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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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    What do you mean by "Corange"?2012-11-07
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    @martini Corange is the transpose of the range of A. so corange = $A^T$ range.2012-11-07
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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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    I got for 1.) that a vector x is in the nullspace/kernal of A iff $Ax=0$. Then the components of Ax are dot products of the row vectors of A with x. All the components must be 0, so x is in the nullspace of A iff it is orthogonal to all the rows of A. Thus, the the nullspace of A is the orthogonal complement of the rowspace of A. For 2.) I do not understand the symbol $\oplus$? For 3.) is that similar to 1.)? and for 4.) I still don't understand what image of A means, does it mean in the range of? 5.) to show that I just have to show that ker= {0}?2012-11-08
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    I will elaborate on the definitions a bit in my answer. Your part 1 seems fine, although I would elaborate a bit more. For 2, the symbol $\oplus$ is called a direct sum. It's ok if you don't know what it is but it's not a difficult concept so I would google it if I were you. Basically part 2 simply shows that the set we build in part 3 is actually a basis. For part 3 you don't do anything, it's a free step. For 4, the image of $A$ is just the range of $A$. And for 5, I'm not too sure what you mean. Kernel of what? If you're talking about $\ker(A)$ then no, it's not necessarily $\{0\}$.2012-11-08
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    Thank you! I am going to go over it again using the details you mentioned.2012-11-08
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    Can you verify if I am on the right track? Let A be a $mxn$ matrix then a vector $x\in \mathbb{R^n}$ lies in kerA iff Ax=0. The $i^{th}$ entry of Ax equals the vector product of the $i^{th}$ row $r^T_i$ of A and x thus $r^T_i(x) = r_i *x=0$ iff x is orthogonal to $r_i$. Therefore, $x\in kerA$ iff x is orthogonal to all the rows of A.Since the rows span $corangeA$, this is the same as x lying in its orthogonal complement. Then Ax = b has a sol'n iff the right belongs to the range of the matrix $b\in rangA$ which means b is orthgonal to its cokernal. thus $y*b = 0$ for y satisfying $A^Ty =0$.2012-11-08
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    A few things. First the kernel is a subspace of $\mathbb{R}^m$ not $\mathbb{R}^n$. When you say "vector product", people typically mean "cross product" in contrast to the "dot product" which people sometimes call the "scalar product". Everything else seems fine although I'm not sure why the last sentence is needed (it seems irrelevant to this question).2012-11-08
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    Thank you very much, EuYu!2012-11-08