Induced norm of identity matrix plus rank one matrix

Question

Let $\alpha\in(0,1)$ and $\hat{g}$ be a unit vector. I would like to bound/find an inequality for the norm of $I - \alpha\hat{g}\hat{g}^\top$. In particular, for the induced norm. $$ \parallel I - \alpha \hat{g}\hat{g}^\top\parallel_I = \sup_{u\neq 0} \frac{||(I - \alpha \hat{g}\hat{g}^\top) u||}{||u||} $$ All I could think of was $$ \parallel I - \alpha \hat{g}\hat{g}^\top\parallel_I \leq ||I||_I + \alpha||\hat{g}\hat{g}^\top||_I = 1 + \alpha ||\hat{g}\hat{g}^\top|| $$ But this seems quite loose since we are not using the fact that we are subtracting the outer product?

Answer 1

If the underlying vector norm is the Euclidean norm then one can estimate $$ \|(I-\alpha gg^T)x\|_2 \le \|x\|_2 + \alpha \|g\|_2 \cdot |g^Tx| \le (1+\alpha)\|x\|_2. $$ To get a better estimate, one can decompose $x = sg + u$ with $u\perp g$. This implies $\|x\|_2^2 = \|u\|_2^2 + s^2$. Then $$ \|(I-\alpha gg^T)x\|_2^2 = \|x - \alpha sg\|_2^2 = \|sg+u-\alpha sg\|_2^2 = \|u\|_2^2 + (1-\alpha) s^2 \le \|x\|_2^2. $$ Hence $\|I-\alpha gg^T\|\le 1$. In addition $\|I-\alpha gg^T\|=1$ if $n>1$. To see this take $x\perp g$. If $n=1$, i.e., the vector space is one-dimensional, then $I-\alpha gg^T = (1-\alpha)I$.