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Row echelon form : Can the first column contain only zeros?

mathmari

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Apr 14, 2013
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Hey!! :giggle:

Let \begin{equation*}A=\begin{pmatrix}0 & -2 & 2 & 0 & 0 & -6 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 1 & -1 & 0 & 2 & 7\\ 0 & 3 & -3 & 1 & 2 & 14\end{pmatrix}, \ b_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix} , \ b_2=\begin{pmatrix}-2 \\ 1 \\ -1 \\ 3 \\ 5\end{pmatrix}, \ b_3=\begin{pmatrix}-2 \\ 2 \\ 0 \\ 4 \\ 7\end{pmatrix}\end{equation*}

(a) Determine the row echelon form of $A$.

(b) Calculate for all $1\leq i\leq 3$ with $L(A,b_i)\neq \emptyset$ a solution $x_i$.

(c) Give a basis of $L(A, 0_{\mathbb{R}^5})$.

(d) Give $L(A,b_i)$ for all $1\leq i\leq 3$ using the basis of (c).



For question (a) : At the matrix $A$ the first column contains only zeros. Can the echelon form contain only zeros at the first column? Or do we have to exchange the first column with an other one to get a non-zero element at the upper left position? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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Hey mathmari !!

It is fine if the first column - and any other columns as well - will contain only zeroes in the row echelon form. 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
It is fine if the first column - and any other columns as well - will contain only zeroes in the row echelon form. 🤔
So we have the following?
\begin{align*}\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & -2 & 2 & 0 & 0 & -6 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 1 & -1 & 0 & 2 & 7\\ 0 & 3 & -3 & 1 & 2 & 14\end{matrix}
\end{matrix}\right|\begin{matrix}0 & -2 & -2 \\ 0 & 1 & 2\\ 0 & -1 & 0 \\ 0 & 3 & 4\\ 0 & 5 & 7\end{matrix}\end{pmatrix} &\ \overset{R_1:\frac{1}{-2}\cdot R_1}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 1 & -1 & 0 & 2 & 7\\ 0 & 3 & -3 & 1 & 2 & 14\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -1 & 0 \\ 0 & 3 & 4\\ 0 & 5 & 7\end{matrix}\end{pmatrix} \\ & \ \overset{R_4:R_4-R_1}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 2 & 4\\ 0 & 3 & -3 & 1 & 2 & 14\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -1 & 0 \\ 0 & 2 & 3\\ 0 & 5 & 7\end{matrix}\end{pmatrix} \\ & \ \overset{R_5:R_5-3\cdot R_1}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 2 & 4\\ 0 & 0 & 0 & 1 & 2 & 5\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -1 & 0 \\ 0 & 2 & 3\\ 0 & 2 & 4\end{matrix}\end{pmatrix} \\ & \ \overset{R_3:R_3-R_2}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 & 2 & 4\\ 0 & 0 & 0 & 1 & 2 & 5\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -2 & -2 \\ 0 & 2 & 3\\ 0 & 2 & 4\end{matrix}\end{pmatrix} \\ & \ \overset{R_4:R_4+R_3}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 2 & 5\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -2 & -2 \\ 0 & 0 & 1\\ 0 & 2 & 4\end{matrix}\end{pmatrix} \\ & \ \overset{R_5:R_5-R_2}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 2\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -2 & -2 \\ 0 & 0 & 1\\ 0 & 1 & 2\end{matrix}\end{pmatrix} \\ & \ \overset{R_5:R_5+\frac{1}{2}\cdot R_3}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -2 & -2 \\ 0 & 0 & 1\\ 0 & 0 & 1\end{matrix}\end{pmatrix} \\ & \ \overset{R_3:\frac{1}{-2}\cdot R_3}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & 1 & 1 \\ 0 & 0 & 1\\ 0 & 0 & 1\end{matrix}\end{pmatrix}\end{align*}
That means that $L(A, b_3)=\emptyset$ and $L(A, b_1)\neq\emptyset$ and $L(A, b_1)\neq\emptyset$.

Right? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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It looks correct to me. (Nod)
 

mathmari

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Apr 14, 2013
4,713
It looks correct to me. (Nod)
So for question (b) we have the following :

$x_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}\in L(A,b_1)$.

Now $x_2$ is not unique, but since we are asked to calculate a solutoion, I suppose we give one of the solutions, right?
Considering $b_2$ we get the following equations : \begin{align*}b-c+3f=&1 \\ d+e+3f=&1 \\ e+2f=&1\end{align*}
So to get one solution we have to take arbirtray values for the free parameters, right?
From last equation we have $e=1-2f$, for $f=0$ we get $e=1$.
From the second equation we getthen $d=0$.
From the first equation we get $b=1+c$ and for $c=0$ we get $b=1$.

So $x_2=\begin{pmatrix}0\\ 1 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix}\in L(A,b_2)$.

Is that correct so far?

:unsure:


For question (c) we have that $L(A,0_{\mathbb{R}^5})=L(A,b_1)=\left \{\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}\right \}$. Is the zero vector the basis? :unsure:


For question (d) why do we need (c) can we not just use (b) ? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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So for question (b) we have the following :

$x_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}\in L(A,b_1)$.

Now $x_2$ is not unique, but since we are asked to calculate a solution, I suppose we give one of the solutions, right?
That should indeed be fine. It will come back in (c) though. (Smirk)

So $x_2=\begin{pmatrix}0\\ 1 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix}\in L(A,b_2)$.

Is that correct so far?
If I fill it in, I get indeed $b_2$, so it's correct. (Nod)

For question (c) we have that $L(A,0_{\mathbb{R}^5})=L(A,b_1)=\left \{\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}\right \}$. Is the zero vector the basis?
A basis must be a set of independent vectors that span the desired space.
The zero vector cannot be an independent vector, so it cannot be an element of a basis either. (Shake)

For question (d) why do we need (c) can we not just use (b) ? :unsure:
We've found the null space in (c).
If we add one of those vectors to a solution, it will again be a solution.
Either way, in (b) we found a single solution. Now we have to find a basis for all solutions.
The question asks to include the basis for the null space in the basis for all solution, instead of coming up with an unrelated basis. 🤔
 

mathmari

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Apr 14, 2013
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A basis must be a set of independent vectors that span the desired space.
The zero vector cannot be an independent vector, so it cannot be an element of a basis either. (Shake)
So no matter which basis we take we getthe zero vector, but in this case the only solution is the zero vector, or not?
So what basis do we take?

:unsure:
 

Klaas van Aarsen

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So no matter which basis we take we get the zero vector, but in this case the only solution is the zero vector, or not?
So what basis do we take?
We didn't actually solve $Ax_1=b_1=0$ yet, did we? :unsure:
 

mathmari

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Apr 14, 2013
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We didn't actually solve $Ax_1=b_1=0$ yet, did we? :unsure:
Ah yes!

As previously we get the equations\begin{align*}b-c+3f=&0 \\ d+e+3f=&0 \\ e+2f=&0\end{align*}
From the last equation we get $e=-2f$.
From the second equation we get $d-2f+3f=0 \Rightarrow d+f=0 \Rightarrow d=-f$.
From the first equation we get $b-c+3f=0 \Rightarrow b=c-3f$.

So the solution vector is \begin{equation*}\begin{pmatrix}a\\ b\\ c\\ d\\ e\\ f\end{pmatrix}=\begin{pmatrix}a\\ c-3f\\ c\\ -f\\ -2f\\ f\end{pmatrix}=\begin{pmatrix}a\\ 0\\ 0\\ 0\\ 0\\ 0\end{pmatrix}+\begin{pmatrix}0\\ c\\ c\\ 0\\ 0\\ 0\end{pmatrix}+\begin{pmatrix}0\\ -3f\\ 0\\ -f\\ -2f\\ f\end{pmatrix}=a\begin{pmatrix}1\\ 0\\ 0\\ 0\\ 0\\ 0\end{pmatrix}+c\begin{pmatrix}0\\ 1\\ 1\\ 0\\ 0\\ 0\end{pmatrix}+f\begin{pmatrix}0\\ -3\\ 0\\ -1\\ -2\\ 1\end{pmatrix}\end{equation*}
So is the basis teh set of these three vectors?

:unsure:
 

Klaas van Aarsen

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mathmari

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Apr 14, 2013
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Klaas van Aarsen

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So we get $L(a,b_i)=x_i+B$ with $B$ the basis above, right?
Aren't there more solutions than just the arbitrary $x_i$ that we picked before? :unsure:
 

mathmari

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Apr 14, 2013
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Aren't there more solutions than just the arbitrary $x_i$ that we picked before? :unsure:
We have that $L(a,b_1)=B$.

As for $b_2$ :
\begin{align*}b-c+3f=&1 \\ d+e+3f=&1 \\ e+2f=&1\end{align*}
We get $e=1-2f$, $d=-f$, $b=1+c-3f$.
But how do we need here $L(a,0)$ ? :unsure:

We have that $L(a, b_3)=\emptyset$.
 

Klaas van Aarsen

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Mar 5, 2012
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So we get $L(a,b_i)=x_i+B$ with $B$ the basis above, right?
As for $b_2$ :
\begin{align*}b-c+3f=&1 \\ d+e+3f=&1 \\ e+2f=&1\end{align*}
We get $e=1-2f$, $d=-f$, $b=1+c-3f$.
But how do we need here $L(a,0)$ ?
You were right after all.
The $x_1$ you is a solution and we can add any vector that belongs to the null space.
So the solution is indeed $x_1+B$. 🤔
 

mathmari

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Apr 14, 2013
4,713
You were right after all.
The $x_1$ you is a solution and we can add any vector that belongs to the null space.
So the solution is indeed $x_1+B$. 🤔
Great!! Thanks a lot!! (Handshake)