Let $X$ be an $n \times p$ matrix with rank$(X)=k$. Let $\epsilon \sim N(0,I_p)$ be a vector with the i.i.d. Gaussian components.

(a) Show that $$E\|\ X\epsilon \|_2^2 = \|\ X \|_F^2 = \sum_{j=1}^k \sigma_j^2$$ where $\sigma_1 \geq \sigma_2 \geq \dots \geq \sigma_k > 0$ are positive signular values of $X$ and $\|\ X \|_F$ is the Frobenius norm of $X$.

(b) Show that if $X$ is a projection matrix, i.e. $X=A(A^TA)^{-1}A^T$ for some matrix $A : n \times p$ with rank$(A)=p$, then $$E \|\ X\epsilon \|_2^2 = \sigma^2p.$$

By the singular value decomposition theorem we know that $X$ can be decomposed in the form: $$X=U\Sigma V^H,$$ where $U$ is an $n \times n$ unitary matrix, $\Sigma$ is a diagonal $n \times p$ matrix with non-negative real numbers on the diagonal, and $V$ is an $p\times p$ unitary matrix, and $V^H$ denotes the conjugate transpose of $V$. The diagonal entries $\sigma_i$ of $\Sigma$ are known as the singular values of $M$, in other words $\Sigma$ has $\left\{\sigma_1, \dots, \sigma_k\right\}$ on the diagonal, as rank$(X)=k$. Now, we have that from definition of Frobenius norm, \begin{align*} \|\ X \|_F^2 &= \operatorname{Tr}(XX^H) \\ &= \operatorname{Tr}((U\Sigma V^H)(U\Sigma V^H)^H) \\ &= \operatorname{Tr}(U\Sigma V^H V \Sigma^H U^H) \\ &= \operatorname{Tr}(U\Sigma\Sigma^H U^H) \\ &= \operatorname{Tr}(\Sigma\Sigma^H)\\ &= \sum_{j=1}^k \sigma_j^2. \end{align*} since $V^HV=I$ and $U^HU=I$, where we also utilized the cyclic property of trace. All that remains to be shown is that $E\|\ X\epsilon \|_2^2 = \sum_{j=1}^k \sigma_j^2$. I am not sure about this part of (a) and, hoping that once I figure that part out of (a), I'll have a better idea of (b). Any help would be appreciated!

  • $\begingroup$ Have you considered just writing out $||X\epsilon||^2$? It is the sum of $Y_{ij}\epsilon_i\epsilon_j$ where $Y_{ij}$ is the $ij$ entry of $X^\prime X.$ Taking expectations reduces everything to (very simple) algebraic properties of $Y.$ If you know them you're done and otherwise you have some exercises in linear algebra. $\endgroup$
    – whuber
    Sep 18, 2018 at 21:48
  • $\begingroup$ @whuber Is my approach correct? $\endgroup$
    – Maxtron
    Sep 18, 2018 at 22:55

2 Answers 2


I think you have almost solved it. $\mathbb{E}( ||X\epsilon||^2_2 ) = \mathbb{E}[ \mathrm{Tr} ( X \epsilon \epsilon^H X^H) ] = \mathrm{Tr} [ \mathbb{E} ( X \epsilon \epsilon^H X^H)]$ since trace and expectation are interchangeable.

\begin{align} \mathrm{Tr} [ \mathbb{E} ( X \epsilon \epsilon^H X^H)] = \mathrm{Tr}[ X \mathbb{E} \{\epsilon \epsilon^H \} X^H ] = \mathrm{Tr} [X I_p X^H] = \sum_{j=1}^k \sigma_j^2 \end{align}

  • 2
    $\begingroup$ I was not aware that trace and expectation are commutative operators. Thank you! $\endgroup$
    – Dragonite
    Sep 19, 2018 at 11:37

Ingeneral, if $\epsilon$ has mean $\mu$ and covariance matrix $\Sigma$, then we have the identity

$$ \mathbb E[\|X\epsilon\|^2] = \text{Trace}(X^TX\Sigma) + \mu^TX^TX\mu. $$

See proof on wikipedia https://en.wikipedia.org/wiki/Quadratic_form_(statistics).

Your particular case is then just a corollary, with have $\mu = 0$ and $\Sigma = I_n$.


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