T(x, y, z) = 0, then x - y = 0, z = 0, y - x = 0, so x =y and the kernel is the
ID: 2938580 • Letter: T
Question
T(x, y, z) = 0, then x - y = 0, z = 0, y - x = 0, so x =y and the kernel is the set of all vectors of the form t(1, -1, 0),t in R.T(x, y, z) = z(0, 1, 0) + (x - y)(1, 0, -1),so the image is the span of (0, 1, 0) and (1, 0, -1).
the above highlighted portion how u come to know that z(0,1,0) andx-y(1,0,1) the answer in my book is (s,t,-s) plz explain how you find the range(image) in suchquestions T(x, y, z) = 0, then x - y = 0, z = 0, y - x = 0, so x =y and the kernel is the set of all vectors of the form t(1, -1, 0),t in R.
T(x, y, z) = z(0, 1, 0) + (x - y)(1, 0, -1),so the image is the span of (0, 1, 0) and (1, 0, -1).
the above highlighted portion how u come to know that z(0,1,0) andx-y(1,0,1) the answer in my book is (s,t,-s) plz explain how you find the range(image) in suchquestions
Explanation / Answer
QuestionDetails: T(x, y, z) = 0, then x - y = 0, z = 0, y - x = 0, so x =y and the kernel is the set of all vectors of the form t(1, -1,0),t in R.
T(x, y, z) = z(0, 1, 0) + (x - y)(1, 0, -1),so the image is the span of (0, 1, 0) and (1, 0, -1).
the above highlighted portion how u come to know that z(0,1,0) andx-y(1,0,1) the answer in my book is (s,t,-s) plz explain how you find the range(image) in suchquestions
FIRSTLY YOU SHOULD MENTION WHAT IS T[X,Y,Z]
FROM WHAT YOU HAVE GIVEN IT APPEARS
T[X,Y,Z]=[(X-Y) , (Y-X) , 0 ]
THEN THE NULL SPACE IS GIVEN BY
T[X,Y,Z]=0=[(X-Y) , (Y-X) , 0 ]
HENCE ....X-Y=Y-X=Z=0...SO X= A ; Y=A ; Z=0
THEN THE KERNEL WILL BE [A,A,0] AND NOT [1,-1,0]
SO T[X,Y,Z] HAS TO BE
T[X,Y,Z]=[(X+Y) , (Y+X) , 0 ] ...THEN HOW DID YOU PUT
x - y = 0, z = 0, y - x = 0, so x = y
PLEASE MENTION WHAT IS T[X,Y,Z] AND CHECK FOR TYPO...
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