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!
Best Answer
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.