Find $\det \left( \lambda_1X_1X_1^T + \lambda_2X_2X_2^T + \cdots + \lambda_nX_nX_n^T \right)$

Question

Let $X \in \mathbb{R}^{n \times n}$, $X_i$ denote the $i$-th column of $X$ and let $\lambda_i \in \mathbb{R}$. Find

$$\det \left( \lambda_1X_1X_1^T + \lambda_2X_2X_2^T + \cdots + \lambda_nX_nX_n^T \right)$$

I calculated that the matrix inside determinant will be symmetrical and its elements will be:

$\lambda_1x_{i1}^2 + ... \lambda_nx_{in}^2$ - for diagonal elements

$\lambda_1x_{i1}x_{j1} + ... + \lambda_nx_{in}x_{jn}$ - everything else

From there I am stuck on what to do next

(this task also asks for which $\lambda_i$ determinant will be $\ge 0$, maybe this gives some clue)

Let $M$ be the matrix inside the determinant, let $Y$ be the inverse of $X$. What is the matrix $YMY^T$ ? — kimchi lover, Aug 30 '19 at 18:02
@kimchilover it took time but I got $YMY^T = diag(\lambda_1,..,\lambda_n)$. — Назар Петровский, Aug 30 '19 at 18:21

Gabriel Romon · Accepted Answer · 2019-08-30T18:42:06.963

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Note that $\lambda_1X_1X_1^T + \lambda_2X_2X_2^T + ... + \lambda_nX_nX_n^T=XDX^T$ where $D$ is a diagonal matrix with diagonal $(\lambda_1,\ldots,\lambda_n)$.

Indeed $(XDX^T)_{ij} = \sum_k X_{ik}D_{kk}X_{jk} = \sum_k \lambda_k (X_kX_k^T)_{ij}=(\sum_k \lambda_k X_kX_k^T)_{ij}$.

Hence $$\begin{aligned}\det(\lambda_1X_1X_1^T + \lambda_2X_2X_2^T + ... + \lambda_nX_nX_n^T) &= \det(XDX^T)=\det(X)\det(D)\det(X^T)\\ &=\left(\prod_k \lambda_k\right) \det(X)^2\end{aligned}$$

edited Aug 30 '19 at 18:42

answered Aug 30 '19 at 18:23

Gabriel Romon

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Thanks, that's exactly what I got by using @kimchilover advice – Назар Петровский Aug 30 '19 at 18:31
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But I think you should expand it farther because $det(AB) = det(A)det(B)$ and $det(A^T)=det(A)$ for any $A,B$. So it will be $(...\lambda_i...)(det(X))^2$ – Назар Петровский Aug 30 '19 at 18:34
@НазарПетровский I got distracted IRL sorry – Gabriel Romon Aug 30 '19 at 18:45
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Another way to see the decomposition is to express the column vectors of $X$ via matrix multiplication, i.e., $X_k = X e_k$. One immediately has $$\sum_k \lambda_k X_k X_k^\top = \sum_k (\lambda X e_k) (e_k^\top X^\top) = X\left(\sum_k \lambda e_k e_k^\top\right)X^\top=XDX^\top.$$ – Semiclassical Aug 30 '19 at 20:35

Find $\det \left( \lambda_1X_1X_1^T + \lambda_2X_2X_2^T + \cdots + \lambda_nX_nX_n^T \right)$

1 Answers1