Finding linear dependence relations among vectors [closed]

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I have a set of row vectors:

$$\begin{matrix} 3 &6 & 0 \\ 4 & 7 &0 \\ 5 &8 & 0 \\ \end{matrix}$$

I know that if we're able to create a zero row using elementary row operations, then we know the rows of the matrix will be linearly dependent. I was able to create a zero row pretty easily:

$$\begin{matrix} 3 &6 &0 \\ 1& 1 &0 \\ 0 &0 & 0 \\ \end{matrix}$$

We are considering the matrix as row vectors.

I'm a bit confused with how to find a dependence relationship however. Are looking for a relationship such that:

$$x[3, 6, 0] + y[4, 7, 0] + z[5, 8, 0] = [0, 0, 0]$$

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2 Answers

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Since you have row vectors, performing row operations is not correct. Instead, you should proceed to solve the last equation in your post which is in the correct direction. So, to solve for $x,y,z$ in $$x[3,6,0]+y[4,7,0]+z[5,8,0]=0$$ just write out the system for each coordinate: \begin{align}3x+4y+5z&=0\\6x+7y+8z&=0\\0x+0y+0z&=0\end{align} (where the last one is trivial, i.e. is satisfied for any possible choice of $x,y,z$ and so you can omit it in the first place). Now, the first two equations give $$\begin{cases}3x+4y+5z&=0\\6x+7y+8z&=0\end{cases} \overset{(2)-(1)}\implies \begin{cases}3x+4y+5z&=0\\x+y+z&=0\end{cases}\overset{1-3\cdot(2)}\implies \begin{cases}y+2z&=0\\x+y+z&=0\end{cases}$$ which yields the solution (set of infinitely many solutions actually) $$(x,y,z)=(z,-2z,z), \quad z\in \mathbb R$$ For example choose $z=1$, then $(x,y,z)=(1,-2,1)$ is a vector of scalars that satisfies your initial equation (and shows therefore that the given vectors are linearly dependent).

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I assume that you obtained the zero row using elementary row operations. If you also write down what operations you did, you will also get the combination your looking for.

In this particular example: $$\left( \begin{array}{ccc|ccc} 3 & 6 & 0 & 1 & 0 & 0 \\ 4 & 7 & 0 & 0 & 1 & 0 \\ 5 & 8 & 0 & 0 & 0 & 1 \end{array} \right)\sim \left( \begin{array}{ccc|ccc} 3 & 6 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 &-1 & 1 & 0 \\ 2 & 2 & 0 &-1 & 0 & 1 \end{array} \right)\sim \left( \begin{array}{ccc|ccc} 3 & 6 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 &-1 & 1 & 0 \\ 0 & 0 & 0 & 1 &-2 & 1 \end{array} \right)$$

What exactly those numbers mean?

In the first step I subtracted the first row from other two rows. So I did the same operation on the right matrix.

Similarly, second step was subtracting the second row twice from the third row. And I did this operation on both part of the matrix.

Notice that in each step the numbers on the right are precisely the coefficients such that the linear combination of the original rows with this coefficients give the rows of the new matrix.

For example, from the last row I get $$1\cdot(3,6,0)-2\cdot(4,7,0)+1\cdot(5,8,0)=(0,0,0).$$

This is an illustration of a general fact that that if you do some row operations on a matrix, the rows of the new matrix are linear combinations of the rows of the original matrix. But, as you see above, if you also want the coefficients of these linear combinations, you need to do some additional work.

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