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What is an easy way to see that $\det(\operatorname{id}_n+aa^t)=1+|a|^2$ for $a\in \mathbb{R}^n$ ?

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up vote 7 down vote accepted

For any orthogonal matrix $U$, $\det(I + a a^t) = \det(U (I + a a^t) U^t) = \det(I + (Ua)(Ua)^t)$. Now $Ua$ could be any vector of norm $\|a\|$. So try it for $a = |a| (1,0,\ldots,0)^t$...

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This is a different approach. +1 :) – Tapu Feb 3 '13 at 8:53

May be the eigenvalues of $aa^t$ are $|a|^2$ and $0$. So, the eigenvalues of $id+aa^t$ are precisely $1+|a|^2$ and $1$ ($n-1$ times). So, the determinant is product of eigenvalues...

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HINT 1: Your matrix $A$ is $$I + a a^T$$

Can you now compute the eigenvalues?

Move your mouse over the gray area below for another hint.

HINT 2: Make use of the fact that $\text{eigen}(\lambda A) = \lambda \text{eigen}(A)$

Move your mouse over the gray area below for another hint.

HINT 3: $\text{eigen}(I + a a^T)$ are $1 + \Vert a \Vert_2^2$ and $1$($n-1$ times).

Move your mouse over the gray area for the complete answer.

We can easily find the eigenvalues of $I + aa^T$. Note that $aa^T$ is a rank one matrix and its eigen values are $a^Ta$ and $n-1$ zeros. If $\lambda$ is an eigen value of $I + aa^T$, then $$\det (I + aa^T - \lambda I) = \det \left(aa^T + (1-\lambda)I \right) = 0$$ This means that $(\lambda-1)$ are the eigenvalues of $ee^T$. Hence, we get that $$\lambda - 1 = \Vert a\Vert_2^2 \text{ or }0 \text{ ($n-1$ times)}.$$ Hence, we get that $$\lambda = 1 + \Vert a\Vert_2^2, 1 \text{ ($n-1$ times)}$$ Hence, the eigenvalues of the initial matrix are $$\lambda = 1 + \Vert a\Vert_2^2, 1 \text{ ($n-1$ times)}$$ The determinant is nothing but the product of all eigenvalues and hence equals $$(1+\Vert a \Vert_2^2)$$

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