Conjugacy

**Deadstar** · November 22nd 2009, 02:46 PM

Let $H = \Bigg{(} \begin{array}{cccc} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{array} \Bigg{)}<br />$

Show that $[1,0,0,0], [1,2,0,0], [1,2,3,0], [1,2,3,4]$ are conjugate with respect to H.

**Deadstar** · November 23rd 2009, 02:04 AM

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 $a^1, a^2, a^3, \dots , a^n$
for $j = 1..n\textrm{, }a_j^i = j \textrm{ if } j \leq i \textrm{ and } 0$ , otherwise.

**tonio** · November 23rd 2009, 02:12 AM

Originally Posted by Deadstar

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 $a^1, a^2, a^3, \dots , a^n$
for $j = 1..n\textrm{, }a_j^i = j \textrm{ if } j \leq i \textrm{ and } 0$ , otherwise.

It could probably help if you defined what is 'vectors conjugate under matrix H"...
The new definition of the vectors isn't crystal clear, either: does it mean $a^1=(1,0,...,0)\,,\,a^2=(1,2,0,...,)$ , etc.?

Tonio

**Deadstar** · November 23rd 2009, 02:22 AM

Yes thats right about the vectors, not sure about the conjugate thing you were saying...

Basically I want to show (although Im sure you know this) that $(a^p)^T H a^q = 0$ for $p \neq q$

**tonio** · November 23rd 2009, 02:33 AM

Originally Posted by Deadstar

Yes thats right about the vectors, not sure about the conjugate thing you were saying...

It wasn't me talking about conjugate stuff but you, in your first post

Basically I want to show (although Im sure you know this) that $(a^p)^T H a^q = 0$ for $p \neq q$

Well, no: I didn't know this, I'm not sure about its meaning, either algebraic or geometric, but it's rather easy to show the given relation.
Nice lil' thing.

Tonio

**Deadstar** · November 23rd 2009, 02:38 AM

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 $(a^p)^T H a^q = 0$ ? I really don't know where to start...[/COLOR]

EDIT: this is for all vectors, not just any two.

**tonio** · November 23rd 2009, 03:39 AM

Originally Posted by Deadstar

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 $(a^p)^T H a^q = 0$ ? I really don't know where to start...[/color]

EDIT: this is for all vectors, not just any two.

Let $a^i\,,\,a^j\,,\,\,i\neq j$ be any two $1\times n$ vectors as you defined them before, and suppose first that $i<j$ , then:

$\left(1\,,2\,,...\,i\,,0\,,...\,,0\right)\left(\be gin{array}{rrrrrr}2&-1&0&0&...&0\\-1&2&-1&0&...&0\\...&...&...&...&...&...\\0&...&0&-1&2&-1\\0&0&...&0&-1&2\end{array}\right)\left(\begin{array}{c}1\\2\\. ..\\j\\0\\...\\0\end{array}\right)=$ $\left(0\,,0\,,...\,,1+i\,,-i\,,0\,,...\,0\right)\left(\begin{array}{c}1\\2\\. ..\\j\\0\\...\\0\end{array}\right)=\left(0\,,0\,,. ..\,,0\right)$

Why? Let us denote by $\alpha_k=\left(\begin{array}{r}\\0\\...\\-1\\2\\-1\\0\\...\\0\end{array}\right)$ the k-th COLUMN of the matrix, with the first -1 in the $(k-1)$ -th positiion.

Now, while $k<i$ we get $\left(1\,,2\,,...,i\,,0\,,...,0\right)\left(\begin {array}{r}\\0\\...\\-1\\2\\-1\\0\\...\\0\end{array}\right)=m\cdot (-1)+(m+1)\cdot 2+(m+2)\cdot (-1)=0$ , for some $m<i$

If $k=i$ , we get $\left(1\,,2\,,...,i\,,0\,,...,0\right)\left(\begin {array}{r}\\0\\...\\-1\\2\\-1\\0\\...\\0\end{array}\right)=-(i-1)+2i=i+1$ , and if $k=i+1$ then $\left(1\,,2\,,...,i\,,0\,,...,0\right)\left(\begin {array}{r}\\0\\...\\-1\\2\\-1\\0\\...\\0\end{array}\right)=-i$ .

For $k\geq i+1$ 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 $a^iH=(0\,,...\,,0\,,i+1\,,-i\,,0\,,...,0)$ , and since the first non-zero entry is in the i=k-th position and $i>j$ , we'll get $\left(a^i\right)^tHa^j=(0,...,0)$ .

Now you try to do the case $i<j$ following the above lines.

Tonio

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