Product of a vector and its transpose norm I'm having trouble with the following.
Let $u,v \in \mathbb{R}^n$, $||u||_2=\frac{1}{2}$ and $||v||_2=1$, where $||\cdot||_2$ denotes the euclidean L2 norm of a vector. Also, for a $A\in\mathbb{R}^{n\times n}$, let $||A||_2=\rho(A^TA)^{1/2}$ be the euclidean L2 induced norm of a matrix, and $\rho(A)$ the spectral radius of $A$.
I encountered the following result.
$$||uv^T||_2^2=\rho\big((uv^T)^T(uv^T)\big)=\rho(vu^Tuv^T)=\frac{1}{4}\rho(vv^T)=\frac{1}{4}$$
I don't understand why the last equality is true. I can see that $u^Tu=||u||_2^2=\frac{1}{4}$, thus $\rho(vu^Tuv^T)=\frac{1}{4}\rho(vv^T)$. However, why is it true that $\rho(vv^T)=1$? As far as I know, $vv^T$ is a matrix and we cannot use $v^Tv=1$ (directly, at least).
 A: For this, you need to compute the spectral radius of $vv^t$ directly. Fortunately that's easy: let $v, b_1, b_2, \ldots, b_{n-1}$ be any orthonormal basis. Then for $i = 1, \ldots n-1$, we have
$$
(vv^t)b_i = v (v^t b_i) = v (v \cdot b)_i) = 0= 0 b_i
$$
so $0$ is an eigenvalue $n-1$ times. But
$$
(v v^t ) v = v (v^t v) = 1 v.
$$
so $1$ is an eigenvalue once, and the spectral radius is $1$.
A: If $\lambda $ is an eigen value with eigen vector $x$ then  $\sum_j v_iv_jx_j=\lambda x_i$ for all $i$ and  (multiplying by $v_i$ and summing over $i$)  $\sum_i v_i^{2} \sum_j v_jx_j=\lambda \sum v_ix_i$, so $\lambda =1$ unless $\sum v_ix_i=0$. But $\sum v_jx_j=0$ gives $\lambda x_i =0$  so $\lambda =0$. Thus only possible eigen values are $0$ and $1$. Note that $1$ is an eigen value corresponding to the eigen vector $v$. Hence the spectral radius is $1$.
A: Here's another way. We have
$$vv^T= \left[ \begin{array}{c}v_1 \\ v_2 \\ \vdots \\v_n \end{array} \right] \left[\begin{array}{cccc}v_1 & v_2 & \ldots & v_n \end{array} \right] = \left[\begin{array}{cccc} v_1^2 & v_1v_2 & \ldots & v_1v_n \\ v_1v_2 & v_2^2 & \ldots & v_2v_n \\ \vdots & \vdots & \ddots & \vdots \\ v_1v_n & v_2v_n & \ldots & v_n^2 \end{array} \right].$$
We see easily that
$$\operatorname{Tr}(vv^T) = v_1^2+v_2^2+\cdots+v_n^2 = 1,$$
since $||v||=1.$ Since the trace of a matrix is equal to the sum of its eigenvalues, we have $\lambda_1 +\cdots+ \lambda_n=1.$ Further, since the matrix $vv^T$ is symmetric, it is diagonalisable, and the rank of a diagonalisable matrix is equal to the number of its non-zero eigenvalues. Since the above matrix $vv^T$ is clearly rank $1$, it has only one non-zero eigenvalue. Therefore $\lambda_i=1$ for some $1 \le i \le n.$ Hence, the spectral radius is 1.
