Largest singular values Given the positive semi-definite, symmetric matrix $A = bb^T + \sigma^2I$ where b is a column vector is it possible to find the singular values and singular vectors of the matrix analytically? I know that it has real eigenvalues since it's symmetric and positive semidefinite but not sure about solving directly for those values and their corresponding vectors.

The singular values are the eigenvalues of $A.$ By definition, when there exists a nonzero vector $\mathbf x$ for which $A\mathbf{x}=\lambda \mathbf{x},$ $\lambda$ is an eigenvalue and $\mathbf{x}$ is a corresponding eigenvector.
Note, then, that
$$A\mathbf{b} = (\mathbf{b}\mathbf{b}^\prime + \sigma^2I)\mathbf{b} = \mathbf{b}(\mathbf{b}^\prime \mathbf{b}) + \sigma^2 \mathbf{b} = (|\mathbf{b}|^2+\sigma^2)\mathbf{b},$$
demonstrating that $\mathbf{b}$ is an eigenvector with eigenvalue $\lambda_1 = |\mathbf{b}|^2 + \sigma^2.$
Furthermore, whenever $\mathbf{x}$ is a vector orthogonal to $\mathbf{b}$ -- that is, when $\mathbf{b}^\prime \mathbf{x} = \pmatrix{0},$ we may similarly compute
$$A\mathbf{x} = (\mathbf{b}\mathbf{b}^\prime + \sigma^2I)\mathbf{x} = \mathbf{b}(\mathbf{b}^\prime \mathbf{x}) + \sigma^2 \mathbf{x} = (0+\sigma^2)\mathbf{x},$$
showing that all such vectors are eigenvectors with eigenvalue $\sigma^2.$
Provided these vectors are in a finite dimensional vector space of dimension $n$ (say), a straightforward induction establishes that the vectors $x$ for which $\mathbf{b}^\prime \mathbf{x}=0$ form a subspace $\mathbf{b}^\perp$ of dimension $n-1.$  Let $\mathbf{e}_2, \ldots, \mathbf{e}_n$ be an orthonormal basis for this subspace.  It extends to an orthonormal basis $\mathscr{E} = (\mathbf{\hat  b}, \mathbf{e}_2, \ldots, \mathbf{e}_n)$ of the whole space where $\mathbf{\hat b} = \mathbf{b}/|\mathbf{b}|$.  In terms of this basis the matrix of $A$ therefore is
$$\operatorname{Mat}(A, \mathscr{E}, \mathscr{E}) = \pmatrix{|\mathbf{b}|^2+\sigma^2 & 0 & 0 & \cdots & 0 \\
0 & \sigma^2 & 0  & \cdots & 0 \\
0 & 0 & \ddots & \vdots & \vdots \\
\vdots & \vdots & \cdots & \ddots & 0 \\
0 & 0 & \cdots & 0 & \sigma^2
}$$
Whether or not every step of this derivation was clear, you can verify the result by setting
$$Q = \left(\mathbf{b}; \mathbf{e}_2; \ldots; \mathbf{e}_n\right)$$
to be the matrix with the the given columns and computing
$$Q\,\operatorname{Mat}(A, \mathscr{E}, \mathscr{E})\,Q^\prime = \mathbf{b}^\prime + \sigma^2I = A.$$
This is explicitly a singular value decomposition of the form $U\Sigma V^\prime$ where $V=Q,$ $\Sigma= \operatorname{Mat}(A, \mathscr{E}, \mathscr{E}),$ and $U=Q^\prime.$
The Gram Schmidt process provides a general algorithm to find $\mathscr{E}$ (and therefore $Q$): its input is the series of vectors $\mathbf{\hat b}$, $(1,0,\ldots,0)^\prime,$ and so on through $(0,\ldots,0,1)^\prime.$  After $n-1$ steps it will produce an orthonormal basis including the starting vector $\mathbf b.$

As an example, let $\mathbf{b} = (3,4,0)^\prime.$ With $\sigma^2 = 1,$ compute
$$\mathbf{b}\mathbf{b}^\prime + \sigma^2 I = \pmatrix{10&12&0\\12&17&0\\0&0&1}$$
Here, $|\mathbf{b}|^2 = 3^2+4^2+0^2=5^2,$ so that $\mathbf{\hat b} = \mathbf{b}/5 = (3/5,4/5,0)^\prime.$  One way to extend this to an orthonormal basis is to pick $\mathbf{e}_2 = (-4/5,3/5,0)^\prime$ and $\mathbf{e}_3 = (0,0,1)^\prime.$  Thus
$$Q = \pmatrix{3/5&4/5&0\\-4/5&3/5&0\\0&0&1}$$
and we may confirm that
$$\begin{align}
Q\,\operatorname{Mat}(A, \mathscr{E}, \mathscr{E})\,Q^\prime &= \pmatrix{3/5&4/5&0\\-4/5&3/5&0\\0&0&1}\pmatrix{5^2+1^2&0&0\\0&1&0\\0&0&1}\pmatrix{3/5&-4/5&0\\4/5&3/5&0\\0&0&1}\\
&=\pmatrix{10&12&0\\12&17&0\\0&0&1}
\end{align}$$
as intended.
