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I used to believe that there is no difference between $\mathbf{v}=(2,3)$ which is a vector lying in $xy$ plane and $\mathbf{u}=(2,3,0)$ which is a three dimensional vector but still lying in $xy$ plane. So, for me both the vectors $\mathbf{v}$ and $\mathbf{u}$ were same. The analogy I used was that both the tail and head coincide for $\mathbf{v}$ and $\mathbf{u}$.

But on the second page of the textbook I am using as reference (Gilbert Strang-Introduction to linear algebra), I found something different . The author says,

The vector $(x,y)$ in a plane is different from vector $(x,y,0)$ in $3-$space.

That's it. Since this statement is not well explained there so I am here. Hoping for help.

Since I am beginning the Linear algebra course so Please show some tolerance. Thanks.

José Carlos Santos
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Vidyanshu Mishra
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3 Answers3

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They're different because they're members of different vector spaces. After all, the vector $(2,3)$ is defined with respect to the standard basis $e_1, e_2$; what's to say that the vector $(2,3,0)$ has been defined wrt a basis whose first two vectors are $e_1, e_2$ and whose third vector is orthogonal to those two?

If you wanted to abuse notation, you could write that $(2,3) = (2,3,0)$. But this is so non-standard that it would confuse everyone who read it. There's no good reason, for instance, that you should miss out the third component rather than the second: why don't you have $(2,3) = (2,0,3) = (2,0)$?

Vector spaces often don't canonically embed into other vector spaces, so there's often no canonical way of identifying a vector in one space with a vector in another.

Patrick Stevens
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  • I think I m confusing things here.. what does vector (2,3) means?? Doesn't that mean x component equals 2, y component equals 3 and z component is 0?? – Vidyanshu Mishra May 18 '17 at 07:41
  • $(2,3)$ is a vector in the space of real pairs. It does not have a $z$ component. Read what @Patrick has said again. – ancient mathematician May 18 '17 at 07:45
  • I can't have (2,3,0)=(2,0)=(2,3) as I have to take care of equality of components which I followed while saying (2,3,0)=(2,3). – Vidyanshu Mishra May 18 '17 at 07:45
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    You didn't follow an equality. $(2,3)$ does not have a $z$-component to even compare! – Nij May 18 '17 at 07:50
  • @VidyanshuMishra But why were you justified in removing the final zero component rather than, say, the first zero component? That is, why is $(0,2,3)$ not equal to $(2,3)$? (That's just what happens if you rotate the space through 90 degrees, and I'm sure you agree that equality shouldn't change under rotation.) – Patrick Stevens May 18 '17 at 07:50
  • Well yes, then what I m doing wrong?? I am considering a component equal to 0 which doesn't even exist? – Vidyanshu Mishra May 18 '17 at 07:53
  • @VidyanshuMishra That is what you are doing wrong, yes. – Patrick Stevens May 18 '17 at 07:53
  • @VidyanshuMishra There are several ways one can view a plane as being part of the greater space $\mathbb{R}^3$, and no one of those ways is mathematically distinguished over the others. One could take the $x,y$-plane, the $y,z$-plane, the plane perpendicular to the vector $(1,3,2)$ through the origin, and so on. All these ways are basically equivalent, and in all of them, $(2,3)$ refers to a different vector in the greater $\mathbb{R}^3$. – Patrick Stevens May 18 '17 at 08:01
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The vector $(2, 3)$ is an element of the two-dimensional vector space $\mathbb R^2$. The vector $(2, 3, 0)$ is an element of the three-dimensional vector space $\mathbb R^3$. What is the difference between the two?

There are infinitely many ways of embedding $\mathbb R^2$ in $\mathbb R^3$. By "embedding," we mean an injective isomorphism of vector spaces. The image of an embedding $\mathbb R^2 \to \mathbb R^3$ is just a plane in $\mathbb R^3$ which contains the origin $(0, 0, 0)$, so the choice of an embedding is (sort of) equivalent to the choice of a plane through the origin. There are infinitely many such planes, and each plane can be thought of as a "copy" of $\mathbb R^2$ sitting inside of $\mathbb R^3$.

When you think of $(2, 3)$ as the same as $(2, 3, 0)$, what you are really doing is thinking of $\mathbb R^2$ as the $(x, y)$-plane in $\mathbb R^3$, as you said. In this case you have chosen to embed $\mathbb R^2$ in $\mathbb R^3$ via the map $(x, y) \mapsto (x, y, 0)$. But there are infinitely many other embeddings you could have chosen.

I think the point the author of your textbook is trying to make is this: elements of two-dimensional vector spaces are not elements of three-dimensional vector spaces. If you want to consider elements of $\mathbb R^2$ as elements of $\mathbb R^3$, you have to choose an embedding.

manthanomen
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They are different because for example we can't add or subtract them. Try ;)

Widawensen
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  • Well, I can think of a way to add and subtract them. It is just not the usual addition in $\Bbb R^2$ or in $\Bbb R^3$ (though it could be quite similar to the latter). I don't think this kind of reasoning is very helpful. I can add $3\in \Bbb R$ to a polynomial, to a function, and even to an $n\times n$ matrix, but that does not make $3$ a polynomial, a function, or a matrix. – Marc van Leeuwen May 18 '17 at 08:40
  • @MarcvanLeeuwen Hmm, such addition would be far beyond the standard definition of addition in vector spaces. If we use non-standard definitions of operations we could do in fact almost everything, it would be interesting for example how we could define, proposed by you, addition of a number to a matrix ? I think, if possible, it could be made with several ways. – Widawensen May 18 '17 at 10:06
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    You did not say "add in a specific vector space", just "add", and that is what I was criticising. One can add (the imaginary unit) $\mathbf i$ to $-2$, although one cannot do this using the addition of the real vector space $\Bbb R$ in which $-2$ lives; making the distinction is needed. As for adding numbers to square matrices, they are interpreted, as in any (unitary) ring, as that multiple of the multiplicative unit, here the identity matrix$~I_n$. Such a convention is needed for identities like the binomial formula (in commutative rings): its coefficients are numbers, not ring elements. – Marc van Leeuwen May 18 '17 at 10:16
  • @MarcvanLeeuwen Interesting procedure,... thank you for the explanation. – Widawensen May 18 '17 at 10:19