# Determinant question

1. Jan 23, 2010

### talolard

Hello all. I have a queston about determinants. The question is from an exam and solutions were not published. I would like to know if my solution is correct. Please excuse me for the imperfect formatting, I am struggling with the interface. Espeacially the superscripts were supposed to be subscripts.
1. The problem statement, all variables and given/known data

The question is to compute the determinant of a matrix of polynomals of degree n-2 or less that looks like this:

F$$_{1}$$(X$$_{1}$$) F$$_{1}$$(X$$_{2}$$) ... F$$_{1}$$(X$$_{n}$$)

F$$_{2}$$(X$$_{1}$$) F$$_{2}$$(X$$_{2}$$) ... F$$_{2}$$(X$$_{n}$$)

$$\cdots$$
$$\cdots$$
$$\cdots$$
F$$_{n}$$(X$$_{1}$$) F$$_{n}$$(X$$_{2}$$) ... F$$_{n}$$(X$$_{n}$$)

I assumed n=3 and showed that in this case the determinant is 0.
I then assumed correctness for n and want to show that this is true for n+1 thus proving by induction.
Writing the same matrix but with n+1 rows and columns, I compute the determinant using expansion by row. I "crossed out" the n+1 row and n+1 column and so have a sum of sclars (which are polynomials) multiplying the NXN matrix that I assumed has a 0 determinant. So the sum is 0 hence conculding the proff.
Is this correct?
Thank you
Tal

2. Jan 23, 2010

### jambaugh

Sounds right. The polynomials in question were given? It would seem not to be the case that det = 0 for the 1x1 case unless the first polynomial was given to be zero. Could you give more details on the problem?

BTW w.r.t. formatting, the inline latex type doesn't always line up vertically with normal text so you should avoid mixing. Rather put all your equation inside a single tex block.

The editor now has a latex reference ($$\Sigma$$ button in advanced editor)
You can also use pmatrix, bmatrix, Bmatrix and vmatrix environments for respectively matrices in (),[],{},||. Also the array environment for more control.

try: e.g.

\begin{vmatrix}
F_1(x_1) & F_1(x_2) &\cdots & F_1(x_n)\\
F_2(x_1) & F_2(x_2) &\cdots & F_2(x_n)\\
\vdots & \vdots & \ddots & \vdots\\
F_n(x_1) & F_n(x_2) & \cdots &F_n(x_n)
\end{vmatrix}

For:

$$\begin{vmatrix} F_1(x_1) & F_1(x_2) &\cdots & F_1(x_n)\\ F_2(x_1) & F_2(x_2) &\cdots & F_2(x_n)\\ \vdots & \vdots & \ddots & \vdots\\ F_n(x_1) & F_n(x_2) & \cdots & F_n(x_n) \end{vmatrix}$$

Also see the AMS's "ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf"[/URL] for a good reference on latex commands (not all implemented here, I don't think, but nothing missing that I've every tried to use.)

Last edited by a moderator: Apr 24, 2017
3. Jan 23, 2010

### talolard

Thanks for the reply. The only information about the Polynomials is that they are of degree n-2 at most. From this I assumed that n greater than or equal to 3. We never encountered polynomals of a negative degree and intuitvely i imagine that there is "no such thing". Was the assumption wrong? My logic is that for inדtance X-2=(1/x)2
Thanks
Tal

4. Jan 23, 2010

### jambaugh

I see. That means e.g. for n=2 you have degree zero, i.e. constants. Yes, thus you have rows with the same value and thus multiples of each other and hence det=0.

I now see a possible problem with your inductive proof. Given the polynomials are of degree at most n-2, as you let n increase you must also let the degree of the submatrix entries increase. Your assertion about say n = K no longer holds when n=K+1 since you haven't shown, and can't assume it is the case that the determinants of the submatrices is zero. They are not restricted to being degree K-2 but rather (K+1)-2.

Hmmm.... tricky. Ahhh yes. The space of polynomials of degree n-2 will be n-1 dimensional. That's easy enough to show since you have a basis x^k for k = 0 to n-2.

I would then expand each polynomial as a row vector of coefficients times a column vector of powers of x. So:

$$F_k(x_j)= \begin{pmatrix} c_{k0}& c_{k1} &\cdots & c_{km}\end{pmatrix} \begin{pmatrix}x_j^0\\x_j^1\\ \vdots \\ x_j^m\end{pmatrix}$$
where m = n-2.

Do this for each row or your matrix and for each variable you can then write:
$$\begin{pmatrix} F_1(x_1) & F_1(x_2) &\cdots & F_1(x_n)\\ F_2(x_1) & F_2(x_2) &\cdots & F_2(x_n)\\ \vdots & \vdots & \ddots & \vdots\\ F_n(x_1) & F_n(x_2) & \cdots &F_n(x_n) \end{pmatrix} = \begin{pmatrix} c_{1,0}& c_{1,1} &\cdots &c_{1,m}\\ c_{2,0}& c_{2,1} &\cdots &c_{2,m}\\ \vdots & \vdots & \ddots & \vdots\\ c_{n,0}& c_{n,1} &\cdots &c_{n,m}\end{pmatrix} \begin{pmatrix} x_1^0 & x_2^0& \cdots &x_n^0\\ x_1^1 & x_2^1& \cdots &x_n^1\\ \vdots & \vdots & \ddots & \vdots\\ x_1^m & x_2^m& \cdots &x_n^m\end{pmatrix}$$

The dimensions correspond to: $$[n\times n] = [n\times(n-1)]\cdot[(n-1)\times n]$$.

Looking at the matrix of coefficients and recalling that the space the rows span is only n-1 dimensional I would then argue that one row is a linear combination of the others, thus so too is one row of the original polynomial matrix and hence its determinant is zero.

(Since all the columns of the variable powers matrix are equal a given linear combination of rows of the coefficient matrix will correspond to a linear combination of rows of the polynomial matrix.)

There, that's how I'd prove it.

5. Jan 24, 2010

### talolard

Thanks. That seems like an even simpler proof. It took me a few minutes to see where my proof didn't work. It was a little subtle. Thanks for pointing that out.
Tal