Confusion about row reduction and linear independence of a column vector.

Question

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!

Answer 1

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.