Suppose a linear transformation $ t : \mathbb{R}^n \rightarrow \mathbb{R}^n $ has the property that $ t(u) = t(v) $ for some pair of distinct vectors $ u $ and $ v $ in $\mathbb{R}^n$.

Can $ t $ map $\mathbb{R}^n$ onto $\mathbb{R}^n$, i.e., is $ t $ onto? Why or why not?

In mathematics, a linear transformation t : Rn → Rn is said to be onto (or surjective) if every vector in the codomain Rn is the image of at least one vector in the domain Rn. However, in your case, there exist two distinct vectors u and v in Rn such that t(u) = t(v). This means that two different input vectors are mapping to the same output vector through this transformation. However, if two different input vectors map to the same output vector, this suggests that there may be output vectors that are not the image of any input vector, and thus the transformation is not onto.

This violates the criterion for a transformation to be onto, since it suggests that there may be vectors in the codomain Rn that are not the image of any vector in the domain. Hence, if t(u) = t(v) for some unique vectors u and v, then the transformation t is not onto.

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Suppose a linear transformation \( t : \mathbb{R}^n \rightarrow \mathbb{R}^n \) has the property that \( t(u) = t(v) \) for some pair of distinct vectors \( u \) and \( v \) in \(\mathbb{R}^n\).Can \( t \) map \(\mathbb{R}^n\) onto \(\mathbb{R}^n\), i.e., is \( t \) onto? Why or why not?

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Suppose a linear transformation \( t : \mathbb{R}^n \rightarrow \mathbb{R}^n \) has the property that \( t(u) = t(v) \) for some pair of distinct vectors \( u \) and \( v \) in \(\mathbb{R}^n\).

Can \( t \) map \(\mathbb{R}^n\) onto \(\mathbb{R}^n\), i.e., is \( t \) onto? Why or why not?