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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 ?

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  • $\begingroup$ You have to take out the correct $n-r$ entries, of course. $\endgroup$ Jul 14, 2020 at 19:13
  • $\begingroup$ @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 ? $\endgroup$
    – Sooraj S
    Jul 14, 2020 at 19:40
  • $\begingroup$ You need to have $r$ pivots in echelon form, so the rows in which they appear are the ones you want to keep. $\endgroup$ Jul 14, 2020 at 19:41
  • $\begingroup$ @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 ? $\endgroup$
    – Sooraj S
    Jul 14, 2020 at 19:44
  • $\begingroup$ 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. $\endgroup$ Jul 14, 2020 at 19:45

2 Answers 2

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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

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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 $Ax = 0$.

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  • $\begingroup$ I have tried to update my OP could you please have a look ? $\endgroup$
    – Sooraj S
    Jul 14, 2020 at 18:58

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