How is OLS estimator converging in quadratic mean equivalent to its variance matrix converging to $0$? $\newcommand{\E}{\mathbb{E}}$ $\newcommand{\Var}{\text{Var}}$ $\newcommand{\b}{\beta}$
Sorry that my title is not clear (if there is any better suggestions, I will edit it as soon as I can)
I want to show that under some conditions, $\hat{\b}\overset{q.m.}{\to}\b$, i.e. $\E[|\hat{\b} - \b|^2]\to 0$ is equivalent to $\Var(\hat{\b})\to 0$, where $\hat{\b}, \b$ are vectors.
($\hat{\b}$ is OLS estimator)
And I found the relation as follows: $$
\begin{align*}
 \E|\hat{\b} - \b|^2 &= \E [\hat{\b}^2 - 2\hat{\b}\b + \b^2] \\
   &= \E[\hat{\b}^2] - 2\b\cdot \E\hat{\b} + \b^2 \\
   &= \E[\hat{\b}^2] - (\E\hat{\b})^2 \;\;\text{since $\E\hat{\b} = \b$} \\
   &= \Var(\hat{\b})
\end{align*}$$
However, it only does make sense if all variables are univariate, i.e. $\hat{\b}, \b\in\mathbb{R}$ (or $\mathbb{C}$) since $\Var(\hat{\b})$ is a matrix.
How can I improve this and show that $\hat{\b}\overset{q.m.}{\to}\b$ is equivalent to $\Var(\hat{\b})\to 0$ in multivariate case?
Any help will be appreciated!
 A: Couple facts:

*

*In general, if $v$ is a random vector, where each entry has finite second moments, then
$$
E[ \|v\|_2^2] = E[v'v] = E[\mbox{trace}(vv')] = \mbox{trace} (E[vv'])
$$
If $v$ has mean zero, then $E[vv']$ is the variance-covariance matrix.


*Suppose $Q_n$ is a sequence of positive semidefinite matrices. Then $Q_n \rightarrow 0$ (in any one of the equivalent matrix norms), if and only if $\mbox{trace}\, Q_n \rightarrow 0$.
Then your calculation in the single variable case extends essentially verbatim, after replacing $v $ by $\hat{\beta} - \beta \,(= \hat{\beta} - E[\hat{\beta}]$): By Fact 1, the quadratic-mean (squared) distance between $\hat{\beta}$ and $\beta$ is
\begin{align*}
E[\|\hat{\beta} - \beta\|^2_2] = \mbox{trace} (E[(\hat{\beta} - \beta)(\hat{\beta} - \beta)']). 
\end{align*}
Since it's assumed that
$$
E[(\hat{\beta} - \beta)(\hat{\beta} - \beta)'] \rightarrow 0,
$$
Fact 2 implies
$$
\mbox{trace} (E[(\hat{\beta} - \beta)(\hat{\beta} - \beta)']) \rightarrow 0.
$$
Notice we have made the assumption (as you did) that $\hat{\beta}$ is unbiased, $E[\hat{\beta}] = \beta$. (This is true, for example, under the Gauss-Markov type assumption $E[\epsilon|X] = 0$.) In general, vanishing variance-covariance matrix just means
$$
E[\|\hat{\beta} - \beta\|^2_2] - \| E[\hat{\beta}] - \beta \|_2^2 \rightarrow 0.
$$
A: The problem here is that it is not (entirely) clear how you want to generalise the squared error to the multivariate case.  There are two obvious ways you can do this depending on what is of interest to you.  One generalisation is to use the inner product, leading to the squared norm of the estimation error.  The other generalisation is to use the outer product, leading to the variance matrix.
I will illustrate each of these generalisations below.  To start with, we note that in either case the OLS estimator in the multiple linear regression model leads to the form:
$$\begin{equation} \begin{aligned}
\hat{\boldsymbol{\beta}} 
&= (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \mathbf{Y} \\[6pt]
&= (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} (\mathbf{X} \boldsymbol{\beta} + \mathbf{\epsilon}) \\[6pt]
&= \boldsymbol{\beta} + (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \mathbf{\epsilon}. \\[6pt]
\end{aligned} \end{equation}$$
From this equation we see that the estimation error is the vector:
$$\hat{\boldsymbol{\beta}} - \boldsymbol{\beta} = (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \mathbf{\epsilon}.$$

Generalising to the squared norm (using the inner product): Using the above form for the OLS estimator you can write the squared norm of the estimator error by the inner product of the estimation error:
$$\begin{align}
||\hat{\boldsymbol{\beta}} - \boldsymbol{\beta}||^2
&= (\hat{\boldsymbol{\beta}} - \boldsymbol{\beta})^\text{T} (\hat{\boldsymbol{\beta}} - \boldsymbol{\beta}) \\[6pt]
&= ((\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \mathbf{\epsilon})^\text{T} ((\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \mathbf{\epsilon}) \\[6pt]
&= \mathbf{\epsilon}^\text{T} \mathbf{x} (\mathbf{x}^\text{T} \mathbf{x})^{-1} (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \mathbf{\epsilon} \\[6pt]
&= \mathbf{\epsilon}^\text{T} \mathbf{x} (\mathbf{x}^\text{T} \mathbf{x})^{-2} \mathbf{x}^\text{T} \mathbf{\epsilon}, \\[6pt]
\end{align}$$
which is a quadratic form of the error vector $\mathbf{\epsilon}$.  Using the standard rule for expectation of a quadratic form you then have:
$$\begin{align}
\mathbb{E}(||\hat{\boldsymbol{\beta}} - \boldsymbol{\beta}||^2)
&= \sigma^2 \text{tr}(\mathbf{x} (\mathbf{x}^\text{T} \mathbf{x})^{-2} \mathbf{x}^\text{T}). \\[6pt]
\end{align}$$
Obviously this is just a scalar value, and it does not correspond to the variance matrix of the estimator.  Nevertheless, there are various sufficient conditions you can impose on the limit of the design matrix to ensure that this value converges to zero.

Generalising to the variance matrix (using the outer product): Using the above form for the OLS estimator you can write the variance of the estimator by the outer product of the estimation error:
$$\begin{align}
(\hat{\boldsymbol{\beta}} - \boldsymbol{\beta}) (\hat{\boldsymbol{\beta}} - \boldsymbol{\beta})^\text{T} 
&= ((\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \mathbf{\epsilon}) ((\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \mathbf{\epsilon})^\text{T}  \\[6pt]
&= (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} (\mathbf{\epsilon} \mathbf{\epsilon}^\text{T}) \mathbf{x} (\mathbf{x}^\text{T} \mathbf{x})^{-1}.  \\[6pt]
\end{align}$$
Using standard moment rules you then have:
$$\begin{align}
\mathbb{V}(\hat{\boldsymbol{\beta}}) 
&= \mathbb{E}((\hat{\boldsymbol{\beta}} - \boldsymbol{\beta}) (\hat{\boldsymbol{\beta}} - \boldsymbol{\beta})^\text{T}) \\[6pt]
&= (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} (\mathbf{\epsilon} \mathbf{\epsilon}^\text{T}) \mathbf{x} (\mathbf{x}^\text{T} \mathbf{x})^{-1} \\[6pt]
&= \sigma^2 (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \boldsymbol{I} \mathbf{x} (\mathbf{x}^\text{T} \mathbf{x})^{-1} \\[6pt]
&= \sigma^2 (\mathbf{x}^\text{T} \mathbf{x})^{-1} (\mathbf{x}^\text{T} \mathbf{x}) (\mathbf{x}^\text{T} \mathbf{x})^{-1} \\[6pt]
&= \sigma^2 (\mathbf{x}^\text{T} \mathbf{x})^{-1}. \\[6pt]
\end{align}$$
If your regression model has $m$ terms (so that the design matrix $\mathbf{x}$ is an $n \times m$ matrix) then this variance matrix is an $m \times m$ matrix.  It is possible to impose conditions on the limit of this matrix form so that the variance "converges to zero" in an appropriate sense.
