Problem with matrix and vector norms I already try to multiply by a orthogonal matrix in both sides, multiply by $q$ and $d$, factor, expand...nothing works. This problem comes from Demmel's book, Applied Numerical Linear Algebra.
Let $q$ be a unit vector and $d$ be any vector orthogonal to $q$. Show that $||(q+d)q^T-I||_2=||q+d||_2$.
The first norm is the induced matrix 2-norm and the second norm is the euclidean vector 2-norm.
Thanks.
 A: There is probably a neat trick to do this, but I don't see it. Here is a somewhat ugly approach.
For the induced matrix $2$-norm, I am using the definition
$$\|A\|_2 = \sqrt{\lambda_{\max}(A^*A)}.$$
First, the easy part:
$$\|q+d\|_2 = \sqrt{(q+d)^T(q+d)} = \sqrt{1 + d^Td}.$$
Let us now find $\|(q+d)q^T - I\|_2$. Denote $X := (q+d)q^T - I$ and note that $X^TX$ has the same eigenvalues as $XX^T$, so we compute the latter:
\begin{align*}
XX^T &= ((q+d)q^T - I) ((q+d)q^T - I)^T = ((q+d)q^T - I) (q(q^T+d^T) - I) \\
&= (q+d)(q^T+d^T) - (q+d)q^T - q(q^T+d^T) + I \\
&= qd^T + dd^T - qq^T - qd^T + I \\
&= dd^T - qq^T + I.
\end{align*}
Note that $dd^T qq^T = 0 = qq^T dd^T$, so $dd^T$ and $qq^T$ commute, which means that there is a common orthonormal eigenvector basis for these vectors, i.e., there is an orthonormal $U$ such that $U^T dd^T U = D_1$ and $U^T qq^T U = D_2$ are diagonal matrices. We can assume that
$$D_1 = \mathop{\rm diag}(d^Td, 0, \dots, 0),$$
because we know that $dd^T$ is of rank $1$, so it has only one nonzero eigenvalue, and that one is equal to the only nonzero eigenvalue of $d^Td$.
Let $u_1$ and $u_2$ be eigenvectors associated with nonzero eigenvalues $\lambda_1 = d^Td$ and $\lambda_2$ of $dd^T$ and $qq^T$, respectively. With the same argumentation as for $\lambda_1$, we see that $\lambda_2 = q^Tq = 1$. Note that
$$\lambda_1 \lambda_2 u_2^T u_1 = (qq^Tu^2)^T (dd^Tu_1) = u_2^T q q^T d d^T u_1 = 0,$$
so $u_1$ and $u_2$ are orthogonal. This means that the top left element of $D_2$ is zero (otherwise, the first column of $U$ would be the common eigenvector of $dd^T$ and $qq^T$ associated with their non-zero eigenvalues).
So, the eigenvalues of $XX^T$ are $\lambda_1 + 1 = d^Td + 1$, $-\lambda_2 + 1 = -q^Tq + 1 = 0$ and $1$ (with multiplicity $n-2$).
Since $d^Td \ge 0$, we see that the largest eigenvalue of $XX^T$ is $d^Td + 1$, so
$$\|(q+d)q^T - I\|_2 = \sqrt{d^Td + 1} = \sqrt{(q+d)^T(q+d)} = \|q+d\|_2.$$
A: I get that $$\|(q-d)q^T - I\| = 1.$$
This is not the answer suggested, so perhaps someone can point out my mistake. $(q+d)q^T$ is a rank 1 matrix, and it has a nonzero eigenvalue $\lambda =1$ because $$(q+d)q^T(q+d) = qq^T(q+d) = q+d.$$
So since $(q+d)q^T$ has eigenvalues $1, 0, \dotsc, 0$, we have that $(q+d)q^T-I$ has eigenvalues $0, -1 , \dotsc, -1$, so it has radius (and norm) equal to 1.
Perhaps I've made a mistake, or perhaps the question really was about the Frobenius norm, as a commenter suggested.
