Sunday 4 August 2013
Variable x, Ex-variable.
Is the variable $x$ still a variable once a value has been assigned to it? Because then, $x$ is fixed at the assigned value and can no longer vary, and that which cannot vary is not variable.
Saturday 3 August 2013
Coefficient Operation
The trace of a matrix satisfies the following equation
$$tr(A+B)=tr(A)+tr(B)$$
The trace is also the coefficient of the second highest power of the characteristic polynomial (up to a factor of $-1$).
The determinant of a matrix satisfies the following equation
$$det(AB)=det(A)det(B)$$
The determinant is also the constant term of the characteristic polynomial (up to a factor of $-1$).
Thus, for both the trace and the determinant functions, there is a binary operation that is defined for both matrices and numbers, and the functions distribute over the corresponding binary operations. Furthermore, both functions appear in the characteristic polynomial.
If the matrix is 3x3 or larger, then there are many coefficients between that of the determinant and that the second highest power. We can think of these as generalizations of the trace and determinant.
That is, suppose that $f_k(A)$ is the coefficient of $\lambda^k$ in the characteristic polynomial $p(\lambda)$ of the $n\times n$ matrix $A$. Then $tr(A)=f_{n-1}(A)$ and $det(A)=f_0(A)$.
That leads to the following question. For each of the $k$, does there exist a binary operation, say $\circ_k$, that is defined for both square matrices and numbers such that $f_k(A\circ_k B)=f_k(A)\circ_k f_k(B)$ (up to a factor of $-1$)? Note that we want $A\circ_k B$ to be a matrix and $a\circ_k b$ to be a number.
We can take $\circ_{n-1}$ to be the trace function and $\circ_0$ to be the determinant function, so the answer is yes for $k=n-1$ and $k=0$.
The coefficient of $\lambda^n$ is always $\pm 1$. Define $\circ$ by $a\circ b=1$ and $A\circ B=I$. Then $f_n(A\circ B)=f_n(I)=1$ and $f_n(A)\circ f_n(B)=1\circ 1=1$. Thus,
$$f_n(A\circ B)=f_n(A)\circ f_n(B)$$
So we could take $\circ_n$ to be the binary operation $\circ$ defined here. So the answer is yes for $k=n$ as well. Note that $a\circ b$ must be defined to be 1 in order to satisfy the desired property, but $A\circ B$ could be defined to be any matrix, since $f_n$ is always $\pm 1$.
What can be said about $\circ_k$ for $k\in\{1,\ldots,n-2\}$?
There was some freedom in how we defined $\circ_n$. It is not unique. Is this true about the other $\circ_k$, if they even exist?
$$tr(A+B)=tr(A)+tr(B)$$
The trace is also the coefficient of the second highest power of the characteristic polynomial (up to a factor of $-1$).
The determinant of a matrix satisfies the following equation
$$det(AB)=det(A)det(B)$$
The determinant is also the constant term of the characteristic polynomial (up to a factor of $-1$).
Thus, for both the trace and the determinant functions, there is a binary operation that is defined for both matrices and numbers, and the functions distribute over the corresponding binary operations. Furthermore, both functions appear in the characteristic polynomial.
If the matrix is 3x3 or larger, then there are many coefficients between that of the determinant and that the second highest power. We can think of these as generalizations of the trace and determinant.
That is, suppose that $f_k(A)$ is the coefficient of $\lambda^k$ in the characteristic polynomial $p(\lambda)$ of the $n\times n$ matrix $A$. Then $tr(A)=f_{n-1}(A)$ and $det(A)=f_0(A)$.
That leads to the following question. For each of the $k$, does there exist a binary operation, say $\circ_k$, that is defined for both square matrices and numbers such that $f_k(A\circ_k B)=f_k(A)\circ_k f_k(B)$ (up to a factor of $-1$)? Note that we want $A\circ_k B$ to be a matrix and $a\circ_k b$ to be a number.
We can take $\circ_{n-1}$ to be the trace function and $\circ_0$ to be the determinant function, so the answer is yes for $k=n-1$ and $k=0$.
The coefficient of $\lambda^n$ is always $\pm 1$. Define $\circ$ by $a\circ b=1$ and $A\circ B=I$. Then $f_n(A\circ B)=f_n(I)=1$ and $f_n(A)\circ f_n(B)=1\circ 1=1$. Thus,
$$f_n(A\circ B)=f_n(A)\circ f_n(B)$$
So we could take $\circ_n$ to be the binary operation $\circ$ defined here. So the answer is yes for $k=n$ as well. Note that $a\circ b$ must be defined to be 1 in order to satisfy the desired property, but $A\circ B$ could be defined to be any matrix, since $f_n$ is always $\pm 1$.
What can be said about $\circ_k$ for $k\in\{1,\ldots,n-2\}$?
There was some freedom in how we defined $\circ_n$. It is not unique. Is this true about the other $\circ_k$, if they even exist?
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