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The solution to self-quiz 14 in math 310, which involves finding the kernel of a linear map t : r4 → r3 and checking if the vector (1, 3, 2, −2)t belongs to the subspace spanned by (1, 2, −1, 1)t, (3, 1, 1, 0)t, and (1, 3, 0, 1)t. The solution uses matrix operations and the gauss-jordan elimination method.
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Self-quiz 14
4 → R
3 with
T ((x, y, z, w)
T ) = (x + y + z + w, 2 x + 3y + 2z + w, 3 x − y + 2z + w)
T
T belongs to the subspace of R
4 spanned
by the vectors (1, 2 , − 1 , 1)
T , (3, 1 , 1 , 0)
T and (1, 3 , 0 , 1)
T .
Self-quiz 14
T ((x, y, z, w)
T ) = (x + y + z + w, 2 x + 3y + 2z + w, 3 x − y + 2z + w)
T
Solution: The kernel of the map is given by the subspace:
{(x, y, z, w)
T : T ((x, y, z, w)
T ) = (0, 0 , 0 , 0)
T }
or
{(x, y, z, w)
T : x + y + z + w = 0, 2 x + 3y + 2z + w = 0, 3 x − y + 2z + w = 0}
In other words, we need to solve the system:
x + y + z + w = 0
2 x + 3y + 2z + w = 0
3 x − y + 2z + w = 0
We will use the Gauss-Jordan elimination method and work with its aug-
mented matrix:
and therefore the basis of ker(T ) is:
Incidentally this implies that the kernel is 1-dimensional which also means
that the image has to be 3-dimensional i.e. all of R
3 .
T belongs to the subspace of R
4 spanned
by the vectors (1, 2 , − 1 , 1)
T , (3, 1 , 1 , 0)
T and (1, 3 , 0 , 1)
T .
Solution: This is equivalent to asking if there exists real numbers λ, μ and
ν such that:
T = λ(1, 2 , − 1 , 1)
T
T
T
The above equation is equivalent to the system:
λ + 3μ + ν = 1
2 λ + μ + 3ν = 3
−λ + μ = 2
λ + ν = − 2
We will solve this system by using the Gauss-Jordan elimination method on
the augmented matrix of the system:
