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The task is that I have to prove this statement:

Given $(m+1)\times(n+1)$ matrix $A$ like this: $$ A=\left[\begin{array}{c |cc} 1 & \begin{array}{ccc}0 & \cdots & 0\end{array} \\ \hline \begin{array}{c}0\\ \vdots\\0\end{array} & {\Large B} \\ \end{array}\right] $$

where $B$ is a $m\times n$ sub-matrix of $A$.

Show: $\operatorname{rank}(A) = r$ implies $\operatorname{rank}(B) = r - 1$.

Here are the steps that I constructed:

(1) First I already proved this claim: if $m\times n$ matrix $S$ has rank $r$, then $r\le m$ and $r\le n$. And I let $\operatorname{rank}(B) = k$.

(2) Use (1), I say the following 2 statements:

  • for matrix $A$, $r\le m+1$ and $r\le n+1$. Thus $r-1\le m$ and $r-1 \le n$
  • for matrix $B$, $k\le m$ and $k\le n$.

(3) Now, this is the part that I feel shaky about. I plan to say that using (2), I have $r-1\le k$ and $k\le r - 1$. Thus $r - 1 = k$. But I'm not certain whether this argument is valid. After noticing that $r - 1\le m$ and $k \le m$ at a same time, I come up with the above relation between $k$ and $r - 1$.

Would someone please help me check if there is anything wrong or missing in this proof? Thank you.

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up vote 1 down vote accepted

In analogy, you're arguing: "$5 \leq 7, 4 \leq 7 \Rightarrow 5 \leq 4$", which, when put this way, is clearly not true.

The way to prove this exercise depends on how your class introduced these concepts. Absent that knowledge, I would argue as follows:

When reduced to row-echelon form, you can read off the rank of a matrix by the number of non-zero rows. So reduce your matrix $B$ to $1 \leq k \leq \min(n, m)$ non-zero rows (which equals the rank of $B$), and note that the elementary matrices multiplied to the left and right of $A$, to achieve this, will not change column 1 (this is quite obvious as all but the first element of column $1$ and row $1$ are $0$). So you still have the $1$ in position $a_{11}$ (and $0$ in the other entries of the first column), and the entire matrix has $k+1$ non-zero rows in row-echelon form which equals the rank of $A$.

So $\operatorname{rank}(A) = \operatorname{rank}(B)+1,$ which was to be shown.

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I'm sorry but would you please clarify your arguments a bit more? How do the multiplication of elementary matrices have any relation to the number of nonzero rows ?? The only definition that my teacher teaches me about the rank = the max number of linearly independent columns (or rows). But then I don't know how I should use this definition on my proof – Cecile Feb 16 '13 at 5:23
Multiplication with elementary matrices amounts to writing rows as linear combinations of others. // For your proof though, maybe then argue as follows: rank B = rank ($0$ | B), a $0$ column added as first column (this should be clear/easy to show if you prefer). But rank A = rank ($0$ | B) + 1: if not, then $(1 0 \dots 0)$ is in the span of ($0$ | B) - but how could it? How to get the 1 in $a_{11}$ of $A$ if all entries below it are zero? – gnometorule Feb 16 '13 at 7:38

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