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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 (3x5) matrix has a rank inferior to 3.

My question is : Is my reasoning false ?

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2 Answers

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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.

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Rank = 3

I removed the 3 linearly independent matrices where I should have removed only two. It leads us to a 3x3 matrix with 3 linearly independant columns

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