare that you are dealing with matrices, then they must be different: one is $1$ x $4$ and the other is $4$ x $1$. But there is an obvious isomorphism between them: $[a, b, c, d]\rightarrow \begin{bmatrix} a\\b\\c\\d\\ \end{bmatrix}$

answered May 21 '15 at 03:24

Matematleta

29,139

Are a row vector and a column vector the same thing?

3 Answers3