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$\DeclareMathOperator{\nullity}{nullity}\DeclareMathOperator{\rank}{rank}$

Theorem (3.34). The dimension theorem of matrices (rank-nullity theorem): If $A$ is a matrix with n columns (number of unknowns) then $$\nullity(\mathbf{A}) + \rank(\mathbf{A}) = n$$

My attempt:

Consider arbitrary matrix $\bf A$ with $n$ columns.


Let $\mathbf R = \operatorname{rref}(\mathbf A)$. Suppose $\rank(\mathbf A) = n$. It follows that $\bf R$ won't have columns without leading ones. Hence the only vector $\bf x$ that satisfies $\bf Rx = O$ is zero vector. Since $\bf Rx = O$ is equivalent to $\bf Ax = O$, solution set must be the same. And that means that null space of $\bf A$ consists of a single vector, $\bf O$. Hence $\nullity(\mathbf{A}) = 0$

We have $\rank(\mathbf A) + \nullity(\mathbf{A}) = n + 0 = n$


Let $\mathbf R = \operatorname{rref}(\mathbf A)$. Suppose $\rank(\mathbf{A}) = m$ where $m < n$. It implies that the linear system $\bf Rx = O$ will have $n-m$ free variables. Since $\bf Rx = O$ is equivalent $\bf Ax = O$, then $\bf Ax = O$ will have $n-m$ free variables too. And because number of free variables equals to nullity of the $\bf A$ (I tried to prove this proposition here), it follows that $\nullity(\mathbf A) = n-m$

We have $$\rank(\mathbf{A}) + \nullity(\mathbf{A}) = m + n - m = n$$ $\Box$

Is it correct?

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1 Answer 1

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It looks correct. It also looks like you don't need to consider the case where the rank is $n$ separately.

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