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I'm just starting to learn linear algebra and I understand Gaussion elimination but I don't understand this proof:

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Is the heart of this the fact that all the while when I've been row-reducing matrices to reduced row-echelon form... I've actually been multiplying a matrix by a series of a elementary matrices?

What does the notation $[A\ \ 0]$ mean?

How is the equation $A = E_1^{-1}...E_k^{-1}$ arrived at?

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When we apply an elementary row operation $E$ to a matrix $A$, what we actually have is $EA$ algebraically. So when we apply a sequence of elementary row operations to $A$, it means multiplying them by $A$ as you said.

The notation $[A\ |\ 0]$ or $[A\ \ 0]$ is equivalent to $$ \left[ \begin{array}{cc|c} 1&2&0\\ 3&4&0 \end{array} \right] $$

where $A = \begin{bmatrix}1&2\\3&4\\ \end{bmatrix}$; and the result is called "augmented matrix" (This is an example of course).

The equation $A=E_1^{-1}E_2^{-1}\cdots E_k^{-1}$ can be obtained by multiplying both sides of the equation $E_k\cdots E_2E_1A=I$ by $E_k^{-1}$, ..., $E_1^{-1}$ from the left in order.

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    $\begingroup$ Is it right to say that I is row equivalent to A when A is invertible? Is that how we're able to rewrite [A 0] as [I 0] $\endgroup$ – Jwan622 Aug 30 '18 at 17:18
  • $\begingroup$ Yes, that's right. When a square matrix $A$ is invertible, $I$ is row equivalent to $A$ (we can also say that $I$ is the reduced row echelon form of $A$). If you are familiar with the "determinant" concept, then a non-invertible matrix $A$ having a reduced row echelon form different than $I$ will make sense to you. $\endgroup$ – ArsenBerk Aug 30 '18 at 17:25
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$[A\,0]$ is so-called block matrix notation, where a large matrix is written by putting smaller matrices ("blocks") next to one another (or above one another). In this case it means the matrix you get if you put $A$ right next to a (suitably tall) $0$ column.

Yes, every operation you've done when row-reducing a matrix may be realized by multiplying by a suitable elementary matrix.

For the final question, we start with $$ E_k\cdots E_1A=I $$ Then we multiply both sides from the left by $E_k^{-1}$ to get $$ E_{k-1}\cdots E_1A=E_k^{-1} $$ Then we multiply both sides from the left by $E_{k-1}^{-1}$ to get ... (and so on, until finally) Then we multiply both sides from the left by $E_1^{-1}$ to get $$ A=E_1^{-1}\cdots E_k^{-1} $$

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  • $\begingroup$ Is there a type in the second equation? Shouldn't the right side by the inverse elementary matrix for k? $\endgroup$ – Jwan622 Aug 30 '18 at 17:10
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    $\begingroup$ @Jwan622 Certainly. Thank you. $\endgroup$ – Arthur Aug 30 '18 at 17:11
  • $\begingroup$ Is it right to say that I is row equivalent to A when A is invertible? Is that how we're able to rewrite [A 0] as [I 0] $\endgroup$ – Jwan622 Aug 30 '18 at 17:17
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"I've actually been multiplying a matrix by a series of a elementary matrices?" Yes.

"What does the notation [A0] mean?" It means augmenting a matrix A by the matrix O. If you have a matrix $((1,1),(1,1))$ and you augment it by the matrix $((2,2))$, then you get $((1,1),(1,1),(2,2))$. It is the core of Gaussian elimination (written in a more elegant way).

"How is the equation A=E(1−1)...E(k−1) arrived at?" Multiply by the inverse of the k elementary matrices (they exist and you can prove it) the previous equation.

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