# Matrix entries

1. May 26, 2009

### jeff1evesque

Question:
Can someone explain the following to me: "... the entries of the matrix $$A - tI_n$$ are not scalars in the field F. They are, however, scalars in another field F(t), the field of quotients of polynomials in t with coefficients from F."

I asked someone earlier today, and I got an explanation that went something like: Entries of a matrix are in the field but since t is a variable, the diagonals of $$A - tI_n$$ are polynomials F(t). And we know the ring of polynomials doesn't necessarily have an inverse. So it follows that $$A - tI_n$$ are not scalars in the field

Questions: Can someone explain to me how the ring of polynomials not having inverses implies that $$A - tI_n$$ are not scalars in the field? Or even how the diagonals being polynomials implies such a conclusion.

Thanks,

JL

2. May 26, 2009

### jbunniii

It's true that you can make a distinction between a polynomial as an object, namely something of the form

$$p(t) = a_n t^n + \ldots + a_0$$

versus the VALUE you get when you evaluate p at some specific $$t = t_0$$:

$$p(t_0) = a_n t_0^n + \ldots + a_0$$

In the first case, we say that p(t) is a polynomial in t with coefficients from F. We think of "t" as some object distinct from any member of F. The set of all such polynomials forms a ring but not a field. (The inverses look like 1/p(t) where p(t) is a polynomial. 1/p(t) is not itself a polynomial in general.) The smallest field containing this ring is the so-called "field of fractions" whose elements look like p(t)/q(t) where p(t) and q(t) are both polynomials. We call this field F(t), which is a "field extension" of F by the element t, i.e., the smallest field that contains both F and t.

In the second case, we are EVALUATING the polynomial at a specific $$t = t_0$$. What sort of object is $$t_0$$? Well, it can be ANY object for which the expression

$$p(t_0) = a_n t_0^n + \ldots + a_0$$

makes sense. In particular, $$t_0$$ could be a member of F, or any subring of F, in which case $$p(t_0)$$ will also be a member of F. Or, $$t_0$$ could be a member of any field G that CONTAINS F, in which case $$p(t_0)$$ will be an element of G but not necessarily of F.

Now, if you aren't going to do anything with your polynomials other than evaluate them at some $$t = t_0$$, then distinguishing between the polynomial as an object versus the VALUE obtained by evaluating the polynomial at $$t_0$$ may be rather pointless.

It looks like you're dealing with eigenvalues, in which case I presume that your t will be taking values in $$\mathbb{C}$$. If $$F = \mathbb{C}$$, then I don't see what is gained by distinguishing between F and F(t).

On the other hand, if F is some other field, then it makes sense to maintain that distinction.

3. May 26, 2009

### HallsofIvy

Staff Emeritus
That's not the way I would have put it. The entries in the matrix $A- tI_n$ are polynomial functions of t not, "scalars in the field". It doesn't really have anything to do with "the ring of polynomials not having inverses", it simply that functions of t are not scalars.

4. May 26, 2009

### jeff1evesque

Jbunni thanks so much for a detailed explanation. Along with HallsofIvy's definition, I understand that the entries of the matrix are polynomial functions of t, and not of the scalars of the field. But how is this the field of quotients of polynomials in t with coefficients from F?

5. May 26, 2009

### jbunniii

If I am interpreting the original quote correctly, all the person was saying is that a polynomial p(t), for the reasons I gave in my last post, is not an element of F. It is an element of "the ring of polynomials with coefficients that are elements of F."

Furthermore, every ring that meets certain conditions (e.g., commutativity, which polynomials satisfy because p(t)q(t) = q(t)p(t) for any two polynomials) can be embedded in a field. What does a field have that a ring doesn't necessarily have? It has multiplicative inverses.

The inverse of a polynomial p(t) is 1/p(t). This is not an element of the ring of polynomials, because it's not a polynomial. But it is an element of the "field of fractions" aka "field of quotients", which is nothing more than the field whose elements look like p(t)/q(t), where p(t) and q(t) are polynomials. Every polynomial lives in this field (just take q(t) = 1), but the field is larger than the ring of polynomials because it also contains inverses.

The field of fractions of polynomials whose coefficients come from F is usually given the standard notation F(t).

I think your friend abused the terminology a bit by saying that a polynomial is a "scalar" in F(t). It's more correct, I think, to say that it is an element or member of F(t), and that F(t) is the smallest field containing all such polynomials.

That's all true, but I'm not sure what point your friend was trying to make, or in what context within linear algebra he/she felt it was necessary to make this distinction. This stuff is important in abstract algebra, specifically in field/Galois theory, but for the purpose of solving the characteristic equation to find the eigenvalues of a linear map (which I'm guessing is what you are doing), I don't see what value it adds.

It is as though you were trying to add 2 + 2, and someone told you, "by the way, did you know that 2 is an element of the ring of integers, which is embedded in the field of rationals, which is embedded in the field of reals, which is embedded in the field of complex numbers?" Sure, but so what?

6. May 27, 2009

### HallsofIvy

Staff Emeritus
In order to think of the set of such matrices as a vector space itself, their entries must be from some field. The set of all polynomials itself is not a field. The smallest field the set of all polynomials belongs to is "the field of quotients of polynomials in t with coefficients from F."