Expressing non-basis columns as a linear combination

[mattyk]

Homework Statement

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}

Homework 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

The question then asks
"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?

 

[Mark44]

Homework Statement

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}

Homework 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

The question then asks
"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?

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.

 

[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?

 

[mattyk]

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.

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?

 

[Mark44]

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?

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.

 

[HallsofIvy]

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)

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

and

d₁(1, 2, 1, 1) + d₂(1, 3, 0, 2) + d₄(2, -1, 0, 1) = (5, -5, -1, 2).

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

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

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