Question

Construct a 3x3 matrix, not in echelon form, whose columns span R^3 . Show that the matrix you construct has the desired property.

Answer 1

3x3 matrix ======> 1 2 -1
-1 5 7
3 1 3

Answer 2

@eliassaab

Answer 3

the matrix you have written above is given or you constructed it?

Answer 4

I made it up

Answer 5

To span R^3, you want to show that the three vectors are linearly independent right?, not sure if \(\LaTeX\) works for you, but mathematically that means if \(v_1=(1,-1,3), v_2=(2,5,1), v_3=(-1,7,3)\) then you want to show that the only solution to:
\[ \lambda_1 v_1 + \lambda_2 v_2 + \lambda_3v_3=0 \implies \lambda_1=\lambda_2=\lambda_3 \]

Answer 6

You can also just row reduce your matrix, if you find 1 pivot element in every column then your vectors are linearly independent.

Answer 7

to make them linearly independent i could make them not multiples of the other vectors?

Answer 8

true, the easiest way to make linearly independent vectors is to introduce a zero where others have a number. It might be a bit of a cheat, because it will then be near to row reduced echelon form. However it's what most do.

Answer 9

so the matrix 5 7 9 would be an example of linear independence
0 2 4
0 -6 -8

Answer 10

indeed.

Answer 11

it spans R^3

Answer 12

how do yo know what matrix to construct to obtain an answer like the one i posted above?

Answer 13

after rref it?

Answer 14

what you posted above is in my book. That is actually discussed in a later chapter.

Answer 15

there are algorithms which guarantee you that your column are guaranteed to be linearly independent and therefore span the column space. An example of such an algorithm would be gram-schmidt orthogonalization. However this can get lengthy.

If you're given a set of vectors, or some that you made up, you can always verify that they are linearly independent by putting the matrix in rref, if you get a full rank, meaning pivots in every column, then your vectors are linearly independent, if you get a free variable, then they are linearly dependent.

Answer 16

If you don't know yet about such a method, then you can always (for low dimensions only) remember yourself that linear independence means that the vectors point in different directions. This statement is fine and correct, however you can of course no longer do that when you work in R^4 or in R^n for that sake of the argument.

However, since you work in R^3 you can also just throw the x-y-z room and pick any three vectors in different directions there (draw a few) then you're guaranteed already that they are linearly independent.

Answer 17

so rref the matrix i constructed gave me 1 0 0 | -9
0 1 0 | 3
0 0 1 | -3

Answer 18

So this would be linearly dependent?

Answer 19

perfect yes, linearly independent because the matrix has full rank (3 times 3 matrix, 3 pivots, implies the rank is full)

Answer 20

Ohhhh okay I see

Answer 21

you augmented the matrix with something as I see, you could have left that away, it doesn't change your result.

Answer 22

in real you augment the matrix with zero, by the definition I have given you above you will see why, seems like that comes later for you. However augmentation by 0 is usually left blank, because you can multiply, add linear multiples of zero and divide zero by anything (except for division through 0) and you will still get zero.

Answer 23

Okay thanks.

Answer 24

you're welcome