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I am trying ti prove that rk(A+B)≤rkA + rkB, where rkA means rank of matrix A, and A is mxn, as well as B.

By the fundamental theorem of linear algebra we know that rkA≤n and rkB≤n, so that we have rkA + rkB≤2n. On the other hand we know that rk(A+B)≤n, and I was trying to use that to arrive at a contradiction by assuming the opposite of what I want to prove, but I could not get any contradiction, could you help me point in the right direction?

Carlos N
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    the rank is the dimension of the image. Try to think about images – Exodd Aug 23 '18 at 19:25
  • The $rk(A+B)=rk(A)+rk(B)-rk(A\cap B)$. So, $rk(A+B)+rk(A\cap B)=rk(A)+rk(B)$. So, $rk(A+B)+j=rk(A)+rk(B)$ for some $j\in \mathbb{W}$. Hence, $rk(A+B)\leq rk(A)+rk(B)$. – W. G. Aug 23 '18 at 19:35

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