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Prove that if $\{\mathbb{v_1},\ \cdots,\ \mathbb{v_r}\}$ are a basis of $\rm{col}(A^\mathrm{T})$ then their images $\{Av_1,\ \cdots,\ Av_r\}$ are a basis for $\rm{col}(A)$.

How will I be able to prove this and what do they mean by image? Does it mean projections?

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    @diimension I've changed rng to col to denote the columnspace for clarity. By image, we simply mean the vector that $\mathbb{v_i}$ is mapped to under the action of $A$ as a mapping.2012-11-07

1 Answers 1

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This follows from a very fundamental fact of linear algebra.

Let me give you a sketch on how to proceed. Let $A$ be an $m\times n$ matrix.

  1. Prove that the nullspace and the rowspace are orthogonal complements.
  2. Conclude that $\mathbb{R}^m = \rm{col}(A^\mathrm{T}) \oplus \ker(A)$
  3. Form a basis of $\mathbb{R}^m$ as a union of the basis of the rowspace and the nullspace. Suppose that we have $\mathcal{B}\cup\mathcal{C}$ where $\mathcal{B}=\{\mathbb{v_1},\ \cdots,\ \mathbb{v_r}\},\ \ \ \ \ \mathcal{C}=\{\mathbb{u_1},\ \cdots,\ \mathbb{u_{m-r}}\}$ such that $\mathcal{B}$ is a basis for the rowspace and $\mathcal{C}$ is a basis for the nullspace.
  4. Show that the set $\{A\mathbb{v_1},\ \cdots,\ A\mathbb{v_r}\}$ spans the image of $A$.
  5. Show that the same set above is linearly independent.

Edit: A bit of elaboration on the different definitions.

When we say the image of a matrix, we really mean the set of outputs of that matrix as a linear mapping. To be specific, the image of an $m\times n$ real matrix $A$ is the set $\rm{Im}(A) = \{\mathbf{x} \in \mathbb{R}^n\mid \mathbf{x} = A\mathbf{y}\ \text{for some }\mathbf{y}\in\mathbb{R}^m\}$ This is sometimes called the range, although image seems more common in my experience. In either case, in this context range and image (and sometimes co-domain) are all taken as synonyms. You can show that the image of $A$ is just the columnspace of $A$, so the columnspace is sometimes taken as another synonym.

Image is sometimes also used for a single vector. We sometimes say the image of $\mathbf{x}$ under $A$ and what this term means is that we are considering the vector that $\mathbf{x}$ is mapped to using $A$; this is the $A\mathbf{x}$.

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    Thank you very much, EuYu!2012-11-08