Find a constant so matrix is invertible I am doing some exercises from my Linear Algebra textbook and i have come across an exercise which I don't quite understand. Every exercise is graded with numbers from [1] to [5]. [1] is meant to be really easy and [5] is meant to be really hard. This exercise has grade [4], so it should be hard. I don't really care about the grades though, since my goal is to do every exercises in this book, no matter how "hard" they are.
Notice: This is my first course (introduction course) in Linear Algebra and this exercise is a part of the first chapter. So I don't have too advanced tools yet :)

Problem: Let $X$, $Y$ be column-vectors. For which constant c is the matrix $I+cXY^T$ invertible? Find the inverse matrix for such c.

So, here's my argue: $I$ is obviously invertible since it's inverse is $I$. So, let's say $c=0$ then $0*XY^T=0$-matrix. Hence $I+0=I$ is invertible and the inverse is $I$? Since the question is "for which constant $c$" then i assume there's only one value for $c$ which makes the matrix invertible and $c=0$ surely makes the matrix invertible? Like i said, there must be something I don't understand. Any clarification and/or hint would be appreciated so i can proceed, thanks :)
 A: Probably your book should not have used the singular "constant", since this matrix will be invertible for most values of$~c$. If you know about the determinant criterion for invertibility, this is clear: the expression $\det(I-cXY^T)$ is a polynomial function of $c$, and it is not identically zero (since it is $1$ for $c=0$) so it will be zero (indicating non-invertibility) for at most $n$ values of $c$, where $n$ is the size of your vectors. So you should read the question as asking to describe precisely for which values the matrix fails to be invertible, and to give the inverse (in terms of $c$) in all other cases.
I turns out you can see that the matrix is invertible for all $c$ with at most one exception (which given the above is somewhat unexpected), by fairly elementary reasoning. In order to fail to be invertible, a matrix must annihilate (i.e., give a zero product with) at least one nonzero column vector $v$. Supposing $v$ is such a column vector, one then has
$$
  0=(I+cXY^T)v=v+cX\langle Y,v\rangle = v+c\langle Y,v\rangle X
$$
where the expression in angle brackets denotes the scalar (or dot) product of $Y$ with $v$. Since $v$ was assumed nonzero, this can only happen if $v$ is a scalar multiple of $X$. Moreover, if it happens for $v$ it will also happen for scalar multiples of $v$, among which one finds $X$, so one can conclude that $I+cXY^T$ is non-invertible if and only if it annihilates $X$ (and $X\neq0$). Now putting $v=X\neq0$, the above equation gives $X(1+c\langle Y,X\rangle)=0$, or equivalently (since $X\neq0$) the equation $c\langle Y,X\rangle=-1$. So a necessary condition for $1+cXY^T$ to not be invertible is that $c\langle Y,X\rangle=-1$ (you already saw that $c=0$ implies invertibility, but this is stronger). This condition is also sufficient condition, and one can say that if $\langle Y,X\rangle\neq0$, the matrix fails to be invertible precisely for $c=-\frac1{\langle Y,X\rangle}$ (and if $\langle Y,X\rangle=0$ it is invertible for all$~c$).
To actually describe the inverse still requires some more work. I doubt whether you are expected to be able to do it as follows, but this would be my approach. The case $\langle Y,X\rangle=0$, that is $X\perp Y$ still needs some special consideration, since we then need a solution for all$~c$, so one cannot divide by anything involving$~c$, while for $\langle Y,X\rangle\neq0$ it seems plausible that the solution should involve division by $1+c\langle X,Y\rangle$, since that was required to be nonzero.


*

*Case $\langle Y,X\rangle\neq0$. Now the subspace $Y^\perp$ perpendicular to $Y$ does not contain the nonzero vector $X$, so every vector $u$ can be decomposed uniquely as a component $u_X$ that is a multiple of $X$ and a component $u_\perp$ that is perpendicular to $Y$. An equation for the inverse of $1-cXY^T$ amounts to solving $w=v+c\langle Y,v\rangle X$ for $v$ in terms of $w$. This equation now splits into a trivial equation $w_\perp=v_\perp$ for the component perpendicular to $Y$, and an equation $w_X=v_X+c\langle Y,v_X\rangle X$ along the direction of $X$; the former gives $v_\perp=w_\perp$, and the latter can be rewritten $w_X=v_X+c\langle Y,X\rangle v_X$ and has as solution $v_X=\frac1{1+c\langle X,Y\rangle}w_X$ (voilà the expected division). In principle this gives you the inverse, though you still have to glue the pieces together (notably you need to explicitly decompose $w=w_\perp+w_X$).

*Case $\langle Y,X\rangle=0$. Now the above doesn't work, but one has $(cXY^T)^2=0$ for all $c$, which means that $(I+cXY^T)(I-cXY^T)=I$, and we have the explicit inverse $I-cXY^T$.
A: If $I+cXY^t$ is not invertible then exists a column vector $0\neq Z$ such that $(I+cXY^t)Z=0$. 
Show that $Z=dX$ for some $0\neq d$. Therefore, $(I+cXY^t)X=0$. 
Now, you can find $c$.
A: Your argument is nice. And yes, the question seems ambiguously posed, as Timbuc comments.
If you believe that the question implies that only one such value of $c$ exists, try to prove this statement! Here is a starting point:
We will work in $n$ dimensions.
Suppose $c$ is such that $I+cXY^T$ is invertible for all $X, Y \in \mathbb{R}^n$.
Also suppose that $c>0$. (The case $c<0$ is similar.)
Define $X := \left(\sqrt{\tfrac{1}{c}},0,...,0\right)^T$ and define $Y:=\left(-\sqrt{\tfrac{1}{c}},0,...,0\right)^T$.
Then... (can you complete the proof?)

Attempting similar for $I+cXX^T$ (i.e. $Y=X$) seems an interesting extension:
For which constants $c \in \mathbb{R}$ is the matrix $I+cXX^T$ invertible for any $X \in \mathbb{R}^n$?
Or (better):
Given $X \in \mathbb{R}^n$ can you find $c \in \mathbb{R}$ such that $I+cXX^T$ is not invertible?

Further reading: Matrix determinant lemma 
