2
$\begingroup$

While doing real data analysis I came up with a problem. I have given lots of efforts to solve it and could not succeed.

Here is the problem:

Suppose, we have a set of vectors $X_1,X_2,\ldots,X_n$with dimension $1,\ldots,n$, respectively.

I want to define a notion of addition between two vectors of different dimensions which will be consistent with the usual vector addition in a n-dimensional vector space (that is componentwise addition of two same dimensional vector).

Here is some of my try: Suppose $X=x_1$ and $Y=(y_1,y_2)'$ (prime denote transpose) than I define $X+Y=(x_1+y_1,x_1+y_2)'$.

Again if $X_1=x_1$ and $Y=(y_1,y_2,y_3)'$ then $X+Y=(x_1+y_1,x_1+y_2,x_1+y_3)'$

and if $X_1=(x_1,x_2)$ and $Y=(y_1,y_2)'$ then $X+Y=(x_1+y_1,x_2+y_2)'$ (usual vector addition).

But if $X_1=(x_1,x_2)$ and $Y=(y_1,y_2,y_3)'$ then how to define the addition between this two vectors such that this operation is consistent with the usual vector addition? And how to proceed for the other vectors of different dimensions.

Any idea or useful references will be gratefully appreciated.

Thanks in advance.

$\endgroup$
4
$\begingroup$

The easiest way to do this is to extend $X = (x_1, x_2)$ to $X' = (X_1, x_2, 0)$ (and otherwise just always extend vectors with as many zeroes as you need). The resulting space is infinite dimensional, and it is in some sense the "smallest" infinite-dimensional vector space construction there is. Polynomials, for instance, work this way if you ignore multiplication.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Could you please explain the polynomial case for me. $\endgroup$ – Janak Jan 7 '15 at 14:13
  • $\begingroup$ Looking at $1+3x - 28x^2$ as the vector $(1, 3, -28)$, you get the common vector space notion of polynomials. You can also see that if you add two polynomials of different degree (i.e. dimension), the lower degree polynomial gets "padded" with zeroes until they have the same degree, and then you just add them together, like this: $$ (1+3x - 28x^2) + (3 + 7x +2x^2 - x^3)\\ = (1 + 3x - 28x^2 + 0x^3) + (3 + 7x + 2x^2 - x^3)\\ = (1 + 3) + (3 + 7)x + (-28 + 2)x^2 + (0-1)x^3\\ = 4 + 10x - 26x^2 - x^3 $$ $\endgroup$ – Arthur Jan 7 '15 at 14:19
  • $\begingroup$ Thanks for the explanation. Could you please indicate some references for further study in this direction. $\endgroup$ – Janak Jan 7 '15 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.