Column vectors are linearly independent using simple test:Why?
Suppose I've a matrix:
![A = \left[\begin{array}{rrrr} 1 & 3 & 1 & 3\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]](http://latex.codecogs.com/png.latex?A = \left[\begin{array}{rrrr} 1 & 3 & 1 & 3\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right])
Now the book i'm following says that "It is easy to see that in matrix 
, the first, second and fourth column vectors are linearly independent (These columns have the leading 
's)".
I don't understand this.
How did the author find out with simple examination that first, second and fourth column vectors are linearly independent?
And why this process works?
Is it possible for anyone to kindly explain this process?
Re: Column vectors are linearly independent using simple test:Why?
Quote:
How did the author find out with simple examination that first, second and fourth column vectors are linearly independent?
This is because only the fourth column has a nonzero third element; therefore, if any linear combination of the three columns is zero, the coefficient of the fourth column has to be zero. Similarly, of the first and second column, only the second one has a nonzero second element, so in a zero linear combination its coefficient has to be zero.
Re: Column vectors are linearly independent using simple test:Why?
Quote:
Originally Posted by
emakarov 
This is because only the fourth column has a nonzero third element; therefore, if any linear combination of the three columns is zero, the coefficient of the fourth column has to be zero. Similarly, of the first and second column, only the second one has a nonzero second element, so in a zero linear combination its coefficient has to be zero.
Thank you for your answer. I understand now.
