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I am struggling with proving the theorem that if $A$ and $B$ are $n\times n$ matrices, then:

$$\operatorname{rank}(AB)\leq \operatorname{rank}(B)$$

Could anyone suggest me a hint? Any help is greatly appreciated, thank you very much.

Michael Hardy
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marco11
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4 Answers4

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Every row of $AB$ is a linear combination of the rows of $B$. Therefore, the dimension of the row space of $AB$ cannot be bigger than the dimension of the row space of $B$.

To see that every row of $AB$ is a linear combination of the rows of $B$, just look at how matrix multiplication is done: the $k$th row of $AB$ is $A_k B$ where $A_k$ is the $k$th row of $A$. There are as many scalars in that row as there are rows of $B$. The $\ell$th entry in the row $A_k$ gets multiplied by the $\ell$ row of $B$, and then those get added up from $\ell=1$ to $\ell=$ the number of rows of $B$.

Michael Hardy
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HINT:

Look at those matrices as linear transformations and use Rank-nullity theorem.

Daniel
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  • Can I write the following? Matrix $A$ and $B$ corresponds to linear transformations $T$ and $S$ respectively, such that $T,S: V \rightarrow V$. Then $Im(TS(U))=Im(T(S(U))) \subseteq Im(S(U))$, so that $rank(TS) \leq rank(S)$ and so $rank(AB) \leq rank(B)$. Am I right? – marco11 May 10 '15 at 17:47
  • Ah, as $Im(S) \subseteq V$, then $Im(T(S)) \subseteq Im(T)$. But well, this proves that $rank(AB) \leq rank(A)$, right? – marco11 May 10 '15 at 17:54
  • @marco11 Please, be careful with the notation: who is $U$? also, the $Im$ "operator" is only applied to linear transformation, not sets. The argument would be "since $S(V)\subseteq V$, $TS(V)\subseteq T(V)$ and therefore $\operatorname{rank}(TS)\leq \operatorname{rank}(T)$", however, you're trying to prove $\operatorname{rank}(TS)\leq \operatorname{rank}(S)$. Argue with the fact that the rank of a matrix and its transpose are equal. – Daniel May 10 '15 at 17:55
  • @Daniel I guess there is a typo in your comment above. Should not it be $$rank(TS)\leq rank(T)$$ – Shweta Aggrawal Feb 27 '19 at 17:51
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here is another way too look at it. every row $i$ of $AB$ is a linear combination of the rows of $B$ weighted by the row $i$ of $A.$ therefore the row space of $AB$ is contained in the row space of $B.$ now use the fact that rank of matrix is the dimension of its row space.

abel
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let the vecotor x ϵ c(AB). c(AB) denotes the column space of AB. then

x=ABy for some vector y

x=Az ϵ c(AB) ,where z=by hence c(AB) is a subspace C(A) =>dim[c(AB)] ≤dim c(A) rank(AB)≤rank(A)

Michael Hardy
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saudade
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