a matrix rank equality problem when image of one is contained in another $P,Q\in \mathbb{R}^{m\times n}$ 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 $rank(Q)<rank(P)$
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.
 A: 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 $\det(P'+0Q')\neq 0$.)
A: 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, $$Im(P)=Im(P+cQ)$$ 
