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There is a three-dimensional vector $v$. Show that $v$ can be expressed as a linear combination of $v_1$, $v_2$, and $v_3$, where $v_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$, $\quad v_2 = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$, $\quad v_3 = \begin{pmatrix} 11 \\ 1 \\ -14 \end{pmatrix}$.

I know that I am supposed to use eigenvalues ($Mv = kv$) and linear combinations of vectors, like in here: http://www.vitutor.com/geometry/vectors/linear_combination.html, but I'm not sure how to use these terms to solve this problem, or even begin it.

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    $\begingroup$ you don't need eigen values for this. $\endgroup$
    – Anurag A
    Aug 6 '15 at 18:35
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Hint

Let $v=\begin{pmatrix}a\\b\\c\end{pmatrix}$. You need to see if there exists $x,y,z$ (scalars) such that $$v=xv_1+yv_2+zv_3.$$ In other words you need to see if the following system is consistent for all $a,b,c$. $$ \left[\begin{array}{ccc|l} 1 & -1 & 11 &a\\ 1 & 1 & 1 & b\\ 1 & 1 & -14 &c\\ \end{array} \right] $$

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  • $\begingroup$ Can you explain the last matrix you had: Are you stating that the value of that 3x3 matrix must be divisible by $a, b, c$? $\endgroup$ Aug 6 '15 at 19:00
  • $\begingroup$ it is an augmented matrix. Nothing to do with divisibility. For more information might I suggest you look up the following: en.wikipedia.org/wiki/Augmented_matrix $\endgroup$
    – Anurag A
    Aug 6 '15 at 20:21
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Anurag A has shown you how to find how $v$ can be expressed as a linear combination of $v_1,v_2,v_3$, which is basically solving the linear equation system $Ax = v$ for $x$, where $A$ is a matrix with $v_1,v_2,v_3$ as columns.

However, the question is to show that a $v$ can be expressed as a linear combination of $v_1,v_2,v_3$, not what the linear combination is. From the wording of your question I also guess that you don't know what $v$ actually is.

So, what you need to do is to show that $v_1, v_2, v_3$ spans three-dimensional space; this means that every vector in three-dimensional space can be expressed as a linear combination of $v_1,v_2,v_3$ (which is just what you want)!

Since you have exactly three vectors, you know that you have just enough vectors to span three-dimensional space, so the question is if your vectors are linearly independent - if they are not, they will only span a lower-dimensional subspace of three-dimensional space.

So, it actually turns out that you want to show that $v_1,v_2,v_3$ is a basis for three-dimensional space; a basis is a minimal spanning set for the space.

You can go about showing that $v_1,v_2,v_3$ are linearly independent or form a basis in the following ways:

  1. Show that none of $v_1,v_2,v_3$ can be written as a linear combination of the others.
  2. Show that the the equation $\alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3 = 0$, where $\alpha_1,\alpha_2,\alpha_3$ are scalars, only has the trivial solution $\alpha_1 = \alpha_2 = \alpha_3 = 0$.
  3. Show that every $v$ in three-dimensional space can be written as a linear combination of $v_1,v_2,v_3$.
  4. The matrix $A = \left( v_1 ~~ v_2 ~~ v_3 \right)$ has non-zero determinant.
  5. The matrix $A$ is invertible.
  6. The matrix $A$ does not have $0$ as an eigenvalue.

I will you show how to do number 2.

The equation $\alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3 = 0$ can be written on matrix form: $$\begin{pmatrix} \mid & \mid & \mid \\ v_1 & v_2 & v_3 \\ \mid & \mid & \mid \end{pmatrix} \begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ we can write this more compactly as an augmented matrix, like Anurag A did: $$\left(\begin{array}{ccc|l} 1 & -1 & 11 &0\\ 1 & 1 & 1 & 0\\ 1 & 1 & -14 &0\\ \end{array} \right)$$ here the variables $\alpha_1, \alpha_2, \alpha_3$ we are solving for are implicit: Row 1 corresponds to variable $\alpha_1$, row 2 to $\alpha_2$, etc.

We can then do Gaussian elimination on this matrix. We first eliminate all non-zero elements in the first column below the first row, by adding multiples of the first row to the rows below it, and get: $$\left(\begin{array}{ccc|l} 1 & -1 & 11 &0\\ 0 & 2 & -10 & 0\\ 0 & 2 & -25 &0\\ \end{array} \right)$$ we then do the same thing for the second column, removing all non-zeros below the second row, and get: $$\left(\begin{array}{ccc|l} 1 & -1 & 11 &0\\ 0 & 2 & -10 & 0\\ 0 & 0 & -15 &0\\ \end{array} \right)$$ and now we see that the last row says $15 \alpha_3 = 0$, which gives us $\alpha_3 = 0$. We can then use this fact to also see that $\alpha_2 = 0$ from row 2, and then use both these facts to see that $\alpha_1 = 0$ from row 1.

So, we see that the equation $\alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3 = 0$ has only the solution $\alpha_1 = \alpha_2 = \alpha_3 = 0$, so the vectors $v_1,v_2,v_3$ are linearly independent, so they form a basis, which means that any vector $v$ in three-dimensional space can be written as a linear combination of them.

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  • $\begingroup$ Thank you for the very detailed, helpful answer. I had a question on this part, how to show that $ v_1,v_2,v_3$ are linearly independent or form a basis--for numbers 1-6, would I have to prove/show all six parts, or would part 2 be sufficient? $\endgroup$
    – Grace
    Aug 8 '15 at 22:04
  • $\begingroup$ To show any one of them is sufficient. $\endgroup$
    – Calle
    Aug 9 '15 at 16:29

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