MHB Possible Echelon Form of Matrix

[bwpbruce]

Please check my solution.View image: Possible Echelon Form of Matrix

 

[Evgeny.Makarov]

What is the rank (the maximum number of linearly independent columns) of the original matrix and the one in 2nd possibility? The rank of a matrix and its row echelon form should be the same.

 

[bwpbruce]

Don't really understand what you're asking. If the solution isn't correct, please explain why it isn't correct. I'm supposed to list all possible echelon forms, not just one possible echelon form. There is no "original matrix". All possible forms of $4 \times 3$ matrix apply.

 

[Evgeny.Makarov]

By the original matrix I mean $A$. With respect to the rest of my explanation, you'll have to explain what is unclear to you. I am asking to compare the number of linearly independent columns in $A$ (i.e., $a_1$, $a_2$ and $a_3$) and in the matrix you wrote under "2nd Possibility".

 

[bwpbruce]

I realize $\textbf{a}_1$ was not a linearly independent column. So only the 1st possibility is possible.

 

[Evgeny.Makarov]

I realize $\textbf{a}_1$ was not a linearly independent column. So only the 1st possibility is possible.

How did you make the conclusion that only the first possibility is possible?

We rarely say that a single vector is linearly (in)dependent. A vector $v$ is linearly dependent iff there exists a nonzero scalar $\alpha$ such that $\alpha v=0$. This happens iff $v=0$. So a single vector is linearly dependent iff it is a zero vector.

But my question is still the same: how many vectors among $\mathbf{a}_1$, $\mathbf{a}_2$ and $\mathbf{a}_3$ are linearly independent? Two are independent for sure since it is said explicitly that $\{\mathbf{a}_1,\mathbf{a}_2\}$ is a linearly independent set. What about $\{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\}$?

 

[bwpbruce]

Because of this:
View attachment 3786

Basically, if $\textbf{a}_3$ isn't in span$ \{\textbf{a}_1, \textbf{a}_2\}$, then it is linearly independent of the other vectors.

 

[Evgeny.Makarov]

Yes, $\mathbf{a}_3\notin\operatorname{Span}\{\mathbf{a}_1,\mathbf{a}_2\}$ implies that $\{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\}$ are linearly independent. Therefore, the row echelon form should also have three linearly independent columns. However, possibility 2 has only 2 linearly independent columns, so it is not, in fact, a possibility.

 

[bwpbruce]

BTW, can you clarify what they are referring to when they say $\textbf{w}$ not in span{$\textbf{u,v}$} is linearly independent? According to their graph, $\textbf{u,v}$ are linearly dependent. So why are they implying that {$\textbf{u,v,w}$} is linearly independent?

 

[Evgeny.Makarov]

BTW, can you clarify what they are referring to when they say $\textbf{w}$ not in span{$\textbf{u,v}$} is linearly independent?

The claim that allows deducing $\{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\}$ is linearly independent under the assumptions of post #1 is the claim in https://driven2services.com/staging/mh/index.php?threads/13869/. Note that it is incorrect to say "$\mathbf{w}$ is linearly independent". Only a set of vectors (even if the set contains one vector) can be linearly dependent.

 

[bwpbruce]

What you just said doesn't exactly answer the specific question I was asking. I'm trying to get clarification on what they mean when they say that $\textbf{w} \notin \text{ span } \{\textbf{u,v}\}$ is linearly independent. Are they saying that the set $\{\textbf{u,w,v}\}$ is linearly independent?