# Linear independence of vectors and minor of matrices

Prove that if a minor of order $$k$$ is nonzero, then the corresponding columns of the matrix are linearly independent

"The rank of a matrix is the maximal order of a nonzero minor of $$A$$"

The proof of this statement is based on the fact that "if a minor of order $$k$$ is nonzero, then the corresponding columns of the matrix are linearly independent".

It looks like if we are given $$r$$ vectors each of dimension $$n$$, then $$r$$ vectors are independent if the vectors with $$n-r$$ dimensions taken out are linearly independent ?

My Attempt $$\vec{A_1}=(a_1,a_2,a_3), \vec{A_2}=(b_1,b_2,b_3)\\ x_1\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}+x_2\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} \implies\begin{bmatrix}a_1&b_1\\a_2&b_2\\a_3&b_3\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\\ \text{For }\vec{A_1},\vec{A_2}\text{ to be independent, }\\ \vec{A_1}x_1+\vec{A_2}x_2=\vec{0} \text{ iff } x_1=x_2=0\\ a_1x_1+b_1x_2=0\\ a_2x_1+b_2x_2=0\\ a_3x_1+b_3x_2=0$$ If we find the solutions to any two out of three equations to be $$x_1=x_2=0$$ then it implies the solutions to all three equations must be $$x_1=x_2=0$$. ie., $$\begin{vmatrix} a_1&b_1\\ a_2&b_2\\ \end{vmatrix}\neq 0$$ Thus $$(a_1,a_2)$$ and $$(b_1,b_2)$$ are linearly independent$$\implies\vec{A_1},\vec{A_2}$$ are linearly independent.

Is it what is happening here ?

And how can one write a formal proof of the above statement ?

• You have to take out the correct $n-r$ entries, of course. Jul 14, 2020 at 19:13
• @TedShifrin I think we got to try all $(n-r)$ entries and see if atleast one of them got determinant nonzero. In that case $x_1=x_2=0$ become a solution, thus the given two vectors are independent, right ? Jul 14, 2020 at 19:40
• You need to have $r$ pivots in echelon form, so the rows in which they appear are the ones you want to keep. Jul 14, 2020 at 19:41
• @TedShifrin How can one formally prove it given the independece of vectors obtained by taking out some of the components, implies the idependence of the original vectors ? Jul 14, 2020 at 19:44
• If you know about echelon form, what I just said gives the proof. Removing the rows without pivots doesn't affect independence of the column vectors. Jul 14, 2020 at 19:45

Thanks @TedShifrin for the hint. $$x_1\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}+x_2\begin{bmatrix}b_1\\b_2\\\vdots\\b_n\end{bmatrix}+\cdots+x_r\begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\0\end{bmatrix}\\ AX=B\implies\begin{bmatrix}a_1&b_1&\cdots&c_1\\a_2&b_2&\cdots&c_2\\\vdots&\vdots&\ddots&\vdots\\a_n&b_n&\cdots& c_n\end{bmatrix}_{n\times r}\begin{bmatrix}x_1\\x_2\\\vdots\\x_r\end{bmatrix}_{r\times1}=\begin{bmatrix}0\\0\\\vdots\\0\end{bmatrix}_{n\times 1}\\$$ If $$\vec{A_1},\vec{A_2},...,\vec{A_r}$$ are linearly independent and $$n>r$$, the rref of $$A$$ will be of the form $$rref(A)=\begin{bmatrix}1&0&\cdots&0\\0&1&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots& 1\\0&0&\cdots& 0\\\vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots& 0\end{bmatrix}_{n\times r}$$ Columns with pivot $$1$$ are linearly independent, and eliminating nonpivot rows(with zeros) will not affect the linearity of the column vectors. ie., eliminating rows of $$A$$ corresponding to nonpivot rows of $$rref(A)$$ will not affect the linearity of corresponding column vectors
Hint: Let $$A$$ denote an $$n \times k$$ matrix, with $$n \geq k$$. The columns of $$A$$ are linearly independent if and only if the system $$Ax = 0$$ has a unique solution.
Let $$e_1,\dots,e_n$$ denote the columns of the $$n \times n$$ identity matrix. Let $$J$$ denote an $$n \times k$$ matrix whose columns are taken from the set $$\{e_1,\dots,e_n\}$$. Note that if $$(JA)x = 0$$ has a unique solution, then so does $$Ax = 0$$.