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Suppose there is a four by five matrix A in which three of the five column vectors are linearly independent. Today my professor demonstrated that the the reduced row echelon form of A hints at the position of the linearly independent matrix.

Specifically, if the u th , v th and w th column vector are the three linearly independent column vectors in the reduced row echelon form matrix of A (hence the column vectors that contain the leading ones), then the u th, v th and w th column vectors of A are also linearly independent. He calls it the column correspondence principle. What is the proof for this?

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To answer this, you probably should also tell us a bit about how your professor describes how to transform a matrix to its reduced row echelon form? That way the answer may be a little more meaningful to you.

Here is a sketch of how I would explain it to my students:

Let's say $ A $ is the original matrix and $R $ the reduced echelon form. The reduction to reduced echelon form can be viewed as the application (multiplication) of a sequence of matrices to $ A $: $ D L_{n-1} \cdots L_0 A = R. $ The matrices $ L_i $ are called Gauss transforms and the matrix $ D $ is diagonal. Importantly, those matrices are square and invertable (nonsingular).

Next, let's take the linearly independent columns of $ R $ and create a matrix $ \widetilde R $ with those and let's take the corresponding columns of $ A $ and make a matrix $ \widetilde A $ with those. Then $ D L_{n-1} \cdots L_0 \widetilde A = \widetilde R $. Now, if the columns of $ \widetilde R $ are linearly independent, then we know that $ \widetilde R x = 0 $ implies that $ x = 0 $. Let's see what this tells us about $ \widetilde A $: Assume $ \widetilde A x = 0 $. Then $ \widetilde A x = L_0^{-1} \cdots L_{n-1}^{-1} D^{-1} \widetilde R x $ which implies that $ \widetilde R x = 0 $. But that means that $ x = 0 $. So, if columns in $ R $ are linearly independent, then the corresponding columns in $ A $ are linearly independent. The converse can be proven similarly.

Now, if your professor did not introduce the concept of Gauss transform, or relate the reduction to echelon form to multiplication by matrices, then it becomes a bit harder to answer your question...

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  • $\begingroup$ Thank you very much for your answer! But can this proof be extended to matrices that are not square ? $\endgroup$ – hephaes Mar 27 '18 at 13:26
  • $\begingroup$ There is no assumption in my discussion about the matrices being square. Only $ L_i $ and $ D $ have to be square. $\endgroup$ – Robert van de Geijn Mar 28 '18 at 2:19

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