Does the determinant of a matrix made up of column vectors being non-zero imply that the vectors are independent?

Question

Let's say we have 3 vectors and we make up a matrix where we depict the vectors as the columns of the matrix. If we calculate the determinant of the matrix and we get a non-zero number, does that mean that the vectors are linearly independent?

Answer 1

Yes. If the columns are linearly dependent, then we can show that the determinant is $0$. So if the determinant is not $0$ then the columns are linearly independent.

If the columns of the matrix are $\vec{c}_1, \vec{c}_2, \ldots, \vec{c}_n$ and there are scalars $a_1, a_2, \ldots, a_n$ such that

$$a_1 \vec{c}_1 + a_2 \vec{c}_2 + \cdots + a_n \vec{c}_n = \vec{0}$$

then we see that the vector

$$\langle a_1, a_2, \ldots , a_n\rangle^T$$

is an eigenvector of the matrix with eigenvalue $0$. Since $0$ is an eigenvalue, the determinant is $0$.

Answer 2

Let's say we have 3 vectors

$$ \left( \begin{matrix} a\\ b\\ c \end{matrix} \right), \left( \begin{matrix} d\\ e\\ f \end{matrix} \right), \left( \begin{matrix} g\\ h\\ i \end{matrix} \right) $$

and we make up a matrix where we depict the vectors as the columns of the matrix

$$ \left( \begin{matrix} a & d & g\\ b & e & h\\ c & f & i\\ \end{matrix} \right) $$

If we calculate the determinant of the matrix and we get a non-zero number

$$ \left| \begin{matrix} a & d & g\\ b & e & h\\ c & f & i\\ \end{matrix} \right| =aei+dhc+gbf-gec-dbi-ahf\neq 0 $$

does that mean that the vectors are linearly independent?

Yes.

Look at the 6 summands that make up the determinant. Note how each one is the product of one element of each of the original vectors and also including one element from each "row". Let me illustrate

$$ \left( \begin{matrix} \color{red}{a}\\ \color{green}{b}\\ \color{blue}{c} \end{matrix} \right), \left( \begin{matrix} \color{red}{d}\\ \color{green}{e}\\ \color{blue}{f} \end{matrix} \right), \left( \begin{matrix} \color{red}{g}\\ \color{green}{h}\\ \color{blue}{i} \end{matrix} \right) $$

and

$$\color{red}{a}\color{green}{e}\color{blue}{i}+\color{red}{d}\color{green}{h}\color{blue}{c}+\color{red}{g}\color{green}{b}\color{blue}{f}-\color{red}{g}\color{green}{e}\color{blue}{c}-\color{red}{d}\color{green}{b}\color{blue}{i}-\color{red}{a}\color{green}{h}\color{blue}{f}\neq 0$$

There seems to be a pattern to this. If you allow yourself for a moment to only care about the colors and substitute each summand by $\color{red}{R}\color{green}{G}\color{blue}{B}$, the result is clearly 0

$$\color{red}{R}\color{green}{G}\color{blue}{B}+\color{red}{R}\color{green}{G}\color{blue}{B}+\color{red}{R}\color{green}{G}\color{blue}{B}-\color{red}{R}\color{green}{G}\color{blue}{B}-\color{red}{R}\color{green}{G}\color{blue}{B}-\color{red}{R}\color{green}{G}\color{blue}{B} = 3\color{red}{R}\color{green}{G}\color{blue}{B}-3\color{red}{R}\color{green}{G}\color{blue}{B}= 0$$

And that's exactly what happens when the vectors are linear dependent, which means all of them can be expressed as one vector multiplied with a scalar, solely for artistic reasons that vector shall be $(\color{red}{R},\color{green}{G},\color{blue}{B})^T$ which is equal to $(\color{red}{a},\color{green}{b},\color{blue}{c})^T$

$$ \left( \begin{matrix} \color{red}{a}\\ \color{green}{b}\\ \color{blue}{c} \end{matrix} \right)=\left( \begin{matrix} \color{red}{R}\\ \color{green}{G}\\ \color{blue}{B} \end{matrix} \right), \left( \begin{matrix} \color{red}{d}\\ \color{green}{e}\\ \color{blue}{f} \end{matrix} \right) = x\left( \begin{matrix} \color{red}{R}\\ \color{green}{G}\\ \color{blue}{B} \end{matrix} \right), \left( \begin{matrix} \color{red}{g}\\ \color{green}{h}\\ \color{blue}{i} \end{matrix} \right)=y\left( \begin{matrix} \color{red}{R}\\ \color{green}{G}\\ \color{blue}{B} \end{matrix} \right) $$

Let the transformation begin

$$ \begin{array}{c|ccccccl} & \color{red}{a}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{c}&+\color{red}{g}\color{green}{b}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{c}&-\color{red}{d}\color{green}{b}\color{blue}{i}&-\color{red}{a}\color{green}{h}\color{blue}{f}&\neq 0\\ \hline \color{red}{a}=\color{red}{R}& \color{red}{R}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{c}&+\color{red}{g}\color{green}{b}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{c}&-\color{red}{d}\color{green}{b}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\ \color{green}{b}=\color{green}{G}& \color{red}{R}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{c}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{c}&-\color{red}{d}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\ \color{blue}{c}=\color{blue}{B}& \color{red}{R}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{B}&-\color{red}{d}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\ \color{red}{d}=x\color{red}{R}& \color{red}{R}\color{green}{e}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\ \color{green}{e}=x\color{green}{G}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\ \color{blue}{f}=x\color{blue}{B}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}x\color{blue}{B}&-\color{red}{g}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}x\color{blue}{B}&\\ \color{red}{g}=y\color{red}{R}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+y\color{red}{R}\color{green}{G}x\color{blue}{B}&-y\color{red}{R}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}x\color{blue}{B}&\\ \color{green}{h}=y\color{green}{G}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}y\color{green}{G}\color{blue}{B}&+y\color{red}{R}\color{green}{G}x\color{blue}{B}&-y\color{red}{R}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}y\color{green}{G}x\color{blue}{B}&\\ \color{blue}{i}=y\color{blue}{B}& \color{red}{R}x\color{green}{G}y\color{blue}{B}&+x\color{red}{R}y\color{green}{G}\color{blue}{B}&+y\color{red}{R}\color{green}{G}x\color{blue}{B}&-y\color{red}{R}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}y\color{blue}{B}&-\color{red}{R}y\color{green}{G}x\color{blue}{B}&\\ \end{array} $$

And that last line is (apart from the x&y's) exactly as the one above, which equalled zero

$$xy\color{red}{R}\color{green}{G}\color{blue}{B}+xy\color{red}{R}\color{green}{G}\color{blue}{B}+xy\color{red}{R}\color{green}{G}\color{blue}{B}-xy\color{red}{R}\color{green}{G}\color{blue}{B}-xy\color{red}{R}\color{green}{G}\color{blue}{B}-xy\color{red}{R}\color{green}{G}\color{blue}{B} = 3xy\color{red}{R}\color{green}{G}\color{blue}{B}-3xy\color{red}{R}\color{green}{G}\color{blue}{B}= 0$$

Intuitively speaking, the linear dependency "smears" all 3 distinguishable a, d and g into one homogenous blob R. The same holds true for the other two rows. The scalar factors cancel out as well due to the construction of the determinant.

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tl;dr; got color blind;

Checking vectors if they are linear dependent and assembling them into a matrix and checking if the determinant is 0 is the same thing.