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(1) We are given $A,B \in R^{n\times n}$ diagonal matrices of n rows and n columns with real values.

Show that there are $X \in R^{n\times n}$ and polynomials $q$ and $p$ such that:

$$q(X)=B \text{ and }p(X)=A$$

What I've tried doing:

I've tried a different combinations of X and p,q, for example:

If we say that $X=A$ and we define $q(X) = X-A+B$ and $p(X)=X$ we indeed solve the problem BUT those p and q are not polynomials, because if i use a scalar as an input for those polynomials instead of a matrix, it doesn't work. There is no scalar version of the matrix A (Unlike for example the unit matrix, which the scalar version of which is 1) So that isn't the way to solve the question

(2) Is this correct for any 2 matrices or just for diagonal matrices? Hint: What basic difference is there between the polynomial ring and the matrix ring?

(Haven't started working on this yet cause didn't solve question 1, but it probably involves that the matrix multiplication isnt commutative)

Community
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Oria Gruber
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  • Please try to make the titles of your questions as informative as possible. Your question will get much more attention if people can read the problem it pertains to in the title -- rather than that it's a problem from your homework. – Lord_Farin Nov 11 '13 at 14:48

2 Answers2

2

Hint:

$$X=\begin{pmatrix}x_1&&\\&\ddots&\\&&x_n\end{pmatrix}$$

$$P(X)=\begin{pmatrix}P(x_1)&&\\&\ddots&\\&&P(x_n)\end{pmatrix}$$

xavierm02
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1

Hints:

  1. Lagrange interpolation would be useful if $X$ is a diagonal matrix with distinct diagonal entries.
  2. The polynomial ring over $\mathbb{R}$ is commutative, but the matrix ring is not. So, $p(X)$ always commutes with $q(X)$ but ...

Remarks:

(a) Note that in question 1, a solution $(p,q,X)$ exists not only because $A$ and $B$ are diagonal matrices, but also because the underlying field ($\mathbb{R}$) is infinite. There can be no feasible solution if the field is finite. For instance, consider $A=\operatorname{diag}(1,0,0)$ and $B=\operatorname{diag}(0,1,0)$ over $GF(2)$. There are only $512$ choices of $X$ and eight choices of each of $p(X)$ and $q(X)$. One can perform a brute-force search to verify that none of these choices solve the set of equations $p(X)=A,\,q(X)=B$.

(b) The wording of the question may lead one to (wrongly) believe that two commuting matrices $A$ and $B$ can always be written as polynomials of some matrix $X$. However, in the previous remark, we have already seen that this is not the case even if $A$ and $B$ are diagonal matrices. For more in-depth discussions, see MSE/326293 and MO/34314.

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user1551
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  • Okay, I've tried to solve this with Lagrange interpolation (we haven't learned it) but what's supposed to come out of it is that you get a polynomial that consists of linear combination s.t for each $x_i$ you have an element $x_0\cdot \frac {x-x_i}{p(x_0)- p(x_i)}$ . Putting X in this polynomial gives a certain matrix. I thought to myself maybe this means that you can get n different polynomials (each time choosing $x_0$ as a different $x_i$). Is this true? Something seems still strange to me, because to me this doesn't neccesarily mean you can get EVERY A,B to come out of these polynomials. – jreing Nov 11 '13 at 18:12
  • @user1685224 The polynomials $p$ and $q$ depend on $A,B$ and the choice of $X$. Let's say $A=$diag$(a_1,\ldots,a_n)$ and $X=$diag$(x_1,\ldots,x_n)$. You need to find a polynomial $p$ such that $p(x_i)=a_i$ for all $i$. That is, you need to fit a single polynomial curve $p$ to $n$ data points $(x_1,a_1),\ldots,(x_n,a_n)$. This is what Lagrange interpolation does. The expression you wrote in your comment is not right. See Wikipedia or MathWorld for the correct formula. – user1551 Nov 11 '13 at 19:24