Can someone explain these rules about determining linear independence in matrices that my professor taught?

Question

So the way that I was taught to determine linear independence and come up with a dependence relationship is to get a matrix, put it into reduced row echelon form, and then make that all equal to zero. If there are non zero solutions then it is linearly dependent. There are some shortcuts he taught though which I think could be useful on a test where I have limited time, but I don't entirely get what why they exist or in what situations I could use them.

1. If there are more columns then rows then it is linearly dependent. From experience though this is not necessarily true because the matrix [-6 -4] which has more columns than rows is actually linearly independent. Why is this the case for this matrix, yet this rule seems to work for some other matrices?
2. If there is a column of zeros then it is linearly dependent.
3. If a column is a multiple of another column then it is linearly dependent.

This next rule is a trend that I've been noticing as I solve these problems but I don't know if it is actually a rule about linear independence or if its just a coincidence.

1. If there is a row of zeros at the end then it is linearly dependent unless that row of zeros is for a matrix with more rows than columns. If it has more rows than columns then it can still be linearly independent with a row of zeros in reduced row echelon form. If there are more rows than column without the last row being zeros then it is linearly dependent. Is all this actually true?

Answer 1

1. As pointed out in the comments, $-\frac{3}{2}(-4) + (-6) = 0$, so they are linearly dependent.

2. Label the columns $\{a_1\dots a_n\}$. Without loss of generality, assume that $a_1$ is all zeros. Then we have $(1)a_1 + (0)a_2+\dots+(0)a_n = 0$ and these vectors are linearly dependent.

3. Assume that $a_2 = \alpha a_1$. If $\alpha=0$, we have the previous case. If $\alpha\neq0$ then we have $(\alpha)a_1 + (-1)a_2 + (0)a_3+\dots+(0)a_n = 0$ and they are linearly dependent.

4. If the last entry of all $n$ vectors in $\mathbb{R}^m$ is $0$, then we can consider the vectors $\{b_1\dots b_n\}$ in $\mathbb{R}^{m-1}$ constructed by taking just the first $m-1$ components. If $n \geq m$ (at least as many columns as rows), then consider the $b$ vectors, we have $n$ vectors in $\mathbb{R}^{m-1}$, but $n>m-1$, so they must be linearly dependent. If the $b$'s are linearly dependent, with coefficients $\beta_1\dots\beta_n$, then notice that $\sum_{k=1}^n \beta_k a_k = 0$ since the first $m-1$ components are $0$ by the linear dependence of the $b$'s and the last component is $0$ from our assumption on the $a$'s.

Answer 2

$\begin{pmatrix}-6&-4\end{pmatrix}$ is linearly dependant since your prof means that you have to put the vectors as columns in a matrix. The vectors in this case would be $(-6)$ and $(-4)$ and they are clearly linearly dependant.

What actually happens is that if you put stuff in row echelon form, if all columns of your original matrix are linearly independant, that will also be the case after transforming to row-echelon form. The next two paragraphs explain your second and third point using this.

If there is a zero column than that column is linearly dependant on the other columns, so no linear independance.

Same for the third point. The two columns that differ by a factor are linearly independant.

Note: in the next paragraph I use the fact that the dimension of the span of the rows of a matrix equals the dimension of the span of the columns of that matrix. I don't prove it, since it is something well-known and is probably in every linear algebra course somewhere. That dimension equals the rank of the matrix.

For the last question, we should be a bit more precise with what we are doing. We want to find the rank of the matrix with your vectors as columns, since the rank is the dimension of the column space (and the row space). Now if there are more rows than columns, and you have a zero row you can still have that the dimension of the span of the rows equals the number of columns and therefore the rank can still equal the number of columns and this means that the columns are linearly independant since $n$ vectors that span an $n$-dimensional space clearly have to be linearly independant.