document.write( "Question 378801: let T:R^3 --> R^3
\n" ); document.write( "T(x,y,z) = (z, x + y, x + y + z). determine
\n" ); document.write( "i) rank T and basis of Im T
\n" ); document.write( "ii) nullity T\r
\n" ); document.write( "\n" ); document.write( "what i tried:\r
\n" ); document.write( "\n" ); document.write( "If t is the matrix that will give you a 1 x 3 image then isn't is safe to conclude that the rank of T is 3? also if im T is the new vector, then the basis of im T is a linear combination of the standard basis vectors of R^3 right? but how do i prove all this mathematically? \n" ); document.write( "
Algebra.Com's Answer #269055 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
= (z, x + y, x + y + z). The transformation matrix has rank 2, because column 3 is just the sum of columns 1 and 2. Thus rank T = 2. Now (z, x + y, x + y + z) = z(1,0,1) + (x+y)(0,1,1). Hence {(1,0,1),(0,1,1)} is a basis for Im T. Since rank T + nullity T = 3 (the number of columns of the transformation matrix), nullity T = 1.
\n" ); document.write( "(We found the transformation matrix by inspection. Direct matrix multiplication would verify this. E.g., If \"%28matrix%283%2C1%2Ca%2Cb%2Cc%29%29\" were the 1st column of the transformation matrix then \"%28matrix%281%2C3%2Cx%2Cy%2Cz%29%29%2A%28matrix%283%2C1%2Ca%2Cb%2Cc%29%29+=+z\" implies that a = b = 0, and c = 1.) \n" ); document.write( "
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