Kernel and image of linear transformation

  • Thread starter Locoism
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Homework Statement


For the linear transformation T: R4 --> R3 defined by TA: v -->Av
find a basis for the Kernel of TA and for the Image of of TA where A is
2 4 6 2
1 3 -4 1
4 10 -2 4


Homework Equations



Let v =
a1 b1 c1
a2 b2 c2
a3 b3 c3
a4 b4 c4

The Attempt at a Solution


so v is a 4x3 matrix, and Ker(T) would just be the solution for Av = 0.
I was unsure as to what the Image would be give by. Is it the matrix
2a1+4b1+6c1+2d1, 2a2+4b2+6c2+2d2, 2a3+4b3+6c3+2d3,
1a1+3b1-4c1+d1, ... etc

(just the general solution of the multiplication)
Which generalizes to
2 0 2
0 1 2
so the basis is [1, 2, -1]

How would I find a basis for the Kernel?
 
Last edited:

Answers and Replies

  • #2
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7,289

Homework Statement


For the linear transformation T: R4 --> R3 defined by TA: v -->Av
find a basis for the Kernel of TA and for the Image of of TA where A is
2 4 6 2
1 3 -4 1
4 10 -2 4


Homework Equations



Let v =
a1 b1 c1
a2 b2 c2
a3 b3 c3
a4 b4 c4
You're really heading down the wrong path here. v is a vector in R4.

The Attempt at a Solution


so v is a 4x3 matrix, and Ker(T) would just be the solution for Av = 0.
Ker(T) is the set of all vectors v in R4 such that Tv = 0.
I was unsure as to what the Image would be give by. Is it the matrix
2a1+4b1+6c1+2d1, 2a2+4b2+6c2+2d2, 2a3+4b3+6c3+2d3,
1a1+3b1-4c1+d1, ... etc

(just the general solution of the multiplication)
Which generalizes to
2 0 2
0 1 2
so the basis is [1, 2, -1]

How would I find a basis for the Kernel?
 

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