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If $T : \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$ is a linear transformation with $m > n$, is it guaranteed that $T$ is not injective?

I believe the answer is yes because $\mathbb{R}^{m}$ is, intuitively, "bigger" than $\mathbb{R}^{n}$.

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    You're correct, but your reasoning is wrong. For any finite $n \in \mathbb{N}$, we have that $| \mathbb{R}^n| = | \mathbb{R} |$. Thus, there are injections from $\mathbb{R}^m \rightarrow \mathbb{R}^n$ where $m > n$, but there are no linear injections. Such an injection would require you to have $m$ independent vectors in $\mathbb{R}^n$, which is impossible. – Joe Jul 23 '18 at 20:22
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    Moreover, there cannot exist a $C^1$ injective function $\mathbb{R}^m \to \mathbb{R}^n$ when $m > n$. – mechanodroid Jul 23 '18 at 20:42

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That's correct. Linear transformation is injective only if it's kernel is $\left\{0\right\}$. But since the sum of the dimensions of the kernel and the image of this map must be equal to the dimension of the domain ($\dim(\ker T)+\dim({\mathop {\mathrm {im} }}T)=\dim(\mathbb R^m )$). This means that the image of the map must have dimension $m$. On the other hand the image is contained in the range $\mathbb R^n$ and has dimension at most $n$ which is smaller than $m$.

gcc-6.0
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Let $\{v_1,\ldots,v_m\}$ be a basis for $\Bbb{R}^m.$ If $m>n$ then any transformation $T:\Bbb{R}^m\longrightarrow \Bbb{R}^n$ has the property that $\{T(v_1),\ldots, T(v_m)\}$ is linearly dependent, since the dimension of $\Bbb{R}^n$ is $n,$ and this is the upper bound on the size of a linearly independent subset of $\Bbb{R}^n$. Since this set is linearly dependent, then $T$ is not injective.

In this sense, your intuition is correct, in that $\Bbb{R}^m$ is bigger in dimension than $\Bbb{R}^n.$ However, the two objects as sets have the same cardinality (size/they're equally big as sets). It is precisely this notion of dimension which allows us to distinguish "size" in vector spaces, which we cannot do viewing them as sets alone.

Chickenmancer
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