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Let T be an arbitrary linear transformation from Rn to Rm which is not one-one. Then,
- a)Rank T> 0
- b)Rank T < n
- c)Rank T < 0
- d)Rank T = n - 1
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Let T be an arbitrary linear transformation from Rn to Rm which is not...
We are given that a linear transformation T : Rn --> Rm such that T is not one-one.
Since T is not one-one.
Therefore, ker T ≠ {0}. Hence, Nullity of T≥1. Using Rank nullity theorem.
Rank T = n - nullity T, < n.
Since T is not one-one.
Therefore, ker T ≠ {0}. Hence, Nullity of T≥1. Using Rank nullity theorem.
Rank T = n - nullity T, < n.
Most Upvoted Answer
Let T be an arbitrary linear transformation from Rn to Rm which is not...
The rank of a linear transformation T from Rn to Rm is the dimension of the column space of T.
Since T is not one-one, there must be at least one vector in Rn that is mapped to the same vector in Rm. This means that there are at least two distinct vectors in the domain of T that are mapped to the same vector in the codomain of T.
Let's say that v and w are two such distinct vectors in Rn that are mapped to the same vector in Rm. Then T(v) = T(w).
Now, let's consider the vector u = v - w. Since T is a linear transformation, we have T(u) = T(v - w) = T(v) - T(w) = 0.
This means that the vector u is in the null space (also known as the kernel) of T.
Since u is nonzero (because v and w are distinct), the null space of T is nontrivial. This means that the dimension of the null space of T is at least 1.
Using the rank-nullity theorem, we have rank T + nullity T = n, where n is the dimension of the domain of T.
Since the nullity of T is at least 1, it follows that the rank of T is less than n.
So, in conclusion, the rank of T is less than the dimension of the domain of T.
Since T is not one-one, there must be at least one vector in Rn that is mapped to the same vector in Rm. This means that there are at least two distinct vectors in the domain of T that are mapped to the same vector in the codomain of T.
Let's say that v and w are two such distinct vectors in Rn that are mapped to the same vector in Rm. Then T(v) = T(w).
Now, let's consider the vector u = v - w. Since T is a linear transformation, we have T(u) = T(v - w) = T(v) - T(w) = 0.
This means that the vector u is in the null space (also known as the kernel) of T.
Since u is nonzero (because v and w are distinct), the null space of T is nontrivial. This means that the dimension of the null space of T is at least 1.
Using the rank-nullity theorem, we have rank T + nullity T = n, where n is the dimension of the domain of T.
Since the nullity of T is at least 1, it follows that the rank of T is less than n.
So, in conclusion, the rank of T is less than the dimension of the domain of T.
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Let T be an arbitrary linear transformation from Rn to Rm which is not one-one. Then,a)Rank T> 0b)Rank T < nc)Rank T < 0d)Rank T = n - 1Correct answer is option 'C'. Can you explain this answer?
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