rank of the matrices of the form $A+\lambda B$ Could anyone help me for the following problems?
$1. $ If $A+\lambda B\in M_{m\times n}(\mathbb{R})$ where we can vary $\lambda\in \mathbb{R}$, has rank $r<n$ then what can we say about the rank of $A$ and $B$?
$2.$ If $im(B)\subseteq im(A)$ then what can be said about the rank of $A+\lambda B$, if it is given that rank of $A=p$
$3.$ is it possible that $rank(A)<rank(A+\lambda B)$ for some values of $\lambda$?
Thank you for help and hints and any suggestions
 A: Q1:
Let $\lambda=0$ and have rank(A)=r. We can also reach the conclusion that $rank(B)\leq r$. If rank(B)=r'>r, then we know that B has an submatrix $B_0$ of order r', $det(B_0)\neq 0$. 
We consider the corresponding submatrix of $A+\lambda B$ (we suppose it is $A_0+\lambda B_0$), and we know that $det(A_0+\lambda B_0)$ is a polynomial of degree r', which has at most r' zeros. But your $\lambda$ has infinite values, so there exists a $\lambda_0$ which makes $det(A_0+\lambda_0 B_0)\neq 0$. Thus, $rank(A+\lambda B)\geq r'>r$.
So the conclusion is $rank(A)=r$ and $rank(B)\leq r$ (You can easily find the example that $rank(B)<r$)
Q2:
$\alpha$ is an arbitrary vector in the vector space V. Because $Im(B)\subseteq Im(A)$, we have $A\alpha \in Im(A)$ and $B\alpha \in Im(A)$. So $(A+\lambda B)\alpha \in Im(A)$, and we have $Im(A+\lambda B)\subseteq Im(A)$. The conclusion is $rank(A+\lambda B)\leq p$.
Q3:
If this question is based on the second question, then we will give the negative answer from the proof above.
