If $\Lambda_k = \lambda I + \sum_{i=1}^{k} \phi_i \phi_i^\top $ where $\|\phi_i\|\leq 1$, then $\det(\Lambda_k) \leq (k+\lambda)^d$ Consider a matrix $\Lambda_k = \lambda I + \sum_{i=1}^{k} \phi_i \phi_i^\top$ where $\phi_i \in \mathbb{R}^d$ and $\|\phi_i\|\leq 1$. Prove that $\det(\Lambda_k) \leq (k+\lambda)^d$. Here $\lambda > 0$ and $d \in \mathbb{N}$.
How can I approach this problem?
 A: my sense is you are supposed to follow my hint and then apply $\text{GM}\leq \text{AM}$.  As currently stated, this is a rather loose bound.  I give a somewhat different proof, below.
$\Lambda_k$ is real symmetric and is equal to a PD matrix plus PSD matrices$\implies \Lambda_k\succ \mathbf 0$
1.) $\text{trace}\big(\Lambda_k\big) = \text{trace}\big(\lambda I_d\big) + \sum_{i=1}^{k} \text{trace}\big(\phi_i \phi_i^\top\big)= d\cdot \lambda +\sum_{i=1}^{k} \big \Vert \phi_i \big\Vert_2^2= \eta \leq d\cdot \lambda + k$
note: this is rather misleading notation as a subscript of $k$ typically indicates that the matrix is $k\times k$ -- e.g. $I_d$ does mean it is $d\times d$.  Also $\Lambda$ typically refers to a diagonal matrix with eigenvalues $\lambda_i$ though it doesn't here.
2.) $\Lambda_k$ is orthogonally similar to a matrix with constant diagonal, $C$, i.e.
$c_{i,i}=\frac{1}{d}\cdot \text{trace}\big(\Lambda_k\big)=\frac{\eta}{d}\leq \lambda + \frac{k}{d}$
(this can be done over reals, but it is more convenient to extend $\mathbb R$ to $\mathbb C$, view $\Lambda_k$ as Hermitian PD, diagonalizable by (real) unitary matrix $Q$, then $C$ is similar to $A$ via $\big(FQ\big)$, where $F$ is the unitary matrix known as Discrete Fourier Transform.  If the determinant bound holds over $\mathbb C$ then it must hold over $\mathbb R$ since $\mathbb R\subset \mathbb C$.)
3.) $0\lt \det\big(\Lambda_k\big)=\det\big(C\big)\leq \big(\frac{\eta}{d}\big)^d \leq \big(\lambda + \frac{k}{d}\big)^d \leq \big(\lambda + k\big)^d$
as desired.
The inequalities are (i) Hadamard Determinant Inequality, and (ii) term by term dominance of positive numbers, raised to the $d$th power
