I got this matrix :
For me, since Column 1 = Column 3 = Column 4
It implies that the maximum rank of this matrix should be 2.
Since Column 2 is linearly independent of Column 5, the rank of this matrix is equal to 2.
This is coherent to the definition of the rank and to the fact that a (3×5) matrix has a rank inferior to 3.
My question is : Is my reasoning false ?
Best Answer
No, the rank of the matrix in this case is 3. Firstly the matrix is a short-wide matrix $(m<n)$. So maximum rank is m at the most
The rank depends on the number of pivot elements the matrix. I would say that your statement "Column 1 = Column 3 = Column 4" is wrong. You can say that Columns 1, 2 & 3 are Linearly Dependent Vectors. If you reduce the matrix by performing row operations you will end to at this state
$$ \begin{bmatrix} 1 & 0 & 1 & 1 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{bmatrix} $$
Now we can perform column operation and send either two from columns 1, 3 & 4 to $[0]$ but not all. Because one stays to be a Identity vector no matter what.
One more way to look at is Column Space of the matrix:
$$ C(Matrix) = c1 \begin{bmatrix} 5 \\5 \\4 \\ \end{bmatrix} + c2 \begin{bmatrix} 4 \\5 \\2 \end{bmatrix} + c3 \begin{bmatrix} 2 \\4 \\1 \\ \end{bmatrix} $$
$$ rank(Matrix) = dim(C(Matrix)) $$
Therefore, rank is 3.