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Let $A$, $B$, $C$, and $D$, be $n\times n$ matrices.

a) Show that the block matrix $\begin{bmatrix}A & B\\ C & D \end{bmatrix}$ has rank at least as large as the rank of $A$.

b) Show that the matrices $\begin{bmatrix}A & 0\\ 0 & B \end{bmatrix}$ and $\begin{bmatrix}A+B & B\\ B & B \end{bmatrix}$ are equivalent.

c) Show that the $rank(A+B)\leq rank(A)+rank(B)$.

Does anybody have an idea to prove above statements?

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  • $\begingroup$ Do you know that rank of a matrix can be defined as the dimension of its row space (equiv. dimension of its column space)? $\endgroup$ – hardmath Oct 8 '13 at 1:35
  • $\begingroup$ @hardmath yeah, how would you apply it to the statement? $\endgroup$ – Mark Oct 8 '13 at 1:38
  • $\begingroup$ Seems like homework questions! $\endgroup$ – Ehsan M. Kermani Oct 8 '13 at 1:39
  • $\begingroup$ @EhsanM.Kermani It is. I just need a way to start the proof $\endgroup$ – Mark Oct 8 '13 at 1:41
  • $\begingroup$ Well then, start with the definition of rank for a) $\endgroup$ – Ehsan M. Kermani Oct 8 '13 at 1:45
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Hints

  1. For the first one, Rank of any matrix is equal to the number of independent non-zero rows that a matrix can have.
  2. For the second one, take the 2nd matrix and try row reductions. Row reduced forms of a matrix are equivalent to the original matrix.
  3. I'm not sure of a rigorous way but you can prove that $(\mathbf{A}+\mathbf{B})\mathbf{x}=\mathbf{}\mathbf{Ax}+\mathbf{Bx}$ can span the space at most the space spanned by $\mathbf{A},\mathbf{B}$ together by considering tall matrices. For a wide matrix the equality can never be attained and for square matrix it's obvious.
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  • $\begingroup$ From $(A+B)x = Ax + Bx$ we see that the column space of $A+B$ is contained in the span of the column space of $A$ and the column space of $B$. $\endgroup$ – Robert Israel Oct 8 '13 at 2:03
  • $\begingroup$ For b) or part 2, (block) row reduction alone does not achieve a tranformation of the second matrix into the first; column operations are also needed. $\endgroup$ – hardmath Oct 8 '13 at 11:21

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