Question

Let T1 : R3 −→ R2 be the linear transformation given by the matrix
[ 1 3 2]
[2 0 −1] 
and let T2 : R2 −→ R1 be the linear transformation given by the matrix  [5 3]

The kernel of T1 is one dimensional. Find a vector u1 which spans Ker(T1) (i.e., a
basis for Ker(T1)). 

Answer 1

Let vector v = defined as \[a\\b\\c\]
Then T1(v) =\[a+3b+2c\\2a-c\]
The kernel of a transformation T is defined as the set of all vectors k for which T(k)=0
Hence, the kernel of T1 will be all vectors v such that
a + 3b + 2c = 0
2a - c = 0

Let's use a as the free variable. From the second equation, we get c = 2a.
From the first equation we get
a + 3b + 2c = 0
a + 3b + 4a = 0
5a + 3b = 0
3b = -5a
b = (-5/3)a

Then the kernel of T1 is all vectors v of the form \[a\\(-5/3)a\\2a\]
Since the basis is one-dimensional, any vector satisfying this form will span Ker(T1) and thus be a valid choice for u1, so pick one. For example, let a = 1. Then u1=\[1\\-5/3\\2\]

Answer 2

(You can plug the vector u1 into the transformation and see if you get the 0 vector to test whether it's in the kernel)

Answer 3

wow thank you so so so much