Mathematics Asked by Larry G on January 21, 2021
We have an upper triangular matrix A=(aij)n*n i<=n,j<=n;
if i>=j, aij=1;
else aij=0;
how to find a matrix B,st. AB=BA?
Take a polynomial of any degree, for example $p(x)=a_0+a_1x+a_2x^2$ and evaluate your matrix as $$p(A)=a_01!!1+a_1A+a_2A^2,$$ then these are many which you want, that is, $B=p(A)$ is going to commute with $A$.
Adendum:
What I saw in $3times3$ matrices is things like this $$A-1!!1=left[begin{array}{ccc}1&1&1\0&1&1\0&0&1end{array}right]- left[begin{array}{ccc}1&0&0\0&1&0\0&0&1end{array}right]=left[begin{array}{ccc}0&1&1\0&0&1\0&0&0end{array}right],$$ $$(A-1!!1)^2=left[begin{array}{ccc}0&0&1\0&0&0\0&0&0end{array}right],$$ but also $(A-1!!1)^3=0$.
Since $A-1!!1$ and $(A-1!!1)^2$ commutes with $A$ they are in your $$Z(A)={B :mbox{$B=p(A)$ for some polynomial $p$}}.$$
It seems then, that any polynomial in $A$ can be written $$p(A)=a_2(A-1!!1)^2+a_1(A-1!!1)+a_01!!1,$$ settling that $dim Z(A)=3$, in $dim=9$ of the vector space of matrices $3times 3$.
From here, to say something for $ntimes n$ with $nge 4$, is at reach.
Answered by janmarqz on January 21, 2021
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