17) let T:R3-R3 be a linear transformation defined by T(x,y,z)=(x+2y-z,y+z,x+y-2z) then find
Rank, nullity, range space, null space
Answers
Answer:
Let T:R3→R3 be the linear transformation defined by T(x,y,z)=(x+3y+2z,3x+4y+z,2x+y−z). Find the rank of T2 and T3.
I formed the matrix of linear transformation for T and squared it, and found the rank which is 2. For T3 I found the matrix by multiplying matrix T by T2. I need to calculate the rank of this matrix.
Is this only way to calculate rank?
This question is asked as multiple choice problem, which should not take much time. But finding rank in this way is really time consuming. Is there any other way?
Step-by-step explanation:
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Answer:
The null space of T is the set of vectors of the form (0,-z,z), which are the solutions to the homogeneous equation T(x,y,z) = (0,0,0). These vectors are mapped to the zero vector under T.
Explanation:
The rank of T is 3 because the image of the standard basis vectors under T forms a linearly independent set of size 3, which means that it is a basis for the range space of T.
The nullity of T is 1 because the null space of T is the set of vectors of the form (0,-z,z), which has dimension 1.
The range space of T is the set of all vectors of the form (x,-y,-x) | x,y in R. This can be obtained by taking the span of the image of the standard basis vectors under T and simplifying the resulting set.
The null space of T is the set of vectors of the form (0,-z,z), which are the solutions to the homogeneous equation T(x,y,z) = (0,0,0). These vectors are mapped to the zero vector under T.
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