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Let $a,b,c \in \mathbb{R},$ $\vec{v_1}=\begin{pmatrix}1\\4\\1\\-2 \end{pmatrix},$ $\vec{v_2}=\begin{pmatrix}-1\\a\\b\\2 \end{pmatrix},$ and $\vec{v_1}=\begin{pmatrix}1\\1\\1\\c \end{pmatrix}.$ What are the conditions on the numbers $a,b,c$ so that the three vectors are linearly dependent on $\mathbb{R}^4$? I know that the usual method of solving this is to show that there exists scalars $x_1,x_2,x_3$ not all zero such that \begin{align} x_1\vec{v_1}+x_2\vec{v_2}+x_3\vec{v_3}=\vec{0}. \end{align}

Doing this would naturally lead us to the augmented matrix \begin{pmatrix} 1 & -1 & 1 &0\\ 4 & a & 1 &0\\ 1& b & 1 &0\\ -2 & 2 & c &0\\ \end{pmatrix}

Doing some row reduction would lead us to the matrix

\begin{pmatrix} 1 & -1 & 1 &0\\ 4 & a & 1 &0\\ 0& b+1 & 0 &0\\ 0 & 0 & c+2 &0\\ \end{pmatrix} I'm not quite sure how to proceed after this. Do I take cases on when whether $b+1$ or $c+2$ are zero and nonzero?

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    $\begingroup$ Just calculate their determinant: they're linearly dependent if and only if their determinant is $0$. $\endgroup$ – Bernard May 21 at 11:12
  • $\begingroup$ I think that would only work if I have three vectors in $\mathbb{R}^3.$ $\endgroup$ – Kurome May 21 at 11:16
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    $\begingroup$ Oh! Sorry, I didn't notice the vectors are $\mathbf R^4$. Then you have to check each $3{\times}3$ minor is $0$. $\endgroup$ – Bernard May 21 at 11:18
  • $\begingroup$ That's quite an unfamiliar method for me, is there an article (or text) that you could lead me to where I can check up on that? $\endgroup$ – Kurome May 21 at 11:20
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    $\begingroup$ Don't you know that the rank of a matrix is the maximum size of a square submatrix with nonzero determinant? $\endgroup$ – Bernard May 21 at 11:31
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Just build a matrix with these vectors as rows and perform row reduce. The vectors will be linearly dependent if at least one row is made of zeros. The idea is that the rank of a matrix is the maximum number of linearly independent rows (or columns), hence, the rows will be linearly dependent if and only if $r(A) < 3$.

$$ \begin{pmatrix} 1 & 4 & 1& -2 \\ -1 & a & b & 2\\ 1 & 1 & 1 & c\end{pmatrix}\to \begin{pmatrix} 1 & 4 & 1& -2 \\ 0 & a+4 & b+1 & 0\\ 0 & -3 & 0 & c+2\end{pmatrix}\to \begin{pmatrix} 1 & 4 & 1& -2 \\ 0 & -3 & 0 & c+2\\ 0 & a+4 & b+1 & 0\end{pmatrix} \to $$

$$ \begin{pmatrix} 1 & 4 & 1& -2 \\ 0 & -3 & 0 & c+2\\ 0 & 0 & b+1 & \frac{(c+2)(a+4)}{3}\end{pmatrix} $$

So the vectors are linearly dependent if and only if the last row is filled with zeros, i.e. $b = -1 \wedge (a=-4 \vee c=-2)$.

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  • $\begingroup$ So can I just take $c+2=0$ or $b+1=0$? $\endgroup$ – Kurome May 21 at 11:30
  • $\begingroup$ I think we should be reducing a 4x3 matrix, not a 3x4 matrix like you did. $\endgroup$ – Kurome May 21 at 11:54
  • $\begingroup$ @Kurome It is equivalent. The rank of a matrix is both the maximum number of linearly independent rows and columns. $\endgroup$ – PierreCarre May 21 at 11:57
  • $\begingroup$ I see, I understand now. Is the reason why you did a 3x4 matrix is that it's more tractable? $\endgroup$ – Kurome May 21 at 11:58
  • $\begingroup$ @Kurome Not really, the workload is the same. $\endgroup$ – PierreCarre May 21 at 11:59
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Your start is fine.

Consider in the equation $x_1v_1+x_2v_2 + x_3v_3 =0$.

We have two choices:

  1. $x_3=0$.

(Can you see the similarity between $v_1$ and $v_2$?) $v_1$ and $v_2$ will spann a one dimensional space if and only if $a=-4$ and $b= -1$. That is the vectors will be linear dependent for any choice of $c$.

  1. $x_3\ne0$.

In this case the equation is equivalent to $xv_1 + yv_2=v_3$.

(I leave the fun part of solving that to you. )

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  • $\begingroup$ I think I like this solution more. It doesn't appeal to the rank of the resulting matrix. $\endgroup$ – Kurome May 21 at 12:14
  • $\begingroup$ Well, there is seldom a unique solution, and I thought this approach is more connected to the understanding of linear dependency. $\endgroup$ – AD. May 21 at 14:03
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The matrix $$ \begin{pmatrix} 1 & -1 & 1 &0\\ 4 & a & 1 &0\\ 0& b+1 & 0 &0\\ 0 & 0 & c+2 &0\\ \end{pmatrix} $$ is not in row echelon form. Sum to the second column the first multiplied by $-4$, getting $$ \begin{pmatrix} 1 & -1 & 1 &0\\ 0 & a+4 & -3 &0\\ 0& b+1 & 0 &0\\ 0 & 0 & c+2 &0\\ \end{pmatrix} $$ Still not row echelon form, but “almost”. Anyway, computing the rank is easy.

If $b+1\ne0$, you can swap the second and third row, then continue the Gaussian elimination getting $$ \begin{pmatrix} 1 & -1 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0& 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} $$ which has rank $3$. Thus a necessary condition so the vectors are linearly dependent is $b=-1$. Now the matrix becomes $$ \begin{pmatrix} 1 & -1 & 1 &0\\ 0 & a+4 & -3 &0\\ 0& 0 & 0 &0\\ 0 & 0 & c+2 &0\\ \end{pmatrix} $$ If $a+4\ne0$ and $c+2\ne0$, a row echelon form is (swapping the third and fourth rows) $$ \begin{pmatrix} 1 & -1 & 1 &0\\ 0 & a+4 & -3 &0\\ 0 & 0 & c+2 &0\\ 0& 0 & 0 &0\\ \end{pmatrix} $$ so the matrix has rank $3$. In order the rank is less than $3$ we need either $a=-4$ or $c=-2$.

You can so state that the vectors are linearly dependent if and only if $b=-1$ and either $a=-4$ or $c=-2$.

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Fleshing out a comment by Bernard, the rank of a matrix is equal to the order of its largest nonzero minor. Writing the three vectors as rows to save vertical space, for the three vectors to be linearly dependent we want the matrix $$A=\begin{bmatrix}1&4&1&-2 \\ -1&a&b&2 \\ 1&1&1&c \end{bmatrix}$$ to be rank-deficient. That occurs when all of the determinants of the matrices obtained by deleting one column of $A$ vanish. This generates the following system of equations: $$3b+3 = 0 \\ ac+2a+4c+8 = 0 \\ bc+2b+c+2=0 \\ ac-4bc+2a-2b+6 = 0.$$ This isn’t the most efficient way to solve this particular problem, but this relation between the rank of a matrix and its minors is worth knowing. It can come in handy in other contexts.

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