Expressing non-basis columns as a linear combination

1. Sep 3, 2015

mattyk

1. The problem statement, all variables and given/known data
A= \begin{bmatrix} 1 & 1 & 1 & 2 & 5 \\ 2 & 3 & 1 & -1 & -5 \\ 1 & 0 & 2 & 0 & -1 \\ 1 & 2 & 0 & 1 & 2 \end{bmatrix}

2. Relevant equations
I get this RREF
\begin{bmatrix} 1 & 0 & 2 & 0 & 0 \\ 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}

Rank = 3
I'm then asked to write a column basis which is the 1st, 2nd & 4th column

3. The attempt at a solution

"Express each non-basis column of A as a linear combination of basis elements"

I'm just asking for a little clarity on the question and then exactly how do I express it.

Am I being asked to express the 3rd and 5th column (the non-basis columns) as a linear combination?
And if so what does that look like? All my searching is just making me more confused?

2. Sep 3, 2015

Staff: Mentor

Since your work shows the 1st, 2nd, and 4th columns to be linearly independent, and therefore a basis for the column space (I didn't check your work for accuracy), then it must be the case that the 3rd and 5th columns are a linear combination of the 1st, 2nd, and 4th columns. Write an equation to express the fact that the 3rd column is equal to $c_1<\text{col. 1}> + c_2<\text{col. 2}> + c_4<\text{col. 4}>$, and determine the constants $c_1, c_2,$ and $c_4$. Do the same sort of thing to show that the 5th column is a linear comb. of those three column vectors.

3. Sep 3, 2015

mattyk

Just to make sure I have this right
I go on to work out

c1(1, 2, 1, 1) + c2(1, 3, 0, 2) + c4(2, -1, 0, 1) = (1, 1, 2, 0)

and

c1(1, 2, 1, 1) + c2(1, 3, 0, 2) + c4(2, -1, 0, 1) = (5, -5, -1, 2).

And this shows that c3 and c5 are linear combinations of c1, c2 & c4?

4. Sep 3, 2015

mattyk

Just to make sure I have this right
I go on to work out

c1(1, 2, 1, 1) + c2(1, 3, 0, 2) + c4(2, -1, 0, 1) = (1, 1, 2, 0)

and

c1(1, 2, 1, 1) + c2(1, 3, 0, 2) + c4(2, -1, 0, 1) = (5, -5, -1, 2).

And this shows that c3 and c5 are linear combinations of c1, c2 & c4?

5. Sep 3, 2015

Staff: Mentor

Don't use the same constants in each equation, since c1 in the first equation is not likely to be the same as c1 in the second equation.

The equations above don't show anything until you actually solve for the constants.

6. Sep 4, 2015

HallsofIvy

Staff Emeritus
So (c1, 2c1, c1, c1)+ (c2, 3c2, 0, 2c2)+ (2c4, -c4, 0, c4)= (1, 1, 2, 0)
(c1+ c2+ 2c4, 2c1+ 3c2- c4, c1, c1+ 2c2+ c1)= (1, 1, 2, 0)
c1+ c2+ 2c4= 1
2c1+ 3c2- c4= 1
c1= 2
c1+ 2c2+ c4= 0

Now solve for c1, c2, c3 and c4

I've changed your "c"s to "d"s. They are not necessarily the same coefficients as before.

No! The "c"s (and "d"s) are numbers. It is the vectors that are linear combinations.