Consider the matrix mapping $A : R^4 \rightarrow R^3$, where

$A = \begin{bmatrix} 1 & 2 & 3 & 1\\ 1 & 3 & 5 & -2\\ 3 & 8 & 13 & -3 \end{bmatrix}$

Find a basis and dimension of the image of $A$ and those of the kernel $A$.

Kernel is basis of domain space.

The mapping A(x,y,z,w) =(x+2y+3z+1w,x+3y+5z-2w,3x+8y+13z-3w).

/*Each row of the matrix */

Thus after reducing equations of A:

x + 0y -z + 7w = 0

0x + y +2z -3w= 0

By these 2 equations we get:-

z = (-y+3w)/2

x = (-y-11w)/2

Maximum basis of A is 5. But, by solving we get:

S = ( (-y-11w)/2,y,(-y+3w)/2,w)

Now we know that, S is a subset of A. Conversely, to show A is subset of S.

As S = {y(-1/2,1,-1/2,0) + w(-11/2,0,3/2,1)} is space of A

with elts={(-1/2,1,-1/2,0), (-11/2,0,3/2,1)} /*As calculated above*/

Thus, S = A So, basis of kernel is {(-1/2,1,-1/2,0), (-11/2,0,3/2,1)}.

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Now, basis of image of A:-  Standard basis of R4.......

A(1,0,0,0) = (1,1,3)

A(0,1,0,0) = (2,3,8)

A(0,0,1,0) = (3,5,13)

A(0,0,0,1) = (1,-2,-3)

Reduce RHS from a 4*3 matrix, to get 4*2 matrix  /*row echelon form*/

As maximum non empty rows are 2. Thus, basis is 2.

So, {(1,1,3),(2,3,8)} is basis of image of  A.

answered Jan 8 by (190 points)
edited Jan 8 by big boss