Basis Problem

Patient · **Joined:** Fri, 1 Apr 2011 19:49:14 UTC **Posts:** 3

Hello everyone.

Let

be a 3-D space over a field

with basis B = (v_1,v_2,v_3)

and consider linear operators $x,y,z:V\rightarrow V$ whose matrices relative to

are:

$X=\left(\begin{array}{ccc}0&1&0\\0&0&0\\0&1&0\end{array}\right), Y=\left(\begin{array}{ccc}0&2&0\\1&0&-1\\0&2&0\end{array}\right), Z=\left(\begin{array}{ccc}1&0&-1\\0&0&0\\1&0&-1\end{array}\right)$

Find vectors $u_1,u_2,u_3\inV$ such that

x(u_1)=y(u_1)=z(u_1)=0

$x(u_2),y(u_2),z(u_2)\in \ V_1=ku_1$ with $u_2\notin \ V_1$
$x(u_3),y(u_3),z(u_3)\in \ V_2=ku_1\oplus \ ku_2$ with $u_3\notin V_2]$

I have been asked to find these vectors so that the matrices of x,y,z

with respect to u_1,u_2,u_3

are stricly upper triangular.

My inital procedure was to first find a common eigenvector for x,y,z

and take that as u_1

, but I couldn't find a common eigenvector that worked, when I calculated u_2, u_3

, the respective matrices for x,y,z

were not stricly upper triangular. I then tried $u_1=\left(\begin{array}{c}1\\0\\1\end{array}\right)$ but that didn't work. Any help would greatly appreciated.

outermeasure · **Posted:** Sat, 2 Apr 2011 05:57:26 UTC

Patient wrote:

Hello everyone.

Let

be a 3-D space over a field

with basis B = (v_1,v_2,v_3)

and consider linear operators $x,y,z:V\rightarrow V$ whose matrices relative to

are:

$X=\left(\begin{array}{ccc}0&1&0\\0&0&0\\0&1&0\end{array}\right), Y=\left(\begin{array}{ccc}0&2&0\\1&0&-1\\0&2&0\end{array}\right), Z=\left(\begin{array}{ccc}1&0&-1\\0&0&0\\1&0&-1\end{array}\right)$

Find vectors $u_1,u_2,u_3\inV$ such that

x(u_1)=y(u_1)=z(u_1)=0

$x(u_2),y(u_2),z(u_2)\in \ V_1=ku_1$ with $u_2\notin \ V_1$
$x(u_3),y(u_3),z(u_3)\in \ V_2=ku_1\oplus \ ku_2$ with $u_3\notin V_2]$

I have been asked to find these vectors so that the matrices of x,y,z

with respect to u_1,u_2,u_3

are stricly upper triangular.

My inital procedure was to first find a common eigenvector for x,y,z

and take that as u_1

, but I couldn't find a common eigenvector that worked, when I calculated u_2, u_3

, the respective matrices for x,y,z

were not stricly upper triangular. I then tried $u_1=\left(\begin{array}{c}1\\0\\1\end{array}\right)$ but that didn't work. Any help would greatly appreciated.

What do you mean by " $u_1=\begin{pmatrix}1\\0\\1\end{pmatrix}$ doesn't work"?

Patient · **Joined:** Fri, 1 Apr 2011 19:49:14 UTC **Posts:** 3

Taking $u_1=\left(\begin{array}{c}1\\0\\1\end{array}\right)$ gives Xu_1=Yu_1=Zu_1=0

but then letting $u_2=\left(\begin{array}{c}a\\b\\c\end{array}\right)$ and $x(u_2),y(u_2),z(u_2)\in \ V_1=ku_1$ with $u_2\notin \ V_1$ gives $Xu_2=\left(\begin{array}{c}b\\0\\b\end{array}\right)=\left(\begin{array}{c}k\\0\\k\end{array}\right)$ for some $k\in K$ . $Yu_2=\left(\begin{array}{c}2b\\a-c\\2b\end{array}\right)=\left(\begin{array}{c}m\\0\\m\end{array}\right)$ for some $m\in K$ . $Zu_2=\left(\begin{array}{c}a-c\\0\\a-c\end{array}\right)=\left(\begin{array}{c}n\\0\\n\end{array}\right)$ for some $n\in K$ . This doesn't really get me anywhere to finding out what u_2

is.

I tried to work out u_3

similarly. This method doesn't give u_1, u_2, u_3

so that the respective matrices are upper triangular. BUt that's probably because I'm going about it in entirely the wrong way.

outermeasure · **Posted:** Sat, 2 Apr 2011 14:46:11 UTC

Patient wrote:

Taking $u_1=\left(\begin{array}{c}1\\0\\1\end{array}\right)$ gives Xu_1=Yu_1=Zu_1=0

but then letting $u_2=\left(\begin{array}{c}a\\b\\c\end{array}\right)$ and $x(u_2),y(u_2),z(u_2)\in \ V_1=ku_1$ with $u_2\notin \ V_1$ gives $Xu_2=\left(\begin{array}{c}b\\0\\b\end{array}\right)=\left(\begin{array}{c}k\\0\\k\end{array}\right)$ for some $k\in K$ . $Yu_2=\left(\begin{array}{c}2b\\a-c\\2b\end{array}\right)=\left(\begin{array}{c}m\\0\\m\end{array}\right)$ for some $m\in K$ . $Zu_2=\left(\begin{array}{c}a-c\\0\\a-c\end{array}\right)=\left(\begin{array}{c}n\\0\\n\end{array}\right)$ for some $n\in K$ . This doesn't really get me anywhere to finding out what u_2

is.

I tried to work out u_3

similarly. This method doesn't give u_1, u_2, u_3

so that the respective matrices are upper triangular. BUt that's probably because I'm going about it in entirely the wrong way.

It is the right method (modulo the fact that you are expressing these vectors with respect to the basis B without saying so in your posts), and you should have some idea what u_2

must be like. Of course you cannot determine uniquely what u_2

must be, because you can scale any basis element, as well as adding multiples of u_1

which is in the kernel.

So from Yu_2

you know a-c=0, so $u_2=\begin{pmatrix}a\\b\\a\end{pmatrix}$ . Indeed any $b\in k^\times=k-\{0\}$ and any $a\in k$ would do. Can you see why?

And you can choose u_3

almost freely --- the only constraint is that $u_3\notin\mathop{\mathrm{span}}\{u_1,u_2\}$ . Can you see why?

Patient · **Joined:** Fri, 1 Apr 2011 19:49:14 UTC **Posts:** 3

I think I understand, I previously calculated u_1, u_2, u_3

the same way, so taking $u_1=\left(\begin{array}{c}1\\0\\1\end{array}\right), u_2=\left(\begin{array}{c}a\\b\\a\end{array}\right)$ as in your post, and taking $u_3=\left(\begin{array}{c}\alpha\\\beta\\\gamma\end{array}\right)$ such that $u_3 \notin span\{u_1,u_2\}$ .

My problem comes when I try to calculate the matrices of the linear operators x,y,z

with respect to $\{u_1,u_2,u_3\}$ . I can't understand how these matrices become stricly upper triangular. Another problem I hadn't previously stated was that I am supposed to write down an explicit change of basis matrix between $\{u_1,u_2,u_3\}$ and $\{v_1,v_2,v_3\}$ , which I don't know how to do with such arbitrary $\{u_1,u_2,u_3\}$ and without knowing $\{v_1,v_2,v_3\}$ .

Thanks for your responses, I appreciate the help.

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Basis Problem

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