# How to determine the column rank of the given matrix?

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!

• Do you know what "row reduction" or "Gauss-Jordan elimination" is? Commented Oct 10, 2016 at 11:52
• @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?
– user200918
Commented Oct 10, 2016 at 11:56

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 : $$\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & -4 & 1 \\ 0 & 0 & 2 & 3 & 4 \\ 1 & 1 & -2 & 2 & 4 \\ 1 & 0 & 0 & 9 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & -4 & 1 \\ 0 & 0 & 1 & 3 & 4 \\ 1 & 1 & -1 & 2 & 4 \\ 1 & 0 & 0 & 9 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 3 & 4 \\ 1 & 1 & -1 & 6 & 4 \\ 1 & 0 & 0 & 9 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 4 \\ 1 & 1 & -1 & 2 & 4 \\ 1 & 0 & 0 & 3 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 4 \\ 1 & 1 & -1 & 3 & 4 \\ 1 & 0 & 0 & 3 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 4 \\ 1 & 1 & -1 & 1 & 4 \\ 1 & 0 & 0 & 1 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 4 \\ 1 & 1 & -1 & 4 \\ 1 & 0 & 0 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 4 \\ 1 & 1 & -1 & 3 \\ 1 & 0 & 0 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & -1 & 7 \\ 1 & 0 & 0 & 7 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & -1 & 1 \\ 1 & 0 & 0 & 1 \end{bmatrix}=\operatorname{rank}\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & -1 \\ 1 & 0 & 0 \end{bmatrix}=3$$