# matrix for linear transformation

• Feb 1st 2007, 01:55 PM
PvtBillPilgrim
matrix for linear transformation
Last question ever, I promise.

If T is a linear operator on the space V and the eigenvector of T is a non-zero vector v in V such that Tv=av for some scalar a. Suppose that V is finite-dimensional and that B is an ordered basis of V such that the first vector v is an eigenvector and a is the corresponding eigenvalue. Show that the first column of [T]B has a as its first entry and the rest of those entries are zero.

I have absolutely no idea what to do here.

Any help would be appreciated.
• Feb 1st 2007, 02:15 PM
ThePerfectHacker
Quote:

Originally Posted by PvtBillPilgrim
Last question ever, I promise.

If T is a linear operator on the space V and the eigenvector of T is a non-zero vector v in V such that Tv=av for some scalar a. Suppose that V is finite-dimensional and that B is an ordered basis of V such that the first vector v is an eigenvector and a is the corresponding eigenvalue. Show that the first column of [T]B has a as its first entry and the rest of those entries are zero.

I have absolutely no idea what to do here.

Any help would be appreciated.

This is simple.

We are given a linear operator
$\displaystyle T:V\mapsto V$
Where $\displaystyle V$ is n-dimensional real vector space.
Thus, there is a $\displaystyle n\times n$ standard matrix corresponding to the transformation.

Let $\displaystyle B=(\b{u}_1,...,\b{u}_n)$ be the ordered basis.

We are told that the first vector $\displaystyle \b{u}_1$ is an eigenvector of the transformation. That is,
$\displaystyle T(\b{u}_1)=a\b{u}_1$
Now relative to the ordered basis, its coordinate vector is,
$\displaystyle [T(\b{u}_1)]_B=(a,0,...,0)$
Because, $\displaystyle a\b{u}_1=a\b{u}_1+0\b{u}_2+...+0\b{u}_n$

Finally, the standard matrix for the linear operator is given by,
$\displaystyle [T]_B=\big[[T(\b{u}_1)]_B\big|[T(\b{u}_2)]_B\big|....\big|[T(\b{u}_n)]_B\big]$
But the first coloum in this matrix is,
$\displaystyle [T(\b{u}_1)]_B$
The coordinate vector relative to the ordered basis.
Which we found to be,
$\displaystyle \left[\begin{array}{c}a\\0\\0\\...\\0 \end{array} \right]$
Thus the first entry is $\displaystyle a$ and everything else is zero.