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If I augment a matrix, say v1 v2 v3 = w, and rref the augmented matrix with result being an identity matrix. Does that mean there is no linear combination of v1 v2 v3 that results in w?

ex.

 1    -1     0     3
 2     0     5    -6
 3     1    -6    17
 4     3     8    11

rref

 1     0     0     0
 0     1     0     0
 0     0     1     0
 0     0     0     1
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  • $\begingroup$ Indeed it does. $\endgroup$ – Omnomnomnom Feb 14 '17 at 18:06
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take your first augmented matrix as saying that $w$ is a linear combination of $v_1, v_2$ and $v_3$. then the equivalent system at the second matrix tell you that the system has no solution(inconsistent). that means the first system of equation has no solution and hence $w$ is linearly independent of the $v'$s.

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If you write a linear system in matrix form $Ax=b$, solvability means $b$ is a linear combination of the columns of $A$.

Elementary row operations never change the linear relations between the columns; more precisely, if $B$ is a matrix with columns $b_1,\dots,b_n,b_{n+1}$, $C$ is a matrix with columns $c_1,\dots,c_n,c_{n+1}$, and $C$ is obtained from $B$ by applying an elementary row operation, then $$ \sum_{i=1}^{n+1}\alpha_ib_i=0 \quad\text{if and only if}\quad \sum_{i=1}^{n+1}\alpha_ic_i=0 \tag{*} $$ In particular, $$ b_{n+1}=\sum_{i=1}^{n}\alpha_ib_i \quad\text{if and only if}\quad c_{n+1}=\sum_{i=1}^{n}\alpha_ic_i \tag{**} $$ This is because there exists an invertible matrix $E$ such that $C=EB$, so that $c_i=Eb_i$.

One elementary row operation or several doesn't change the situation.

If, in a row reduced form of the augmented matrix $[A\mid b]$ (the augmented matrix), the last column is a pivot column, then the last column is not a linear combination of other columns, so neither $b$ is a linear combination of the columns of $A$, because of (**). Hence the linear system $Ax=b$ has no solution.

Also the converse is true: if in a row reduced form of $[A\mid b]$ the last column is not a pivot column, then it is a linear combination of the previous columns and therefore also $b$ is a linear combination of the columns of $A$.

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