Let $\displaystyle H = \Bigg{(} \begin{array}{cccc} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{array} \Bigg{)}
$
Show that $\displaystyle [1,0,0,0], [1,2,0,0], [1,2,3,0], [1,2,3,4]$ are conjugate with respect to H.
Let $\displaystyle H = \Bigg{(} \begin{array}{cccc} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{array} \Bigg{)}
$
Show that $\displaystyle [1,0,0,0], [1,2,0,0], [1,2,3,0], [1,2,3,4]$ are conjugate with respect to H.
Actually I've just realized that it should be n-dimensional vectors and matrix...
So the matrix just continues in the same way but the vectors are defined as...
We have n vectors $\displaystyle a^1, a^2, a^3, \dots , a^n$
for $\displaystyle j = 1..n\textrm{, }a_j^i = j \textrm{ if } j \leq i \textrm{ and } 0$, otherwise.
To be honest im not entirely sure either all I have is that definition to go on. Our lecturer tends to SAY most of the important stuff rather than write it down so you tend to miss lots of information...
So, given the vectors defined as you said, and the matrix H as in my first post.
How could I show that $\displaystyle (a^p)^T H a^q = 0$? I really don't know where to start...[/COLOR]
EDIT: this is for all vectors, not just any two.
Let $\displaystyle a^i\,,\,a^j\,,\,\,i\neq j$ be any two $\displaystyle 1\times n$ vectors as you defined them before, and suppose first that $\displaystyle i<j$, then:
$\displaystyle \left(1\,,2\,,...\,i\,,0\,,...\,,0\right)\left(\be gin{array}{rrrrrr}2&-1&0&0&...&0\\-1&2&-1&0&...&0\\...&...&...&...&...&...\\0&...&0&-1&2&-1\\0&0&...&0&-1&2\end{array}\right)\left(\begin{array}{c}1\\2\\. ..\\j\\0\\...\\0\end{array}\right)=$ $\displaystyle \left(0\,,0\,,...\,,1+i\,,-i\,,0\,,...\,0\right)\left(\begin{array}{c}1\\2\\. ..\\j\\0\\...\\0\end{array}\right)=\left(0\,,0\,,. ..\,,0\right)$
Why? Let us denote by $\displaystyle \alpha_k=\left(\begin{array}{r}\\0\\...\\-1\\2\\-1\\0\\...\\0\end{array}\right)$ the k-th COLUMN of the matrix, with the first -1 in the $\displaystyle (k-1)$-th positiion.
Now, while $\displaystyle k<i$ we get $\displaystyle \left(1\,,2\,,...,i\,,0\,,...,0\right)\left(\begin {array}{r}\\0\\...\\-1\\2\\-1\\0\\...\\0\end{array}\right)=m\cdot (-1)+(m+1)\cdot 2+(m+2)\cdot (-1)=0$ , for some $\displaystyle m<i$
If $\displaystyle k=i$ , we get $\displaystyle \left(1\,,2\,,...,i\,,0\,,...,0\right)\left(\begin {array}{r}\\0\\...\\-1\\2\\-1\\0\\...\\0\end{array}\right)=-(i-1)+2i=i+1$ , and if $\displaystyle k=i+1$ then $\displaystyle \left(1\,,2\,,...,i\,,0\,,...,0\right)\left(\begin {array}{r}\\0\\...\\-1\\2\\-1\\0\\...\\0\end{array}\right)=-i$.
For $\displaystyle k\geq i+1$ the above product yields zero (since the first -1 in the column vector mutiplies the i+1 entry of the vector, which already is zero, and the same for the next ones)
We see, thus, that the only contribution in the above product is in the k-th and the (k+1)-th columns, and thus $\displaystyle a^iH=(0\,,...\,,0\,,i+1\,,-i\,,0\,,...,0)$ , and since the first non-zero entry is in the i=k-th position and $\displaystyle i>j$ , we'll get $\displaystyle \left(a^i\right)^tHa^j=(0,...,0)$.
Now you try to do the case $\displaystyle i<j$ following the above lines.
Tonio