2
$\begingroup$

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

enter image description here

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?

|cite|improve this question
$\endgroup$
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

$[A\,0]$ 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 $$ A=E_1^{-1}\cdots E_k^{-1} $$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
0
$\begingroup$

"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.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.