A $k$-dimensional vector subspace of $\mathbb{R}^n$ is a $k$-dimensional manifold
by Hidenori
Proposition
Prove that a $k$-dimensional (vector) subspace of $\mathbb{R}^n$ is a $k$-dimensional manifold.
Solution
Let $V$ denote the subspace and $v_1, \cdots, v_k$ denote a basis of $V$. We can extend it to a basis of $V$. Then $A = \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}$ is a invertible matrix. (Each $v_1, \cdots, v_n$ is a column vector with $n$ entries.)
Let $x = (x_1, \cdots, x_n) = a_1v_1 + \cdots + a_kv_k \in V$ be given. Then
\[\begin{align*} \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix} \begin{bmatrix} a_1 \\ \vdots \\ a_k \\ 0 \\ \vdots \\ 0 \end{bmatrix} &= \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}. \end{align*}\]By multiplying $A^{-1}$ from left, we obtain
\[\begin{align*} \begin{bmatrix} a_1 \\ \vdots \\ a_k \\ 0 \\ \vdots \\ 0 \end{bmatrix} &= A^{-1}\begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}. \end{align*}\]Therefore, $A^{-1}$ maps $V$ into a subspace of $\mathbb{R}^n$ whose $k + 1, \cdots, n$th coordinates are all 0 bijectively. Hence, a $k$-dimensional vector subspace of $\mathbb{R}^n$ is a $k$-dimensional manifold.
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