Sylvester's identity

Question

Hi I need help with the next exercise:

Show that for any arbitrary vectors $a,b\in\mathbb{R}^p$ $$\det(I_p-ab^T)=1-a^Tb$$

I have applied the Sylvester identity that I found on Wikipedia. It states that if you have two matrices $A\in\mathbb{R}^{m\times n}, B\in\mathbb{R}^{n\times m}$ then $$\det(I_m+AB)=\det(I_n+BA)$$

Considering it with $A=a$ and $B=b^T$ I have $m=p$ and $n=1$ then $$\det(I_p-ab^T)=\det(I_1-b^Ta)=1-a^Tb$$ The last equality is OK? I don't know how to justify, can I compute the determinant of a real number?

It is okay, the determinant of a $1\times 1$ matrix is its only entry. The Sylvester identity is doing all the work for you here. The same result has been obtained in different ways for previous Questions, e.g. treating the term $ab^T$ as a "rank one update" to the identity. — hardmath, Oct 17 '17 at 12:47

score 3 · Accepted Answer · answered Oct 17 '17 at 12:47

$I_1 - b^Ta$ is not, technically speaking, a real number. It is a $1\times 1$ matrix with one element that is equal to $1-b^Ta$, and luckilly, the determinant of a $1\times 1$ matrix $A=[a]$ is just the element of that matrix, i.e. $a$.

In your case, the only element of $I_1-b^Ta$ is $1-b^Ta$ and since $b^Ta$ is just the scalar product of $b$ and $a$, that's the same as $1-a^Tb$.

Sylvester's identity

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