Know which variables are linear combinations of others from covariance (correlation) matrix Suppose I have the covariance and correlation matrices for several variables. I know one of the variables is almost a linear combination of the others. Is there any characteristic about the corresponding row in the covariance (correlation) matrix that I can use to pick out this variable?
 A: Let's say you have some random variable $Y$ and you want to know the projection of $Y$ onto a series of random variables $X_1, \ldots, X_k$ and a constant 1. This is a linear regression of the random variable $Y$ on random vector $\mathbf{X} = \begin{bmatrix} X_1 \\ X_2 \\ \ldots \\ X_k \end{bmatrix}$ and 1. Can you do that knowing just covariance matrices? Yes!


*

*Define the inner product of random variables $U$ and $V$ as $\langle U, V \rangle = \operatorname{E}[UV]$. Consequently, covariance is the inner product of demeaned random variables.

*Let $\Sigma_{\mathbf{X}\mathbf{X}} = \operatorname{Cov}\left(  \mathbf{X} \right) = \operatorname{E}[(\mathbf{X} - \operatorname{E}[\mathbf{X}](\mathbf{X} - \operatorname{E}[\mathbf{X}]')$ be the covariance matrix for random vector $\mathbf{X}$.

*Let $\Sigma_{\mathbf{X}Y} = \operatorname{Cov}\left( \mathbf{X}, Y \right) = \operatorname{E}\left[(\mathbf{X} - \operatorname{E}[\mathbf{X}])(Y - \operatorname{E}[Y]) \right] $ be a $k$ by 1 matrix giving the covariance with $X_1$ and $Y$, $X_2$ and $Y$ etc...


Then we can write the orthogonal decomposition:
$$ Y - \operatorname{E}[Y] =  \left( \mathbf{X} - \operatorname{E}[\mathbf{X}] \right) \cdot \mathbf{b} + \epsilon $$
Doing some algebra:
$$ Y = \underbrace{  \left( \operatorname{E}[Y] + \mathbf{b} \cdot \operatorname{E}[\mathbf{X}] \right)}_{\text{intercept}} +    \mathbf{b} \cdot \mathbf{X} + \underbrace{\epsilon}_{\text{error term}} $$
Where: 
$$\mathbf{b} = \Sigma_{\mathbf{X}\mathbf{X}}^{-1} \Sigma_{\mathbf{X}Y}$$ 
And $\epsilon$ is a random variable orthogonal to the linear span of the random variables $X_1, \ldots, X_k$ and 1. (Note, orthogonality of $\epsilon$ to 1 implies that $\operatorname{E}[\epsilon] = 0$.)
Example:
Let the covariance matrix of $ \begin{bmatrix} X_1 \\ X_2 \\ Y \end{bmatrix}$ be given by:
$$ \Sigma = \begin{bmatrix}     1.0010  &  0.5005  &  3.5035\\
    0.5005  &  1.2505 &   4.7529\\
    3.5035  &  4.7529 &  22.2675\end{bmatrix} $$
Then 
$$\Sigma_{XX} =  \begin{bmatrix}    1.0010  &  0.5005 \\ 0.5005 &   1.2505 \end{bmatrix} \quad \quad \Sigma_{XY} = \begin{bmatrix} 3.5035\\4.7529 \end{bmatrix}$$
$$ \mathbf{b} = \Sigma_{XX}^{-1} \Sigma_{XY} \approx \begin{bmatrix} 2 \\ 3 \end{bmatrix} $$
And we could write
$$ Y = \alpha + 2X_1 + 3 X_2 + \epsilon$$
where $\alpha$ is a scalar and $\epsilon$ is a mean zero random variable orthogonal to the $X_1$ and $X_2$.
Maybe you want the null space of your covariance matrix?
From your comment, it sounds like you want to know what vectors span the the null space of the covariance matrix. For example in MATLAB, you would call null(Sigma) where Sigma is your covariance matrix.
Computation of the null space is typically done through a singular value decomposition.
Let $\mathbf{w}$ be a vector and $\mathbf{X}$ be a random vector.. If $\mathbf{w}$ belongs to the null space of $\Sigma = \operatorname{Cov}(\mathbf{X})$ then:
$$ \Sigma \mathbf{w} = \mathbf{0} $$
Hence $\operatorname{Var}\left(\mathbf{X} \cdot \mathbf{w} \right)  = 0$ and $\mathbf{X} \cdot \mathbf{w} = 0$ almost surely.
