If the Nullity of a Linear Transformation is Zero, then Linearly Independent Vectors are Mapped to Linearly Independent Vectors

Problem 722

Let $T: \R^n \to \R^m$ be a linear transformation.
Suppose that the nullity of $T$ is zero.

If $\{\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k\}$ is a linearly independent subset of $\R^n$, then show that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.

Suppose that we have a linear combination
\[c_1T(\mathbf{x}_1)+c_2T(\mathbf{x}_2)+\cdots+c_k T(\mathbf{x}_k)=\mathbf{0}_m,\]
where $\mathbf{0}_m$ is the $m$ dimensional zero vector in $\R^m$.
To show that the set $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is linearly independent, we need to show that $c_1=c_2=\cdots=c_k=0$.

Using the linearity of $T$, we have
\[T(c_1\mathbf{x}_1+c_2\mathbf{x}_2+\cdots+c_k \mathbf{x}_k)=\mathbf{0}_m.\]
Then the vector $c_1\mathbf{x}_1+c_2\mathbf{x}_2+\cdots+c_k \mathbf{x}_k$ is in the nullspace $\calN(T)$ of $T$. Since the nullity, which is the dimension of the nullspace, is zero, we have $\calN(T)=\{\mathbf{0}_n\}$. This yields
\[c_1\mathbf{x}_1+c_2\mathbf{x}_2+\cdots+c_k \mathbf{x}_k=\mathbf{0}_n.\]

Since the vectors $\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k$ are linearly independent, we must have $c_1=c_2=\dots=c_k=0$ as required.

Thus we conclude that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.

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Prove the followings.
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\alpha \\
\beta
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\[ T\left (\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \,\right) = \begin{bmatrix}
x_1+2x_2+3x_3-x_4 \\
3x_1+5x_2+8x_3-2x_4 \\
x_1+x_2+2x_3
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Proof.
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