# Does matrix A belong to span(A_1,A_2, )

1. Mar 17, 2005

Please excuse the lack of latex, I'm in somewhat of a hurry and I do not have the time to lookup the latex syntax for matrices. With that out of the way, here is my question?

For an upcoming test question we were given example questions that may or may not appear on the test.
One question was given as:

Does matrix A belong to span(A_1,A_2,A_3)
A_1 = (1 2)
(3 4)
A_2 = ....
A_3 = ....

or for which values of $$\alpha$$
--------
NOTE:
The syntax of A_1 is just what I'm using it to make it readable, thus:
a_11 = 1
a_12 = 2
a_21 = 3
a_22 = 4
--------

So, as you can see the question is not really detailed. It's just to give us an idea of the question. If the question was something like:

does $$\vec{V}$$ belong to span(A_1,A_2) where A_1=(1,4,3)^T , A_2=(-2,-1,0)^T

Then I definitely understand how to solve the problem. I'm just confused with the comment of, does this matrix belong, and where A_1 will equal something like:
(a b)
(c d)

Would it make sense to solve the problem like this:
* If we use the given problem above.
----
Does matrix A belong to span(A_1,A_2,A_3)
A_1 = (1 2)
(3 4)
A_2 = ....
A_3 = ....
----
* Then write A_n as a vector like:
Does the matrix A (where A=something like (a1,a2,a3,a4)^T ) belong to span(A_1,A_2,A_3)
where A_n = something like (a,b,c,d)^T

Ok, sorry if this is confusing... I just feel like I don't understand well enough what the question is asking. Also, one last thing... when someone says span(A_1,A_2)
where A_1 = (a b)
(c d)
does that mean that A_1 is in the vector space R^4 and is that the same as R^(2x2)?

thank you. and again I apologize for not using tex. (i'm just in a hurry here)
:)
thanks guys

2. Mar 17, 2005

### HallsofIvy

Actually, any 2 by 2 matrix can be considered to be in R^4 which is, of course, the same as R^(2x2) because 2x2= 4! To say that A belongs to the span of A1, A2, A3 means that A is a linear combination of A1, A2, and A3. that is that A= aA1+ bA2+ bA3 for some numbers a, b, c. Since, in this example every matrix has 4 components, this gives 4 equations for the 3 numbers a, b, c. The fact that 4 "independent" equations in 3 unknowns does not necessarily have a solution is what leads to the question about whether that is possible at all.

3. Mar 17, 2005

### AKG

A matrix is a vector in the sense that the set of (m x n) matrices form a vecto space. Just check for yourself (check the rules for addition and scalar multiplication). So, to see if a vector v is in Span{v1, v2, ..., vk}, see if you can find a linear combination of the vi such that v equals it, i.e.

$$v = a_1v_1 + \dots + a_kv_k$$

In your case, the $a_i$ will probably be real numbers, and v and the $v_i$ will be 2x2 matrices. This will give you 4 equations (one equation for the 1,1 element of v = the 1,1 element of the linear combination, one equation for the 1,2 element of v = the 1,2 element of the lin. comb, etc.) with k unknowns (a1, ..., ak). If a solution for the ai exists, it is in the span.

And yes, you could treat the 2x2 matrices as elements of R^4. They are NOT the same, but expressing the matrix:

(a b)
(c d)

with respect to the standard basis will give you the co-ordinates (a, b, c, d) which looks just what you're used to seeing when you talk about vectors of R^4. Note that really, (a, b, c, d) is not a vector of R^4, but an expression of a vector (in some 4-dim vector space) with respect to some basis. And, of course, R^4 is the same as R^(2x2) since 2x2 = 4 ;).

4. Mar 18, 2005

Ok awesome, I definitely understand the problem now. Yeah, one would think that $$\Re^{2 \times 2}$$ would be equal to $$\Re^{4}$$ since 2x2=4 :) But sometimes I think math notation can be confusing, such as:

$$sin^{-1}(x) \neq \frac{1}{sin(x)}$$

- or -

$$y=x^2$$
$$y'=2x=2x^1$$
yet then someone could then use: $$x'$$ as a variable, and it has no relation to the derivative.

- or -

$$\frac{\partial^2}{\partial x^2} \neq \frac{\partial}{\partial x} \times \frac{ \partial}{\partial x}$$

etc... I mean yes you can definitely get used to it, and it works. But I'd just rather ask a question instead of assuming something. :)

so anyways, thank you... I appreciate it.