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I am trying determine the row and column ranks of the matrix $$\begin{bmatrix} 0 & 2 & 3 & -4 & 1 \\ 0 & 0 & 2 & 3 & 4 \\ 2 & 2 & -5 & 2 & 4 \\ 2 & 0 & -6 & 9 & 7 \end{bmatrix}.$$

To determine the row rank I multiply i-th row by $\lambda_i$ then sum the rows and equal it to be zero which results in $\lambda_1 = -\lambda_2$, $\lambda_1 = -\lambda_3$, $\lambda_3 = -\lambda_4$ and $\lambda_1 = \lambda_4$. So for $\lambda_1 = 1$, $$\begin{bmatrix} 0 & 2 & 3 & -4 & 1 \end{bmatrix} \\ + \begin{bmatrix} 0 & 0 & -2 & -3 & -4 \end{bmatrix} \\ + \begin{bmatrix} -2 & -2 & 5 & -2 & -4 \end{bmatrix} \\ + \begin{bmatrix} 2 & 0 & 6 & -9 & 7 \end{bmatrix} \\ =0.$$ So we are right that 4 rows are dependent. Then I eliminate the 4-th row, by guess, and I see that there is no way the rows of $$\begin{bmatrix} 0 & 2 \\ 0 & 0 \\ 2 & 2 \end{bmatrix} $$ to be dependent so the row rank is $3$. By a theorem that I've studied it the row rank and the column rank of a matrix are same. But the book wants the column rank of the given matrix by calculation and I can't find out it column rank.

Here is my attempt: By multiplication of i-th column by $\lambda_i$ then sum the columns and equal the sum to be zero which results in (after some calculations,) $\lambda_1 = -\frac{15}{8}\lambda_4$, $\lambda_2 = \frac{19}{8}\lambda_4$, $\lambda_3 = 0$ and $\lambda_5 = \frac{6}{8}\lambda_4$. Neither all zero nor all non-zero and even if it was too complicated to guess next step.

My questions are:

1- How $\lambda_3 = 0$ can help? It meaning is so ambiguous! Will it appear in the collection of maximum number of independent columns or never?

2- How can I calculate the column rank of the given matrix when I can't simply guess it?

Thank you!

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  • $\begingroup$ Do you know what "row reduction" or "Gauss-Jordan elimination" is? $\endgroup$ Oct 10 '16 at 11:52
  • $\begingroup$ @Omnomnomnom, Yes, actually. I know reduced row echelon form. But I don't know how to use it in this situation. Also, is there a way to determine the column rank by the information I have about $\lambda_i$ s? $\endgroup$
    – Liebe
    Oct 10 '16 at 11:56
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A matrix's rank is the maximum amount of linear independent columns/rows, which is exactly the dimension of the subspace spanned by these. If you perform Gauss-Jordan elimination, you will end up with a set of rows/columns which keep generating the same space, and you will easily see if there are any linearly dependent ones.

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The column rank does not change when you add, substract, multiply by a constant and permute the columns. So we will try to symplify the matrix with such operations (I only gave the explanations of the operations for the first three operations.) Divide C1 and C2 by 2 : $$\operatorname{rank}\begin{bmatrix} 0 & 2 & 3 & -4 & 1 \\ 0 & 0 & 2 & 3 & 4 \\ 2 & 2 & -5 & 2 & 4 \\ 2 & 0 & -6 & 9 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 3 & -4 & 1 \\ 0 & 0 & 2 & 3 & 4 \\ 1 & 1 & -5 & 2 & 4 \\ 1 & 0 & -6 & 9 & 7 \end{bmatrix}$$ Substract $3$ times C2 to C3 : $$\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & -4 & 1 \\ 0 & 0 & 2 & 3 & 4 \\ 1 & 1 & -8 & 2 & 4 \\ 1 & 0 & -6 & 9 & 7 \end{bmatrix}$$ Add $6$ times C1 to C3 : $$\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & -4 & 1 \\ 0 & 0 & 2 & 3 & 4 \\ 1 & 1 & -2 & 2 & 4 \\ 1 & 0 & 0 & 9 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & -4 & 1 \\ 0 & 0 & 1 & 3 & 4 \\ 1 & 1 & -1 & 2 & 4 \\ 1 & 0 & 0 & 9 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 3 & 4 \\ 1 & 1 & -1 & 6 & 4 \\ 1 & 0 & 0 & 9 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 4 \\ 1 & 1 & -1 & 2 & 4 \\ 1 & 0 & 0 & 3 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 4 \\ 1 & 1 & -1 & 3 & 4 \\ 1 & 0 & 0 & 3 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 4 \\ 1 & 1 & -1 & 1 & 4 \\ 1 & 0 & 0 & 1 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 4 \\ 1 & 1 & -1 & 4 \\ 1 & 0 & 0 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 4 \\ 1 & 1 & -1 & 3 \\ 1 & 0 & 0 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & -1 & 7 \\ 1 & 0 & 0 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & -1 & 1 \\ 1 & 0 & 0 & 1 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & -1 \\ 1 & 0 & 0 \end{bmatrix}=3$$

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