I am quite new to studying linear algebra and I am struggling with the concept of writing a set of vectors in column form and finding out if the set of vectors are linear independent (I feel like I might have missed something fundamental).

So here is how I think:

Let's say we have a set of three vectors $\vec{u_1},\vec{u_2},\vec{u_3}$ in some base $\mathbb{B}$ ($\mathbb{R}^3$) with the coordinates: $u_1 = (a_1, b_1, c_1)$, $u_2 = (a_2, b_2, c_2)$, $u_3 = (a_3, b_3, c_3)$. You want to find out if these vectors are linear independent.

The set of vectors are linear independent if: $x\vec{u_1}+y\vec{u_2}+z\vec{u_3} = \vec{0}$ only has the trivial solution: $x = y = z = 0$.

This equation can be written as the the system:

$$\left( \begin{array}{3} a_1x+a_2y+a_3z = 0 \\ b_1x+b_2y+a_3z = 0 \\ c_1x+c_2y+c_3z = 0 \end{array}\right )$$

which can be written as:

$$\left(\begin{array}{3} a_1,a_2,a_3 \\ b_1,b_2,b_3\\ c_1,c_2,c_3 \end{array} \right )\left(\begin{array}{3} x \\ y\\ z \end{array} \right ) = \left(\begin{array}{3} 0 \\ 0\\ 0 \end{array} \right )$$

The coordinates for $\vec{u_1},\vec{u_2},\vec{u_3}$ will then become the columns in the matrix $A$.

$$A = \left(\begin{array}{3} a_1,a_2,a_3 \\ b_1,b_2,b_3\\ c_1,c_2,c_3 \end{array} \right )$$

What I don't understand is:

To me the most obvious way would be to write the coordinates of the vectors in row-form. If I then row reduce the matrix I would easily see what vector ($\vec{u_1},\vec{u_2},\vec{u_3}$) that would be a multiple of another vector. Like this:

$$A = \left(\begin{array}{3} a_1,b_1,c_1 \\ a_2,b_2,c_2\\ a_3,b_3,c_3 \end{array} \right )$$

If I row reduce the coordinate matrix when it is in column form I am confused in what kind of operation I am actually doing (I am comparing coordinates in one "direction" of the base with coordinates in another "direction"?).

I understand mathematically that the determinant is the same even if the matrix is transposed and that it will give the same result. But still the choice to do the operation this way feel "wrong" to me.

So please explain to me what I have missed in my reasoning above.

Thank you!

  • $\begingroup$ First, A has to be a 3x3 matrix. So, you can't write A as 3 rows of $a_i + b_i + c_i$. Now if A is 3x3, then the (x, y, z) vector needs to be a 3x1 matrix by the rules of matrix multiplication. But if you wrote A the way you have, you would not be able to get the original set of equations back. $\endgroup$ – user137481 May 7 '14 at 21:57
  • $\begingroup$ Sorry, that was my mistake, should be correct now. Thank you for your comment! $\endgroup$ – Lukas Arvidsson May 8 '14 at 6:18

I will explain this looking at a much simpler example, that is something in the 2-dimensional case. Say we have the following equations :

\begin{equation} \begin{aligned} 2x + 3y & =5&\text{ (1) } \\ x + 3y &= 4 & \text{ (2) } \end{aligned} \end{equation}

This system can be represented as follows:

$$\begin{pmatrix} 2 & 3 \\ 1 & 3 \end{pmatrix} \begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix} $$

When doing row reduction, I am allowed to do the following operations :

(1) Interchanging two rows

(2) Multiplying a row by a non-zero scalar.

(3) Adding a multiple of one row to another row

All these operations on the matrix translate to the operations we are familiar with when solving a system of linear equations. For example , subtracting equation $2$ from $1$ will result in the equation $x = 1$. On the matrix this means subtracting row $2$ from row $1$ on both sides or on the augmented matrix, which gives

$$\begin{pmatrix} 1 & 0 \\ 1 & 3 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$$ To simplify further, we can subtract row $1$ from $2$ and it follows

$$\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$$

Why are we doing this ? Matrices became more than just a tool for solving linear equations. They became algebraic objects themselves, with their many properties. Read A.CALEY, A memoir on the theory of matrices. Sorry, I digress.

You can also do column operations but then the matrices have to be different

$$\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} 2 &1 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 5 & 4 \end{pmatrix}$$ Why don't we represent it this way ? You tell me. I didn't answer your question directly , but I think with the right motivation you will find your way.

Now coming back to linear independence, say we have the vectors $u_1 = \begin{pmatrix} 2 \\ 0 \end{pmatrix} $ and $u_2= \begin{pmatrix} 1 \\ 2 \end{pmatrix} $. As you mentioned, the vectors are linearly independent if the system of equations has only a trivial solution, that is, $xu_1 + yu_1 = 0 $ if $x=y=0$ which means $$\begin{pmatrix} 2x \\ 0 \end{pmatrix} + \begin{pmatrix} y \\ 2y \end{pmatrix} = \begin{pmatrix} 2x +y \\ 2y \end{pmatrix} = \begin{pmatrix} 2x +y \\ 0x + 2y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ if $$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ Now the problem of finding whether a set of vectors are linearly independent has been reduced to a problem of finding a solution to a system of linear equations. It is to be noted that $$ \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} u_1, u_2 \end{pmatrix}$$ It is just a representation which is convenient.

  • $\begingroup$ thank you very much for your answer! To clarify my question, in your example, lets say we have two vectors $u_1$ = $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ and $u_2$ = $\begin{pmatrix} 8 \\ 6 \end{pmatrix}$ Now this set of vectors is clearly linearly dependant. But if we write it in column form $\begin{pmatrix} 4 & 8 \\ 3 & 6 \end{pmatrix} = \begin{pmatrix} u_1, u_2 \end{pmatrix}$ we cannot easily see that one row is a multiple of another row, but still we use row operations to find out if they are? $\endgroup$ – Lukas Arvidsson May 8 '14 at 7:18
  • $\begingroup$ Yes, but we do see that the columns are linearly dependent. They question of linear dependency boils eventually down to the question of whether the matrix is invertible or not. This matrix is not invertible since the determinant is zero. We can explicitly calculate the determinant to see if it's zero or we can use the fact that the determinant is zero if one row/column is the multiple of another. Keep in mind that if $A$ is invertible , then $Ax=0$ has only a trivial solution. $\endgroup$ – The very fluffy Panda May 8 '14 at 7:30
  • $\begingroup$ Thank you very much! for your time and explanation, makes sense! It is much appreciated! $\endgroup$ – Lukas Arvidsson May 8 '14 at 7:33

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