# Thread: Linear operators and finding bases

1. ## Linear operators and finding bases

Hi, I'm having trouble with a couple of questions.

A linear operator T on R^4 has the matrix
A= [some 4x4 matrix]
with respect to bases Alpha=Beta=E4.

Find the basis for T(R^4).

-I understand what to do when the question reads find T(1,2,-4,0) or something like that, I'm not sure what to do when it asks me to find a basis for all of R^4.

If someone could please explain the steps on how to tackle this problem I would be appreciative of that.

Find the basis of the kernel of T
and
Find the standard basis of the kernel of T.

If anyone could clarify what is the difference between the two and how the calculation procedures are different I would also appreciate that.

Thank you very much.

2. Originally Posted by Torcida1911
Hi, I'm having trouble with a couple of questions.

A linear operator T on R^4 has the matrix
A= [some 4x4 matrix]
with respect to bases Alpha=Beta=E4.

Find the basis for T(R^4).

-I understand what to do when the question reads find T(1,2,-4,0) or something like that, I'm not sure what to do when it asks me to find a basis for all of R^4.
Consider the following 4 vectors in $R^4$

$e_1 = (1,0,0,0)$, $e_2 =(0,1,0,0)$, $e_3=(0,0,1,0)$, $e_4=(0,0,0,1)$

These vectors are linearly independent and span $R^4$. Thus the vectors form a basis of $R^4$ called the usual basis of $R^4$.

If someone could please explain the steps on how to tackle this problem I would be appreciative of that.

Find the basis of the kernel of T
and
Find the standard basis of the kernel of T.

If anyone could clarify what is the difference between the two and how the calculation procedures are different I would also appreciate that.

Thank you very much.
The standard unit vectors $e_1,e_2,...,e_n$ are linearly independent for if we write $c_1e_1+...c_ne_n =0$
The usual or standar basis in component form then we obtain $c_1...c_n = (0,...0)$, which implies the coeficients are zeroes. Furthermore the vectors span $R^n$ because an arbitrary vector $x = (x_1,...,x_n)$ can be expressed as

$x=x_1(1,0,...0)+x_2(0,1,...,0)+...+x_n(0,0,...,1)= x_1e_1+x_2e_2+...+x_ne_n$

We call $e_1, e_2,...,e_n$ the usual or standard basis for $R^n$.

Edit: I think in your problem to find a basis of the kernel of the map T you must set T(v)=0 where v=(x,y,z,t) and then set corresponding components equal to each other to form a homogeneous system whose solution space is the kernel of T.