Proof of determinant formula

Question

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.

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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.

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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).

Answer 1

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, $\det(1 + Y^TX) = \det(I + XY^T)$.

Answer 2

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:

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

2. $x \ne 0$. 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.