Question

Let T: $R^{3} \rightarrow R^{3}$ be defined by T(x,y,z) = (4x+3y+2z, 2x+y—z, −x+y). Find a basis and dimension of kernel of T.

Answer 1

Answer:

T(x,y,z) = (0,0,0)  <=>  (x,y,z) = (0,0,0)     .... (*)

That is, T is non-singular.

To see this, either:

• use row operations on the matrix to find that you don't get a zero row
• or calculate the determinant 0 + 3 + 4 - (-2) - (-4) - 0 = 13 to see that it is non-zero
• or do it from first principles:  -x+y = 0 => x = y, then 2x+y-z = 0 => z = 3x and 4x + 3y + 2z = 0 => 2z = -7x = 2(3x) = 6x => x = 0 => y = z = 0

However you have shown that T is non-singular, the statement in (*) says that the kernel of T is the 0 vector space.  Its dimension is 0.  A basis is the empty set { }.