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Theorem: $rank(A+B) \leq rank (A) + rank(B)$.

Proof: Let $U = Im(A)$ and $W = Im(B)$. By dimension theorem, we know that: $Dim(U+W) = Dim(U) + Dim(W) - Dim (U \cap W)$. By substituting $U$ and $W$ we get: $Dim(Im(A)+Im(B))= Dim(Im(A)) + Dim(Im(B)) - Dim(Im(A) \cap Im(B))$.

I am stuck here. I know that $dim(Im) = Rank$ but I cannot continue from here. Please can someone help me? Thank-you:)

M47145
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Annalise Attard
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1 Answers1

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If you see that knowing that $\text{Im}(A+B)\subseteq \text{Im}(A)+\text{Im}(B)$ is sufficient, then notice that, for any vector $x$, we'd have that $(A+B)x=Ax+Bx$. Note that $Ax$ is in $\text{Im}(A)$ and $Bx$ is in $\text{Im}(B)$. Their sum is therefore in $\text{Im}(A)+\text{Im}(B)$, implying that $\text{Im}(A+B)\subseteq \text{Im}(A)+\text{Im}(B)$.

Milo Brandt
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  • Thank-you. However, please, can you tell me how I can show that Im(A+B)⊆ Im(A) + Im(B)? – Annalise Attard Dec 21 '14 at 18:30
  • That's what the answer is intending to do; any element of Im$(A+B)$ is of the form $(A+B)x$, and I decompose that to a sum of elements from Im$(A)$ and Im$(B)$ - but the sum of an elements from Im$(A)$ and one from Im$(B)$ is in Im$(A)$+Im$(B)$ by definition. – Milo Brandt Dec 21 '14 at 21:23
  • Thank-you! I understand now :) I have mis-read what he said. Could you please give me an example where Im(A+B)< Im(A) + Im (B)? – Annalise Attard Dec 21 '14 at 23:20
  • $B=-A$ would suffice. – Milo Brandt Dec 21 '14 at 23:26