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This might be one of the easiest problems of linear algebra, but just to make sure I am not doing something completely stupid (as usual).

The exercise I have asks us to determine the rank of 2 matrices. The rank of 2 matrices is the number of linear independent rows or column. The rank of the columns and rows is equal.

The first matrix is:

$\left(\begin{matrix} 2 & -1 & 5 & -2 \\ 3 & 6 & -9 & 2 \\ -4 & 3 & 7 & 11\end{matrix}\right)$

Looking just for the rows, I can immediately see that none of the rows is a multiple of the other, so the 3 rows are linear independent. Therefore the rank of the matrix is 3.

The second matrix is:

$\left(\begin{matrix} 3 & 2 & 1 & 5 & 4 \\ 5 & 4 & 3 & 7 & 6 \\ 6 & 4 & 2 & 10 & 8 \\ -10 & -8 & -6 & -14 & -12 \\ 4 & 5 & -2 & -7 & 13 \\ \end{matrix}\right)$

We can see that the 3rd row is the double of the first one. So we remove 1 from the possible maximum rank 5. We can also observe that the 4th row is the negative double of row 2, so we exclude it.

Then the rank of this matrix would be also 3.

Am I missing something?

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    $\begingroup$ It's not just multiples, though; if a row can be written as a linear combination of two other rows, it is also excluded from the rank. $\endgroup$
    – user208649
    Commented Mar 19, 2015 at 0:11
  • $\begingroup$ what exactly is beautiful about this matrix? $\endgroup$
    – abel
    Commented Mar 19, 2015 at 0:35

2 Answers 2

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The fact that none of the rows in the first matrix is a multiple of another does not mean that the rank of the matrix is $3$. For example, consider the matrix whose rows are $(3,5)$, $(1,2)$, $(2,3)$. No row is a multiple of another, yet the sum of the second and third is the first. And indeed this matrix cannot possibly have rank $3$ since it has only two columns. As SRX said, the way to determine the rank is to row-reduce it.

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  • $\begingroup$ The answer is probably too long for a comment. But the short version is: row-reducing a matrix is an attempt to turn some rows into zeros. Any row you can turn into zeros is a row that is linearly dependent on the other rows. So the rows that are left give you the rank. $\endgroup$
    – rogerl
    Commented Mar 19, 2015 at 1:39
  • $\begingroup$ I'm not sure what you mean. It may be cumbersome, but in theory you can row-reduce any matrix. And it's not really that difficult, it would just take a little while. $\endgroup$
    – rogerl
    Commented Mar 19, 2015 at 1:42
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To determine the rank of a matrix, one can compute its row reduced echelon form. The rank of the matrix is just the number of columns which do NOT correspond to a free variable.

Edit: Let's obtain the RREF of the first matrix first.

$$\left(\begin{matrix} 2 & -1 & 5 & -2 \\ 3 & 6 & -9 & 2 \\ -4 & 3 & 7 & 11\end{matrix}\right)$$

Dividing the first row by $a_{11} = 2$ we have

$$\left(\begin{matrix} 1 & -0.5 & 2.5 & -1 \\ 3 & 6 & -9 & 2 \\ -4 & 3 & 7 & 11\end{matrix}\right)$$

Subtracting 3 times and -4 times the first row from the second row and the third row respectively we have

$$\left(\begin{matrix} 1 & -\frac{1}{2} & \frac{5}{2} & -1 \\ 0 & \frac{15}{2} & -\frac{33}{2} & 5 \\ 0 & 1 & 17 & 7\end{matrix}\right)$$

Dividing the second row by $a_{22} = \frac{15}{2}$ yields

$$\left(\begin{matrix} 1 & -\frac{1}{2} & \frac{5}{2} & -1 \\ 0 & 1 & -\frac{11}{5} & \frac{2}{3} \\ 0 & 1 & 17 & 7\end{matrix}\right)$$

Subtracting $\frac{-1}{2}$ times and 1 times the second row from the first row and the third row respectively we have

$$\left(\begin{matrix} 1 & 0 & \frac{7}{5} & -\frac{2}{3} \\ 0 & 1 & -\frac{11}{5} & \frac{2}{3} \\ 0 & 0 & \frac{96}{5} & \frac{19}{3}\end{matrix}\right)$$

Finally, dividing the third row by $a_{33} = \frac{96}{5}$ yields

$$\left(\begin{matrix} 1 & 0 & \frac{7}{5} & -\frac{2}{3} \\ 0 & 1 & -\frac{11}{5} & \frac{2}{3} \\ 0 & 0 & 1 & \frac{95}{288}\end{matrix}\right)$$

and subtracting $\frac{7}{5}$ times and $-\frac{11}{5}$ times the third row from the first row and the second row respectively we have

$$\left(\begin{matrix} 1 & 0 & 0 & -\frac{325}{288} \\ 0 & 1 & 0 & \frac{401}{288} \\ 0 & 0 & 1 & \frac{95}{288}\end{matrix}\right) = U$$

which is the RREF of A. Now what we have done so far is we have found a Gaussian elimination process $M = E_{k}E_{k-1} \ldots E_2E_1$ such that $MA = U$. It is well known that elementary matrices are nonsingular and so is M. Let's see why the above statement

To determine the rank of a matrix, one can compute its row reduced echelon form. The rank of the matrix is just the number of columns which do NOT correspond to a free variable.

is true in this particular example. Note that the fourth column $\vec{u_4}$ of $U$ can be written into a linear combination of the other three columns of $U$. $$\vec{u_4} = -\frac{325}{288}\vec{u_1} + \frac{401}{288}\vec{u_2} + \frac{95}{288}\vec{u_3}$$ Multiplying $M^{-1}$ on both sides yields $$\vec{a_4} = -\frac{325}{288}\vec{a_1} + \frac{401}{288}\vec{a_2} + \frac{95}{288}\vec{a_3}$$ which means $a_4$ can be written into a linear combination of the other three columns of $A$. Therefore the spanning set $\{\vec{a_1},\vec{a_2},\vec{a_3},\vec{a_4}\}$ for $R(A)$ can be pared down to $\{\vec{a_1},\vec{a_2},\vec{a_3}\}$. Also, $\vec{x} \in N(A)$ if and only if $x_1 = \frac{325}{288}x_4, x_2 = -\frac{401}{288}x_4, x_3 = -\frac{95}{288}x_4$ if and only if $\vec{x} \in \text{span}\{(325,-401,-95,288)^T\}$. Therefore $\text{nullity}(A) = 1$ and by rank-nullity theorem $\text{rank}(A) = n - \text{nullity}(A) = 4 - 1 = 3$. We also find a basis $\{\vec{a_1},\vec{a_2},\vec{a_3}\}$ for $R(A)$.

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  • $\begingroup$ Rows do not correspond to variables, columns do. Columns corresponding to free variables are those which are not $\vec{e_i}$. $\endgroup$
    – Empiricist
    Commented Mar 19, 2015 at 0:21
  • $\begingroup$ If you still could not get this, probably you can obtain the RREF of one of these matrices and we would demonstrate how to carry on. $\endgroup$
    – Empiricist
    Commented Mar 19, 2015 at 0:22
  • $\begingroup$ I have added an explanation about why the rule is true for the first matrix. The concepts can be generalized into any matrix: (1) a basis for $N(A)$ is obtained by each time letting only one free variable to be nonzero in $Ax = 0$. (2) columns of $A$ corresponding to free variables can always be written into a linear combination of columns of $A$ which do not correspond to a free variable, and those columns can be pared down in find a spanning set for $R(A)$. (3) apply rank-nullity theorem. $\endgroup$
    – Empiricist
    Commented Mar 20, 2015 at 9:23

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