Some properties of the determinant and the trace

Question

Could anyone please explain steps $\color{red}{(1)}$ and $\color{blue}{(2)}$? :)

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A reference to the background of these guys would also be helpful.

Answer 1

When $X=I+tvv^T$, $YX = Y + tYvv^T = Y + t\lambda vv^T$. Trace is linear, so $$ \operatorname{tr}{(YX)} = \operatorname{tr}{(Y+t\lambda vv^T)} = \operatorname{tr}{Y} + t\lambda \operatorname{tr}{(vv^T)}. $$ Trace is also cyclic, in that $\operatorname{tr}{(AB)}=\operatorname{tr}{(BA)}$, so $\operatorname{tr}{(vv^T)} = \operatorname{tr}{(v^Tv)} = \operatorname{tr}{(\lVert v \rVert_2^2)} = 1$ since this is just a scalar, and the red expression follows.

For the determinant, $\det{(I+tvv^T)}$, since a determinant is invariant under change-of-basis, we can evaluate it using an orthonormal basis where $v$ is the first vector. Then the matrix of $I+tvv^T$ is diagonal with elements $ 1+t , 1,1, \dotsc $, which has determinant $(1+t) \cdot 1 \cdot 1 \cdot \dotsb = 1+t$, which is the blue expression.

Answer 2

Step $\color{blue}{\boxed{2}}$ uses the Weinstein-Aronszajn determinant identity,

$$\det \left( \,\mathrm I + t \, \mathrm v \mathrm v^\top \right) = \det \left( 1 + t \, \mathrm v^\top \mathrm v \right) = 1 + t \, \| \mathrm v \|_2^2 = \color{blue}{1 + t}$$

because $\| \mathrm v \|_2 = 1$.