Consider the plane x1 x2 x3 0 The mapping T R3 rightarrow

Consider the plane x_1 + x_2 + x_3 = 0. The mapping T: R^3 rightarrow R^3 define by T(x) is the point obtained by projecting x onto the plane is a linear mapping. Find the matrix representation of T with respect to the basis U = {[1 1 1], [1 0 -1], [1 -1 0]} Let E stand for the standard basis of R^3. Find the U to E change of basis matrix, and the E to U change of basis matrix. Find the matrix representation of T with respect to the standard basis.

Solution

given plane is x1 +x2 +x3 = 0

let (x,y,z) be any point is R³  ,let (x1,x2,x3) be projection of point (x,y,z) on given plane .

by coordinate geometry ,we know that plane has normal (1,1,1)

hence (x-x1)/1 = (y-x2)/1 = (z-x3)/1 = -t ,

x1 = x +t (1)

x2 = y +t (2)

x3 = z +t (3)

since   x1 +x2 +x3 = 0 ,we get

x +y +z +3t = 0 (4)

if (x,y,z) will be given we can find t, hence the x1,x2,x3 ,let us see part a)

a) the point is (1,1,1)

hence using equation 4 we get , 3 +3 t = 0    t = -1

by using eq. 1 ,2,3 ,we get

x1 = 1-1= 0

x2 =  1-1= 0

x3 =  1-1= 0

hence the projection is (0,0,0)

second point is (1,0,-1) ,clearly this is on the plane   x1 +x2 +x3 = 0 ,hence projection will give same point ,

let us check as we have done for 1st point ,

using eq. 4 ,we get   

1 +0 + -1 +3t = 0

t =0

using eq. 1 ,2,3 ,we get   

x1 =x, x2 =y ,x3 =z

,similarly (1,-1,0) is on the plane itself ,so the projection will be the same.

hence the matrix representation with respect to given basis is

b) E stands for the standard basis,

the U to E change of basis matrix

as let X be required matrix

UX = E

X = U^-1 E   {multiplying identity matrix with any matrix is samematrix}

X = U^-1 which is

the E to U change of basis matrix is U itself

as let X be required matrix

so EX = U

X = U {multiplying identity matrix with any matrix is same matrix}

using eq. 4 we get 1 +3t = 0 c) standard basis of R³ are (1,0,0) (0,1,0) & (0,0,1)

t = -1/3

x1 = 1-1/3 = 2/3

x2 = -1/3

x3 = -1/3

so the point is (2/3,-1/3,-1/3)

similarly for (0,1,0) is (-1/3,2/3,-1/3)

(0,0,1) is (-1/3,-1/3,2/3)

hence the representation of T with respect to the standard basis is

0	1	1
0	0	-1
0	-1	0