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I have this problem:

Determine whether the column vectors $$ \begin{bmatrix}0\\0\\0\\1\end{bmatrix}\, , \begin{bmatrix}0\\0\\2\\1\end{bmatrix}\,, \begin{bmatrix}0\\3\\2\\1\end{bmatrix}\,, \begin{bmatrix}4\\3\\2\\1\end{bmatrix} $$ are independent or dependent, no explanation necessary.

Right away, I thought it was dependent; just by looking at it, we can easily see the reduced form is in a staircase form, which would mean that it was dependent. I.e, divide every row by the coefficient leading, then perform row reductions.

I asked Wolfram, and it said it was independent. I do not quite understand what I have done incorrectly. Perhaps I am looking at the problem the wrong way; maybe they were talking about rows, not columns!

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No; remember that row rank = column rank. Have you actually done row-reduction, aka Gaussian elimination? –  user99680 Mar 29 at 20:38
    
Yeah, and I get rref form but I'm looking through the answers to see why i am wrong. –  A A Mar 29 at 20:44
1  
Let $c_i, i=1..4$ be the four column vectors. Then $|[c_1, c_2, c_3, c_4]|=0\implies$ vectors are dependent, and $|[c_1, c_2, c_3, c_4]|\neq 0\implies$ vectors are independent. $|\cdot|$ denotes the determinant. –  Sujaan Kunalan Mar 29 at 20:44
    
But my book says, 'a set of vectors is linearly dependant if there are coefficents that makes the equation not zero', but after reducing the vectors to rref, we find that there are possible coefficents –  A A Mar 29 at 20:47

4 Answers 4

up vote 8 down vote accepted

I think you are confusing the words dependent and independent. If you set these vectors as rows (or columns) of a matrix and row reduce the matrix, you will indeed find that the matrix is in "staircase form", as you call it. This means that the vectors are linearly independent, not linearly dependent.

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Oh, you are right, that was why I was wrong –  A A Mar 29 at 20:44

Suppose there is a linear combination $$a(0,0,0,1)+b(0,0,2,1)+c(0,3,2,1)+d(4,3,2,1) \\ =(4d,3(c+d),2(b+c+d),a+b+c+d)=(0,0,0,0)$$

So we have $d=0$ (since $4d=0$), so that $c=0$ (since $3c+3d=3c=0$), so that $b=0$ (since $2(b+c+d)=2b=0$), and finally $a=0$ (since $a+b+c+d=a=0$).

Thus, we see that the set of vectors is independent.

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Call the vectors $a,b,c,d$ then:

$a=[0,0,0,1]$

$\frac 12 (b-a)=[0,0,1,0]$

$\frac 13 (c-b)=[0,1,0,0]$

$\frac 14 (d-c)=[1,0,0,0]$

So your four vectors form a basis for the space, since they generate the vectors of the standard basis, and are therefore independent (and also form a spanning set).

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Form a $4\times 4$ matrix with the four vectors, starting from the rightmost one: $$ \begin{bmatrix} 4 & 0 & 0 & 0 \\ 3 & 3 & 0 & 0 \\ 2 & 2 & 2 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix} $$ This is a lower triangular matrix with no zero diagonal elements, so it's invertible and hence has rank $4$.

The row echelon form (staircase form) is really easy to find, isn't it?

Then…

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