# Homework Help: A Question on Notation about Linear Algebra

1. Jan 2, 2012

### GreenPrint

Could I write
v = 4i + 3j -2k as
v =
[ 4 |
| 3 | * I_3
| -2 ]

were I_3 is the 3x3 identity matrix and the other thing it's multiplied by is a 3x1 matrix of values 4,3,-2

f(x) = x^2 + 3
can be written as
f: x |-> x^2 + 3

can I write vectors in this matter as well if instead
v = 4x^2 i + 3x j -2x k
as
v: x |-> 4x^2 i + 3x j - 2x k
or something like
v: x |->
[ 4x |
| 3 | * x * I_3
| -2 ]

I'm sort of confused by
f: x |-> and just want to make sure I'm applying it correctly and am wondering if I can use the identity matrix instead of the unit vectors?

2. Jan 2, 2012

### Staff: Mentor

Sure, but I think it adds needless complexity.
This looks OK. Your notation is saying that v maps a number x to a vector-valued function.
I don't see anything wrong with this, but it seems like overkill.

3. Jan 2, 2012

### GreenPrint

Can I write
v = 4i + 3j -2k
as v =
[ 4 |
| 3 |
| -2 ]
or this way
v = [ 4 3 -2]?

I believe this writing it like this would take less time. All you need is three numbers and two brackets [ ]. Writing it like this 4i + 3j -2k requires you write three numbers, three unit vectors, and any plus signs that might be needed. I'm just not sure that
v = [ 4 3 -2] = 4i + 3j -2k is true or maybe if it was a 3x1 matrix instead of a 1x3 this be true? If either of these are I rather would because it's less writing.

4. Jan 3, 2012

### Staff: Mentor

I agree with you about the i, j, k notation taking more effort to write, and usually write vectors in this form: <4, 3, -2>. If they really need to be column vectors, you can add a T superscript (for transpose), like this: <4, 3, -2>T.

5. Jan 3, 2012

### genericusrnme

I hate the i, j, k mainly because my handwriting is terrible and I find i and k to be incredibly hard to write without making everything look messy

nono, it's symbol, domain, codomain, correspondance
$f: \Re \rightarrow \Re ; x \rightarrow x^2+3$
the script R is both the doman and the codomain (R+, the positive reals is technically the codomain(technically the complex numbers is the domain and the codomain))
the x -> x^2 + 3 is the rule of correspondance

if you really wanted to you could but there isn't really much point since you'd end up writing 6 extra zeros

Also you can write $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ as $\begin{pmatrix} x & y & z\end{pmatrix}^T$ or $\begin{pmatrix} x \\ y \\ z \end{pmatrix}^T$ as $\begin{pmatrix} x & y & z\end{pmatrix}$