# Thread: Linear combinations (from a 5x7 matrix)

1. ## Linear combinations (from a 5x7 matrix)

Sorry this might be a bit messy! I apologize!

The subset S = {v1, v2, v3, v4, v5, v6, v7} of R^5 is given by:
v1 = (-2, -2, -1, 1, 2), v2 = (3, 3, 2, -1, -3), v3 = (5, 5, 4, -1, -5), v4 = (3, 2, 2, -2, -4), v5 = (4, 3, 3, -2, -5), v6 = (-2, -2, -1, 2, 3), v7 = (-3, -4, -1, 3, 4)

The reduced row echelon form of the matrix A is the matrix B where:

A = $\begin{bmatrix} -2 & 3 & 5 & 3 & 4 & -2 & -3 \\ -2 & 3 & 5 & 2 & 3 & -2 & -4 \\ 1 & 2 & 4 & 2 & 3 & -1 & -1 \\ 1 & -1 & -1 & -2 & -2 & 2 & 3 \\ 2 & -3 & -5 & -4 & -5 & 3 & 4 \end{bmatrix}$

AND

B = $\begin{bmatrix} 1 & 0 & 2 & 0 & 1 & 0 & 1 \\ 0 & 1 & 3 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$

(A) Write down a four element subset S1 of S so that every vector which is a linear combination of elements of S is already a linear combination of elements of S1.

I have:
Columns associated with v1, v2, v4 and v6 have leading entries, thus s1 = v1=a, v2 = b, v4 = c, v6 = d
(where a, b, c, d are real values)

(B) Express each of the three elements of S which are not in S1 as a linear combination of elements of S1.

I currently have:
v3 = 2v1 + 3v2 (this is correct)
v5 = v1 + v2 + v3 (this is not correct)
v7 = v1 + v3 + 2v4 (this is not correct)

That took a while to write up!

Help, maybe from scratch?

Thanks!

2. Originally Posted by iExcavate
Sorry this might be a bit messy! I apologize!

The subset S = {v1, v2, v3, v4, v5, v6, v7} of R^5 is given by:
v1 = (-2, -2, -1, 1, 2), v2 = (3, 3, 2, -1, -3), v3 = (5, 5, 4, -1, -5), v4 = (3, 2, 2, -2, -4), v5 = (4, 3, 3, -2, -5), v6 = (-2, -2, -1, 2, 3), v7 = (-3, -4, -1, 3, 4)

The reduced row echelon form of the matrix A is the matrix B where:

A = $\begin{bmatrix} -2 & 3 & 5 & 3 & 4 & -2 & -3 \\ -2 & 3 & 5 & 2 & 3 & -2 & -4 \\ 1 & 2 & 4 & 2 & 3 & -1 & -1 \\ 1 & -1 & -1 & -2 & -2 & 2 & 3 \\ 2 & -3 & -5 & -4 & -5 & 3 & 4 \end{bmatrix}$

AND

B = $\begin{bmatrix} 1 & 0 & 2 & 0 & 1 & 0 & 1 \\ 0 & 1 & 3 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$

(A) Write down a four element subset S1 of S so that every vector which is a linear combination of elements of S is already a linear combination of elements of S1.

I have:
Columns associated with v1, v2, v4 and v6 have leading entries, thus s1 = v1=a, v2 = b, v4 = c, v6 = d
(where a, b, c, d are real values)

(B) Express each of the three elements of S which are not in S1 as a linear combination of elements of S1.

I currently have:
v3 = 2v1 + 3v2 (this is correct)
v5 = v1 + v2 + v3 (this is not correct)
v7 = v1 + v3 + 2v4 (this is not correct)

That took a while to write up!

Help, maybe from scratch?

Thanks!
It looks like you have a mistake in the first column of A.
Maybe fix it and carry out the reduction again.

3. Your picking of 4 linearly independent vectors went well, so now you can form any other vector within $\mathbb{R}^4$

$S_1 = v_1, v_2, v_4, v_6 = \begin{bmatrix} -2 \\ -2 \\ 1 \\ 1 \\ 2\end{bmatrix},\begin{bmatrix} 3 \\ 3 \\ 2 \\ -1 \\ -3\end{bmatrix},\begin{bmatrix} 3 \\ 2 \\ 2 \\ -2 \\ -4\end{bmatrix},\begin{bmatrix} -2 \\ -2 \\ -1 \\ 2 \\ 3\end{bmatrix}$

You now want to make a linear combination of the basis elements of $S_1$ to generate the elements $v_3,v_5,v_7$

So for example:

$v_3 = \begin{bmatrix} 5 \\ 5 \\ 4 \\ -1 \\ -5\end{bmatrix} = a\cdot v_1 + b\cdot v_2 + c\cdot v_4 + d\cdot v_6$

You now get a system of equations you should be able to solve.

4. Sorry, for the previous part, I had written the question itself out incorrectly! In the matrix, the '1' in column A should be '-1', like the brackets! The row-reduction is still correct!

Thanks, also, how do I find 'a, b, c, d' for the linear combinations? I'm a little confused sorry brouwer

5. I'll give you a simple example: I want to express $\begin{bmatrix}1 \\ 0\end{bmatrix}$ as a linear combination of $\begin{bmatrix}2 \\ 3\end{bmatrix},\begin{bmatrix}1 \\ 1\end{bmatrix}$. You'll get the following system:

$a\cdot \begin{bmatrix}2 \\ 3\end{bmatrix} + b \cdot \begin{bmatrix}1 \\ 1\end{bmatrix} = \begin{bmatrix}1 \\ 0 \end{bmatrix}$

This is the same as the equations:

$2a+b = 1$
$3a+b = 0$

Now we have to get $a$ in terms of $b$ and substitute in the other equation so:

$a = \frac{-b}{3}$

from the 2nd equation and plug this in the 1st equation to get

$\frac{-2b}{3} + b = 1$

And we solve for $b$ to get $b = 3$. Filling in this $b$ in either equation will give us our desired $a$. With 4 variables it's the same, but you have to substitute a few times. Good luck.

6. Originally Posted by iExcavate

(B) Express each of the three elements of S which are not in S1 as a linear combination of elements of S1.
Can't you just read the linear combination off the row reduced matrix? I learnt this method, but cannot understand it in terms of this 5 x 7 matrix. I didn't think there should be a need to use simultaneous equations?

v3 = 2v1 + 3v2

How do I read v5 and v7 off the row reduced matrix? Perhaps, that is an easier way?

7. Yes it's also possible to read the linear combination off your reduced matrix. And it's really a lot easier actually, sorry.

$v_3 = 2v_1+3v_2$
$v_5 = v_1+v_2+v_4$

I don't know why you put $v_3$ here again, as it's a linear combination of the elements of $S_1$. Because these are 4 linearly independent vectors and span whole $\mathbb{R}^4$, which basically means that you can form any other vector that looks like $\begin{bmatrix}a & b & c & d & 0\end{bmatrix}^T$ out of these. $v_7$ can be read off like the other vectors but just in terms of $av_1+bv_2+cv_4+dv_6$.