# Can the vectors be written as a linear combination?

• MHB
• mathmari
In summary, the conversation discusses linear dependence and linear combinations of vectors. The given vectors are shown to be linearly dependent through the use of the Gauss elimination algorithm. The possibility of writing vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ as linear combinations of the given vectors is also explored. It is concluded that $\overrightarrow{b}$ can be written as a linear combination, while $\overrightarrow{c}$ cannot. The conversation also touches on the idea of using column elimination instead of row elimination for finding linear dependencies.
mathmari
Gold Member
MHB
Hey!

We have the vectors $\overrightarrow{a_1}=\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}, \overrightarrow{a_2}=\begin{pmatrix}-1 \\0 \\ 2\end{pmatrix}, \overrightarrow{a_3}=\begin{pmatrix}7 \\ 8 \\ 6\end{pmatrix}$.

I have shown that these vectors are linearly dependent:
$\begin{bmatrix} \begin{matrix} 1 & -1 & 7\\ 2 & 0 & 8\\ 3 & 2 & 6 \end{matrix}\left|\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right.\end{bmatrix}\begin{matrix} \\ 2.\text{row}-2\cdot 1.\text{row}\\ 3.\text{row}-3\cdot 1.\text{row} \end{matrix} \longrightarrow \begin{bmatrix} \begin{matrix} 1 & -1 & 7\\ 0 & 2 & -6\\ 0 & 5 & -15 \end{matrix}\left|\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right.\end{bmatrix}\begin{matrix} \\ \\ 3.\text{row}-\frac{5}{2}\cdot 2.\text{row} \end{matrix} \longrightarrow \begin{bmatrix} \begin{matrix} 1 & -1 & 7\\ 0 & 2 & -6\\ 0 & 0 & 0 \end{matrix}\left|\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right.\end{bmatrix}$

right? (Wondering)

Can we write the vectors $\overrightarrow{b}=\begin{pmatrix}-6 \\ -4 \\ 2\end{pmatrix}$ and $\overrightarrow{c}=\begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix}$ as linear combinations of the vectors $\overrightarrow{a_1}, \overrightarrow{a_2}, \overrightarrow{a_3}$ ? (Wondering)

To check this I applied the Gauss elimination algorithm:

$\begin{bmatrix} \begin{matrix} 1 & -1 & 7\\ 2 & 0 & 8\\ 3 & 2 & 6 \end{matrix}\left|\begin{matrix} -6\\ -4\\ 2 \end{matrix}\right.\end{bmatrix}\begin{matrix} \\ 2.\text{row}-2\cdot 1.\text{row}\\ 3.\text{row}-3\cdot 1.\text{row} \end{matrix} \longrightarrow \begin{bmatrix} \begin{matrix} 1 & -1 & 7\\ 0 & 2 & -6\\ 0 & 5 & -15 \end{matrix}\left|\begin{matrix} -6\\ 8 \\ 20 \end{matrix}\right.\end{bmatrix}\begin{matrix} \\ \\ 3.\text{row}-\frac{5}{2}\cdot 2.\text{row} \end{matrix} \longrightarrow \begin{bmatrix} \begin{matrix} 1 & -1 & 7\\ 0 & 2 & -6\\ 0 & 0 & 0 \end{matrix}\left|\begin{matrix} -6\\ 8\\ 0 \end{matrix}\right.\end{bmatrix}$ $\begin{bmatrix} \begin{matrix} 1 & -1 & 7\\ 2 & 0 & 8\\ 3 & 2 & 6 \end{matrix}\left|\begin{matrix} 0\\ 2\\ 1 \end{matrix}\right.\end{bmatrix}\begin{matrix} \\ 2.\text{row}-2\cdot 1.\text{row}\\ 3.\text{row}-3\cdot 1.\text{row} \end{matrix} \longrightarrow \begin{bmatrix} \begin{matrix} 1 & -1 & 7\\ 0 & 2 & -6\\ 0 & 5 & -15 \end{matrix}\left|\begin{matrix} 0\\ 2 \\ 3 \end{matrix}\right.\end{bmatrix}\begin{matrix} \\ \\ 3.\text{row}-\frac{5}{2}\cdot 2.\text{row} \end{matrix} \longrightarrow \begin{bmatrix} \begin{matrix} 1 & -1 & 7\\ 0 & 2 & -6\\ 0 & 0 & 0 \end{matrix}\left|\begin{matrix} 0\\ 2\\ -2 \end{matrix}\right.\end{bmatrix}$

So, in both cases the vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ cannot be written as linear combinations of the vectors $\overrightarrow{a_1}, \overrightarrow{a_2}, \overrightarrow{a_3}$.

Is everything correct? Could I improve something? (Wondering)
I have to give all the possible solutions of the linear equations system.

For the vector $\overrightarrow{b}$:

we get $2\lambda_2-6\lambda_3=8$ and $\lambda_1-\lambda_2+7\lambda_3=-6$.

Solving at the first equation for $\lambda_2$ we get $\lambda_2=4+3\lambda_3$.

From the other equation we get $\lambda_1=\lambda_2-7\lambda_3-6 =4+3\lambda_3-7\lambda_3-6=-2-4\lambda_3$.

So, are all the possible solutions of the linear equations system: $(\lambda_1, \lambda_2, \lambda_3)=(-2-4\lambda_3, 4+3\lambda_3, \lambda_3)=\lambda_3(-4, 3, 1)+(-2, 4, 0)$ ? (Wondering)

Last edited by a moderator:
Yes, these vectors are dependent. In showing whether or not vector $$\vec{b}$$ can be written as a linear combination, you arrived at a reduced matrix in which all 4 entries in the last row equal to 0. since 0x+ 0y+ 0z= 0 for any x, y, z, that shows that $$\vec{b}$$ can be written as a linear combination, in infinitely many ways.

For $$\vec{c}$$ you arrived at a reduce matrix with last row "0 0 0 | -2". Since 0x+ 0y+ 0z is not equal to -2 or any x, y, z, yes, $$\vec{c}$$ cannot be written as a linear combination.

HallsofIvy said:
Yes, these vectors are dependent. In showing whether or not vector $$\vec{b}$$ can be written as a linear combination, you arrived at a reduced matrix in which all 4 entries in the last row equal to 0. since 0x+ 0y+ 0z= 0 for any x, y, z, that shows that $$\vec{b}$$ can be written as a linear combination, in infinitely many ways.

For $$\vec{c}$$ you arrived at a reduce matrix with last row "0 0 0 | -2". Since 0x+ 0y+ 0z is not equal to -2 or any x, y, z, yes, $$\vec{c}$$ cannot be written as a linear combination.

I understand! (Smile)

And are the infinitely many ways for $\vec{b}$ described as follows?

mathmari said:
For the vector $\overrightarrow{b}$:

we get $2\lambda_2-6\lambda_3=8$ and $\lambda_1-\lambda_2+7\lambda_3=-6$.

Solving at the first equation for $\lambda_2$ we get $\lambda_2=4+3\lambda_3$.

From the other equation we get $\lambda_1=\lambda_2-7\lambda_3-6 =4+3\lambda_3-7\lambda_3-6=-2-4\lambda_3$.

So, are all the possible solutions of the linear equations system: $(\lambda_1, \lambda_2, \lambda_3)=(-2-4\lambda_3, 4+3\lambda_3, \lambda_3)=\lambda_3(-4, 3, 1)+(-2, 4, 0)$ ? (Wondering)

Yes. Since your matrix can be row reduced a matrix with one row all 0s, this is equivalent two equations in three variables, a, b, and c. You can solve for two of the variables in terms of the third, or solve for all three in terms of a new parameter. Here you have the single parameter $$\lambda_3$$.

mathmari said:
Could I improve something? (Wondering)

Hey mathmari! (Smile)

I would recommend using column elimination instead of row elimination.
As a result we'll find the minimum set of independent vectors that also makes it easier to find the other dependencies. (Nerd)

Ah ok...

Thank you very much! (Happy)

## 1. Can you explain what a linear combination is?

A linear combination is a mathematical operation that involves multiplying each vector in a set by a corresponding scalar value and then adding those products together. This creates a new vector that can be expressed as a combination of the original vectors.

## 2. How do you determine if a set of vectors can be written as a linear combination?

To determine if a set of vectors can be written as a linear combination, you need to check if there exists a set of scalars that, when multiplied with each vector and added together, equal the desired vector. This can be done using matrix operations or by setting up a system of equations.

## 3. What is the significance of being able to write a vector as a linear combination?

Being able to write a vector as a linear combination allows us to represent complex vectors in a more simplified and organized form. It also allows us to perform operations on vectors more easily, such as addition, subtraction, and finding linearly independent vectors.

## 4. Can any set of vectors be written as a linear combination?

No, not all sets of vectors can be written as a linear combination. For a set of vectors to be written as a linear combination, they must be linearly dependent, meaning that at least one vector in the set can be expressed as a linear combination of the others. If a set of vectors is linearly independent, it cannot be written as a linear combination.

## 5. How does writing a vector as a linear combination relate to solving systems of equations?

Writing a vector as a linear combination is essentially the same as solving a system of equations. The coefficients of the linear combination represent the solutions to the system of equations. Similarly, if a set of vectors can be written as a linear combination, it means that there exists a solution to the system of equations formed by those vectors.

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