I am taking an introductory course to finance in my Master's, and wanted to go further in the topic of portfolio theory (I am an engineering bachelor graduate, but as I just hinted, I am new to advanced maths for statistics). My question might look dumb to some of you, but I think my main problem is simply understanding the notation of the variance-covariance matrix. I watched a video saying that when calculating the optimal vector of weights to obtain the minimum-variance portfolio, in the case of many assets, we have: $$ \sigma_{portfolio}^2 = w^T\Sigma w$$ $$w^T\textbf{1}=1 $$ , where $w$ is a column vector of weights, and $\textbf{1}$ is (I suppose?) a column vector filled with 1s with the same dimension as $w$.

He then goes on to write the Lagrangian as follows: $$ \operatorname{L}=w^T\Sigma w + \lambda(w^T -1) $$ And, by deriving the Lagrangian with respect to $w$ and setting the derivative to 0, and after a couple of other steps, he ends up with: $$ w=\frac{\Sigma^{-1}\textbf{1}}{\textbf{1}^T\Sigma^{-1}\textbf{1}} $$ (If you'd like to see the intermediary steps, please ask).

The reasoning seems pretty clear to me, however I have no idea how to interpret the notation of the variance-covariance matrix in this final equation: does $\Sigma^{-1}\textbf{1}$ mean the inverse variance-covariance matrix of vector $\textbf{1}$ (which wouldn't make much sense according to the calculations I did for that case), or is it implied that $\Sigma^{-1}$ is the inverse variance-covariance matrix of the expected returns of the respective stocks? And, therefore, $\textbf{1}^T\Sigma^{-1}\textbf{1}$ would be the elements of the inverse var-covar matrix of expected returns, but as scalars (which would make sense because we can only divide scalars together and not matrices?).

Am I right in this assumption, or am I still missing the point? Again, maybe this question will look awkward to some of you, but I'm new to the topic, and after some research on the web I couldn't find the var-covar notations well explained anywhere.

Thank you!

  • 2
    $\begingroup$ $\Sigma^{-1}\textbf{1}$ really is just multiplication of the matrix $\Sigma^{-1}$ with the column vector $\textbf{1}$, so standard multiplication rules apply. When multiplying with a column vector of ones, you get the row sums. $\endgroup$ Commented Mar 19, 2021 at 14:42

1 Answer 1


I think what is going on here is the matrix-vector product $\Sigma^{-1} \mathbf{1}$ and quadratic form $\mathbf{1}^T\Sigma^{-1} \mathbf{1}$ are being used as a convenient way of specifying summation. That is, they bridge matrix-vector notation and summation notation.

Assuming $\Sigma^{-1} \in \mathbb{R}^{d \times d}$, then $\Sigma^{-1} \mathbf{1}$ is the product of the matrix $\Sigma^{-1}$ with the column vector $\mathbf{1} \in \mathbb{R}^d$. Hence the matrix-vector $\Sigma^{-1} \mathbf{1}$ product is just a column vector in $\mathbb{R}^d$ where each element is the sum of the rows of $\Sigma^{-1}$.

Without any further context, $\Sigma^{-1} \mathbf{1}$ means take your inverse variance-covariance matrix $\Sigma^{-1}$ and take the sum of each row, to give a column vector.

Further, the quadratic form $\mathbf{1}^T \Sigma^{-1} \mathbf{1}$ is a product of the row vector $\mathbf{1}^T \in \mathbb{R}^{1 \times d}$, the matrix $\Sigma^{-1}$, and the column vector $\mathbf{1} \in \mathbb{R}^{d \times 1}$. Computing this will yield the sum of all elements in $\Sigma^{-1}$.

Similar to above, $\mathbf{1}^T \Sigma^{-1} \mathbf{1}$ means take your inverse variance covariance matrix $\Sigma^{-1}$ and sum all the elements, yielding a scalar.

I guess then what the expression for $w$ means is to take the column vector $\Sigma^{-1} \mathbf{1}$ and divide each element by the scalar $(\mathbf{1}^T \Sigma^{-1} \mathbf{1})$ (or more precisely multiply each element in the column vector by the scalar $(\mathbf{1}^T \Sigma^{-1} \mathbf{1})^{-1}$).

You might wonder why do this when we can just use summation notation - you would be surprised just how useful this is when writing vectorised mathematical pseudocode.

  • $\begingroup$ I see now, things are becoming clearer. However one last thing, you're talking about $\Sigma^{-1}$, but $\Sigma^{-1}$ of what vector quantity exactly? I understand that the diagonal of $\Sigma$ contains the variances and the rest of cells the covariances, but of what vector quantity here exactly? How I see things is that $\Sigma$ alone doesn't calculate the (co)variances of anything unless it's implied somehow $\endgroup$
    – it'syasper
    Commented Mar 19, 2021 at 15:08
  • $\begingroup$ My response only addresses the problematic interpretations of the matrix-vector product and quadratic form in the numerator and denominator respectively of $w$ from the perspective of linear algebra. If you want context specific interpretations concerning $\Sigma$, which I cannot guarantee, but others on the forum may be able to assist with , please supply more info on what is contained in $\Sigma$. $\endgroup$
    – microhaus
    Commented Mar 19, 2021 at 15:27
  • $\begingroup$ Alright thanks for the explanations! All clear now $\endgroup$
    – it'syasper
    Commented Mar 19, 2021 at 16:53

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