# The Rank of the Sum of Two Matrices

## Problem 441

Let $A$ and $B$ be $m\times n$ matrices.
Prove that
$\rk(A+B) \leq \rk(A)+\rk(B).$

## Proof.

Let
$A=[\mathbf{a}_1, \dots, \mathbf{a}_n] \text{ and } B=[\mathbf{b}_1, \dots, \mathbf{b}_n],$ where $\mathbf{a}_i$ and $\mathbf{b}_i$ are column vectors of $A$ and $B$, respectively.

Then the rank of the matrix $A$ is the dimension of the column space of $A$.
That is, we have
$\rk(A)=\dim\left(\,\Span(\mathbf{a}_1, \dots, \mathbf{a}_n) \,\right).$ Similarly, we have
$\rk(B)=\dim\left(\, \Span(\mathbf{b}_1, \dots, \mathbf{b}_n) \,\right)$ and
$\rk(A+B)=\dim\left(\,(\Span(\mathbf{a}_1+\mathbf{b}_1, \dots, \mathbf{a}_n+\mathbf{b}_n)\,\right)$ since $A+B=[\mathbf{a}_1+\mathbf{b}_1, \dots, \mathbf{a}_n+\mathbf{b}_n]$.

We claim that
$\Span(\mathbf{a}_1+\mathbf{b}_1, \dots, \mathbf{a}_n+\mathbf{b}_n) \subset \Span(\mathbf{a}_1, \dots, \mathbf{a}_n)+\Span(\mathbf{b}_1, \dots, \mathbf{b}_n).$ Any vector $\mathbf{x}\in \Span(\mathbf{a}_1+\mathbf{b}_1, \dots, \mathbf{a}_n+\mathbf{b}_n)$ can be written as
\begin{align*}
\mathbf{x}=r_1(\mathbf{a}_1+\mathbf{b}_1)+\cdots +r_n(\mathbf{a}_n+\mathbf{b}_n)
\end{align*}
for some scalars $r_1, \dots, r_n$.

Thus we have
\begin{align*}
\mathbf{x}&=r_1(\mathbf{a}_1+\mathbf{b}_1)+\cdots +r_n(\mathbf{a}_n+\mathbf{b}_n)\\
&=(r_1\mathbf{a}_1+\cdots +r_n\mathbf{a}_n)+(r_1\mathbf{b}_1+\cdots +r_n\mathbf{b}_n)\\
&\in \Span(\mathbf{a}_1, \dots, \mathbf{a}_n)+\Span(\mathbf{b}_1, \dots, \mathbf{b}_n),
\end{align*}
and hence the claim is proved.

Then we have
\begin{align*}
&\rk(A+B)\\
&=\dim\left(\,(\Span(\mathbf{a}_1+\mathbf{b}_1, \dots, \mathbf{a}_n+\mathbf{b}_n)\,\right)\\
&\leq \dim \left(\,\Span(\mathbf{a}_1, \dots, \mathbf{a}_n)+\Span(\mathbf{b}_1, \dots, \mathbf{b}_n) \,\right)\\
&\leq \dim\left(\, \Span(\mathbf{a}_1, \dots, \mathbf{a}_n) \,\right)+\dim \left(\, \Span(\mathbf{b}_1, \dots, \mathbf{b}_n) \,\right) \\
&=\rk(A)+\rk(B).
\end{align*}

Here we used the fact that for two subspaces $U$ and $V$ of a vector space, we have
$\dim(U+V)\leq \dim(U)+\dim(V).$ For a proof of this fact, see the post “Dimension of the sum of two subspaces“.

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1. 07/11/2017

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##### Dimension of the Sum of Two Subspaces

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