Proof of determinant formula I have just started to learn how to construct proofs. That is, I am not really good at it (yet). In this thread I will work through a problem from my Linear Algebra textbook. First i will give you my "solution" and then, hopefully, you can tell me where I went wrong. If my proof strategy for this case is wrong I would love to hear why it's wrong (if it is possible) since I think that is how I will become better :)

My textbook says an often good proof strategy of different determinant formulas is by Mathematical induction and I think it also works in this case, but as i said earlier, I am not too good at constructing proofs yet.

Problem:


Let $X, Y$ be column-vectors. Show that $det(I+XY^T)=1+Y^TX$, where the last product is interpreted as a number.

Ok so here is my attempt to solve the problem:
Proof strategy: Induction
1. Base case:
The statement is true when n=2, since:
$$I=\left( \begin{array}{ccc}
1 & 0 \\
0 & 1 \end{array} \right), XY^T=\left( \begin{array}{ccc}
x_1y_1 & x_1y_2 \\
x_2y_1 & x_2y_2 \end{array} \right)$$
and
$|I+XY^T|=\begin{vmatrix}
x_1y_1+1 & x_1y_2\\ x_2y_1 & x_2y_2+1
\end{vmatrix}$
When we expand the determinant, we get:
$(x_1y_1+1)(x_2y_2+1)-x_1y_2x_2y_1= 1+(x_1y_1x_2y_2+x_1y_1+x_2y_2-x_1y_2x_2y_1)=1+(x_1y_1+x_2y_2)=1+Y^TX$

2. induction hypothesis:
Suppose it's true for the value $n-1$ and now I want to prove it's true for n.  

3. The inductive step: 
$det(I+XY^T)=x_1y_1+x_2y_2+...+x_{n-1}y_{n-1}+x_ny_n + 1$
And here is where i pretty much get stuck. I don't know where to go from here. It's kinda hard for me to grasp the idea behind mathematical induction. I don't really know what to do when I come to this step. What can I do to finish the proof? (well, if what I have done so far is correct, that is).
 A: The "holes-digging" method might be interesting to prove this.
On one hand, dig a hole at the lower-left corner of $A$,
$$A := \begin{bmatrix}I & X \\
-Y^T & 1 \end{bmatrix} = \begin{bmatrix} I & 0 \\
Y^T & 1\end{bmatrix}\begin{bmatrix}I & X \\
0 & 1 + Y^TX\end{bmatrix}$$
Take determinants on both sides to have $\det(A) = \det(I)\det(I + Y^TX) = \det(1 + Y^TX)$.
On the other hand, dig a hole at the upper-right corner of $A$,
$$A = \begin{bmatrix}I & X \\
-Y^T & 1 \end{bmatrix} = \begin{bmatrix} I & X \\
0 & 1\end{bmatrix}\begin{bmatrix}I + XY^T & 0 \\
-Y^T & 1 \end{bmatrix}$$
Take determinants on both sides to have $\det(A) = \det(I + XY^T)\det(1) = \det(I + XY^T)$.
Therefore, $\det(1 + Y^TX) = \det(I + XY^T)$.
A: Your base case is solid. I'd like to suggest that you take a different approach to constructing the proof. Doing this inductively is likely to be pretty messy. Break into cases:


*

*$x = 0$ or $y = 0$. In this case, the theorem's pretty clear (I hope!).

*$x \ne 0$. In this case, you can set $x_1 = x$ and extend this to a basis $$ x_1, \ldots x_n$$ where the $x_i$ are pairwise perpendicular. If you then express $x$ and $y$ in this basis, it's relatively easy to write out the proof. And expressing it in that basis means replacing $I + x^t y$ with $Q (I + x^t y) Q^{-1}$, where $Q$ is a change-of-basis matrix, so it doesn't change the determinant, etc. 
