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$P,Q\in \mathbb{R}^{m\times n}$ with image $(Q)\subseteq$ image(P). Then for allmost all $c\in \mathbb{R}$ we need to show

$(i)rank(P)=rank(P+cQ)$

$(ii)$ $im(P)=im(P+cQ)$

I understand that both are completely equivalent statement.

I thought about (i), since im(Q) is lying inside im(P) so clearly $\dim(im(Q))\le \dim(im(P))$ if the containmaint is strict then this dimension inequlaity also become strict, so then $rank(Q)

Now for almost all real number $c$ , $P+cQ$ is is a linear map and image of $(P+cQ)\subseteq image (P)$ due to $im(Q)\subseteq im(P)$ so $rank(P+cQ)\le rank(P)$

I am not able to prove the other way inequality. thanks for helping.

2 Answers 2

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Let $z=\text{rank}(P)$.

Our problem is clearly equivalent to proving that there is a finite number of values $c$ such that $\text{rank}(P+cQ)\neq z$. (clearly the rank of $P+cQ$ is at most $z$).

To do this we can select a $z\times z$ submatrix of $P$ that has rank $z$ (it is quite easy to show this is possible by first selecting a $z\times m$ submatrix and then a $z\times z$ submatrix of the previous submatrix).

Let $P'$ be the aforementioned submatrix. and let $Q'$ be the corresponding submatrix of $Q$. Clearly, it is sufficient for $P'+cQ'$ to have rank $z$ so that $P+cQ$ has rank $z$.

So it is sufficient to have $\det(P'+cQ')\neq 0$. And this can happen at most $z$ times because $\det(P'+cQ')$ is a non-zero polynomial in $c$ of degree $z$.( to see it is non-zero notice $\det(P'+0Q')\neq 0$.)

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    So basically what we did is make the problem "harder", and then attacked it with some theory. – 2017-01-21
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    what is rank(n) collumns of P ??? – 2017-01-21
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    I simply dont understand your solution please explain more – 2017-01-21
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    it should have been $rank(P)$. – 2017-01-21
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    Can you prove It when P is invertible and n=m? – 2017-01-21
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    If P an $n\times n$ matrix and is invertible and Q is $n\times n$ matrix then – 2017-01-22
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    If P an $n\times n$ matrix and is invertible and Q is $n\times n$ matrix then I understand that there are finite (at most n) c for which det(P+cQ)=0 and for those c rank(P+cQ)$det(P+cQ)\ne 0$ and hence rank(P)=rank(P+cQ) for all most all real c – 2017-01-22
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    where we are using the fact that im(Q) containe in im(P)? and if P,Q are rectangular matrix how can we talk about determinant? – 2017-01-22
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    Exactly, that is why you sobre works with the whole rectangular Matrix, but with a smaller one instead. I am goin to rewrite the argument, I hope It is clear now. – 2017-01-22
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    @miosaki Is it clearer now ${}{}{}{}{}$ – 2017-01-22
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Consider the left multiplication transformations for the given matrices $P$ and $Q$.

Let, $$\;T_1:\Bbb R^n\to \Bbb R^m$$ defined by, $$T_1(x)=Px$$

Let, $$\;T_2:\Bbb R^n\to \Bbb R^m$$ defined by, $$T_2(x)=Qx$$

Given that, $Im(Q)\subseteq Im(P)$ $i.e.\;\;Im(T_2)\subseteq Im(T_1) \;\;i.e.\;\; T_2(\Bbb R^n) \subseteq T_1(\Bbb R^n) $

Also for $c\in \Bbb R$, clearly $cT_2(\Bbb R^n) \subseteq T_1(\Bbb R^n)-------(1)$

Consider the transformation, $T_1+cT_2:\Bbb R^n\to R^m$ which is given by the left multiplication matrix $P+cQ$

Now using (1),

$$(T_1+cT_2)(\Bbb R^n)=T_1(\Bbb R^n)+(cT_2)(\Bbb R^n)=T_1(\Bbb R^n)+cT_2(\Bbb R^n)=T_1(\Bbb R^n)$$

And equivalently, $$Im(P)=Im(P+cQ)$$

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    Wait what?${}{}{}$ – 2017-01-22
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    so you concluded it holds for every $c$? that is false. – 2017-01-22
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    what is one counter example which can show his argument is false or there is a fallacy here – 2017-01-23
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    Let P=Q be a nonzero Matrix and take c=-1 – 2017-01-23
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    $T_1(\Bbb R^n),\; T_2(\Bbb R^n)$ are subspaces of $R^m$ and it is given that, $T_2(\Bbb R^n)\subseteq T_1(\Bbb R^n)$ thus, $T_2(\Bbb R^n)$ is a subspace of $T_1(\Bbb R^n)$. So for any $c\in \Bbb R$ the conditions must be satisfied due to the closure of scalar multiplication. – 2017-01-23
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    thank you @JorgeFernándezHidalgo – 2017-01-23