# Linear Dependence & independence

• Mar 18th 2009, 03:19 PM
Craka
Linear Dependence & independence
The question is to determine whether the following vectors are linearly dependent or linearly independent.

The vectors are
{(1, 2, 3, 4), (2, 3, 4, 5), (3, 4, 5, 6), (4, 5, 6, 7)}

I put into matrix form and do gauss elimination on it

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ \end{array}} \right]$

$\displaystyle = \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & { - 1} & { - 2} \\ 0 & 0 & 0 & 5 \\ \end{array}} \right]$

$\displaystyle = \left[ {\begin{array}{*{20}c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & { - 1} \\ 0 & 0 & { - 1} & { - 2} \\ 0 & 0 & 0 & 5 \\ \end{array}} \right]$

From here I'm not sure how to intepret it, for whether it is dependent or independent. Answer say it is linearly dependent
• Mar 18th 2009, 03:46 PM
arbolis
Hi Craka,
You haven't row-reduced the matrix. For example the last row can be divided by 5 in order to get a 1 instead of a 5. From it you can add multiples of the opposite of this row to the 3 other rows in order to row-reduce the matrix.
By the way if what you made is right, then the answer must be linear independent.
In order to find if the vectors are linear independent, construct a matrix as you did. (that is by putting your vectors as column). Then row-reduce the matrix. If it has at least a null row then the vectors are linear dependent.
EDIT: By the way I don't really understand what you've done. And be aware that the matrices you've put are NOT EQUAL, but "should" be equivalent. ("should", that is, if you don't make mistakes!)
• Mar 18th 2009, 11:04 PM
Craka
Thanks, my working for the matrix as far as row reduction is wrong, I realise now, and yes I shouldn't have used equal signs, they should have been equivalent signs.
• Mar 19th 2009, 01:45 AM
Craka
I've gone through the working again, hopefully this it is correct, could someone check please. I'm really not confident in doing Gauss-Jordan elimination.

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ \end{array}} \right]$

$\displaystyle ~\left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & { - 1} & { - 2} & { - 3} \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ \end{array}} \right]$

$\displaystyle ~\left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & { - 1} & { - 2} & { - 3} \\ 0 & { - 2} & { - 4} & { - 12} \\ 4 & 5 & 6 & 7 \\ \end{array}} \right]$

$\displaystyle ~\left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & { - 1} & { - 2} & { - 3} \\ 0 & { - 2} & { - 4} & { - 12} \\ 0 & { - 3} & { - 6} & { - 9} \\ \end{array}} \right]$

$\displaystyle ~\left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & { - 2} & { - 4} & { - 12} \\ 0 & { - 3} & { - 6} & { - 9} \\ \end{array}} \right]$

$\displaystyle ~\left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & { - 1} & { - 8} \\ 0 & { - 3} & { - 6} & { - 9} \\ \end{array}} \right]$

$\displaystyle ~\left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & { - 1} & { - 8} \\ 0 & 0 & 0 & 0 \\ \end{array}} \right]$

Am I suppose to intepret from here the independency or dependency? Am I also meant to continue working on the matrix to get zeros above the ones aswell?
• Mar 19th 2009, 02:02 AM
Showcase_22
Quote:

$\displaystyle \begin{pmatrix} {1}&{2}&{3}&{4}\\ {2}&{3}&{4}&{5}\\ {3}&{4}&{5}&{6}\\ {4}&{5}&{6}&{7} \end{pmatrix}$ perform Gaussian-Jordan elimination on this matrix. ( was always told these were "elementary row and column operations" =S).
Let's do:
$\displaystyle r_1 \rightarrow r_1$
$\displaystyle r_2 \rightarrow r_2-2r_1$
$\displaystyle r_3 \rightarrow r_3-3r_1$
$\displaystyle r_4 \rightarrow r_4-4r_1$.

$\displaystyle \begin{pmatrix} {1}&{2}&{3}&{4}\\ {0}&{-1}&{-2}&{-3}\\ {0}&{-2}&{-4}&{-6}\\ {0}&{-3}&{-6}&{-9} \end{pmatrix}$

We have accomplished a first column with a 1 at the top and zeroes for the other column entries.

Let's now do:
$\displaystyle r_1 \rightarrow r_1+2r_2$
$\displaystyle r_2 \rightarrow r_2$
$\displaystyle r_3 \rightarrow r_3-2r_2$
$\displaystyle r_3 \rightarrow r_3-3r_2$

$\displaystyle \begin{pmatrix} {1}&{0}&{-1}&{-2}\\ {0}&{-1}&{-2}&{-3}\\ {0}&{0}&{0}&{0}\\ {0}&{0}&{0}&{0} \end{pmatrix}$

We're pretty much done! There is however one more step we need to do:

$\displaystyle r_2 \rightarrow -r_2$

$\displaystyle \begin{pmatrix} {1}&{0}&{-1}&{-2}\\ {0}&{1}&{2}&{3}\\ {0}&{0}&{0}&{0}\\ {0}&{0}&{0}&{0} \end{pmatrix}$

We now have a matrix in row-reduced form with only two rows. This implies the matrix has a rank of two. Since the rank of a matrix is equal to the number of linearly independent rows, only two of those vectors are linearly independent.

$\displaystyle (1,2,3,4)$ and $\displaystyle (2,3,4,5)$ are the only linearly independent vectors.
• Mar 19th 2009, 03:54 AM
Craka
So Showcase with respect to all 4 vectors it would be true to say they are dependent?

Also is my working correct to the point I go to?
• Mar 19th 2009, 05:31 AM
ssj4Gogeta
Quote:

Originally Posted by Craka
Also is my working correct to the point I go to?

There appears to be a calculation mistake in your working.:)
• Mar 19th 2009, 06:55 AM
Showcase_22
Quote:

Originally Posted by Craka
So Showcase with respect to all 4 vectors it would be true to say they are dependent?

Yes, i'd say so. However, I would make it clear that $\displaystyle (1,2,3,4)$ and $\displaystyle (2,3,4,5)$ are linearly independent from each other.

Quote:

Also is my working correct to the point I go to?
As ssj4Gogeta said, there appears to be a mistake with what you've done.
If you look at your third matrix down, it has a -12 in it which shouldn't be there. It should be 6-12=-6.

From first glance that seems to be the only part which is wrong. A nice way to check if two answers are the same is firstly to check the numbers, and secondly to check the rank of the final matrices. Both our final matrices have two different ranks suggesting we've differed somewhere along the way.

I'd recommend working through it again and trying to get the answer i've got.
• Mar 20th 2009, 01:51 AM
Craka
Thanks for all you help guys, I'm hoping for a little more on this if its okay.
For what seems so elementary for some, I seem to find this rather irritably hard.

From just prior to where I made my mistake above, i've redone the working as follows.

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & { - 1} & { - 2} & { - 3} \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ \end{array}} \right]$

my new row3 = old row 3 - 3*row1 for the next below matrix
$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & { - 1} & { - 2} & { - 3} \\ 0 & { - 6} & { - 4} & { - 6} \\ 4 & 5 & 6 & 7 \\ \end{array}} \right]$

my new row4 = old row4 - 4*row1 for next below matrix

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & { - 1} & { - 2} & { - 3} \\ 0 & { - 6} & { - 4} & { - 6} \\ 0 & { - 3} & { - 6} & { - 9} \\ \end{array}} \right]$

my new row2 = old row 2*(-1) for next below matrix

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & { - 6} & { - 4} & { - 6} \\ 0 & { - 3} & { - 6} & { - 9} \\ \end{array}} \right]$

my new row3 = old row3 - 2*row4 for next below matrix

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 8 & {12} \\ 0 & { - 3} & { - 6} & { - 9} \\ \end{array}} \right]$

my new row4 = old row4 + 3*row2 for next below matrix

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 8 & {12} \\ 0 & 0 & 0 & 0 \\ \end{array}} \right]$

Which I think means I can take the row of all zeros out, and have just a 3 by 4 matrix of

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 8 & {12} \\ 0 & 0 & 0 & 0 \\ \end{array}} \right]$

Now here is my first point of confusion, I think I should divide the 8 by one to get a diagonal of 1s going left to right across matrix, and than I think I need to get a zero above the leading ones as per my next working

my new row3 =old row3 divided by 8 for next below matrix

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & {3/2} \\ \end{array}} \right]$

my new row1 = old row1 - 2*row2 for next below matrix

$\displaystyle {\begin{array}{*{20}c} 1 & 0 & { - 1} & { - 2} \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & {3/2} \\ \end{array}}$

my new row2 = old row 2 - 2*row3 for next below matrix

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 0 & { - 1} & { - 2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & {3/2} \\ \end{array}} \right]$

my new row1 = old row1 + row3 for next below matrix

$\displaystyle \left[ {\begin{array}{*{20}c} 1 & 0 & 0 & { - 1/2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & {3/2} \\ \end{array}} \right]$

And here is my second point of confusion I can't see how I can reduce any further without stuffing up my leading ones.
• Mar 20th 2009, 01:54 AM
Showcase_22
There's a mistake in your second matrix down.

The -6 in your second column should actually be a -2 since the calculation is 4-6=-2.

It's also better to leave the row of zeroes in otherwise you have a matrix of different dimension to the one you started with (you have a 4x4 matrix going to a 3x4).
• Mar 20th 2009, 02:08 AM
Craka
Woo Hoo, finally got the same result. Thanks so much for the help showcase, I was getting so frustrated with these, as I didn't think I was doing the correctly. Just need some practice and to stop making stupid little numerical errors and I think I'll be right.
• Mar 20th 2009, 02:39 AM
Showcase_22
Hey, no problem!

If you have any more questions like this, just holler!(Cool)