Thread: 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.

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

Where is n-dimensional real vector space.
Thus, there is a standard matrix corresponding to the transformation.

Let be the ordered basis.

We are told that the first vector is an eigenvector of the transformation. That is,

Now relative to the ordered basis, its coordinate vector is,

Because,

Finally, the standard matrix for the linear operator is given by,

But the first coloum in this matrix is,

The coordinate vector relative to the ordered basis.
Which we found to be,

Thus the first entry is and everything else is zero.