Suppose that we have vector $y_t = (z_t, x_t)'$ for $t = 1, 2, \dots, T$. Let $\Omega_t$ be the information set available at time $t$ and $z_t(h|\Omega_t)$ be optimal $h$-step predictor. $\Sigma_z(h|\Omega_t)$ denotes forecast MSE matrix.

We say that $x_t$ helps to predict $z_t$ (Granger Causality) if $$\Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right) - \Sigma_z(h|\Omega_t)$$ for at least one $h = 1,2, \dots$ is positive semidefinite and $$\Sigma_z(h|\Omega_t) \neq \Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right).$$ The definition is clear for me. $x_t$ helps predicting $z_t$ if the covariance matrix of the prediction without using $x_t$ is bigger then when we use $x_t$ in a matrix sense.

My confusion is more about interpretation of the semidefinite matrix. By definition $m\times m$ matrix $A$ is positive semidefinite iff for any $m\times 1$ vector $x$ $$x'Ax \geq 0.$$

Let $A \equiv \Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right) - \Sigma_z(h|\Omega_t).$ So what is the intuition that if $x'Ax \geq 0$, then I can conclude that $x_t$ helps in predicting $z_t$?

Suppose that we have vector $y_t = (z_t, x_t)'$ for $t = 1, 2, \dots, T$. Let $\Omega_t$ be the information set available at time $t$ and $z_t(h|\Omega_t)$ be optimal $h$-step predictor. $\Sigma_z(h|\Omega_t)$ denotes forecast MSE matrix.

We say that $x_t$ helps to predict $z_t$ (Granger Causality) if $$\Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right) - \Sigma_z(h|\Omega_t)$$ for at least one $h = 1,2, \dots$ is positive semidefinite and $$\Sigma_z(h|\Omega_t) \neq \Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right).$$ The definition is clear for me. $x_t$ helps predicting $z_t$ if the covariance matrix of the prediction without using $x_t$ is bigger then when we use $x_t$ in a matrix sense.

My confusion is more about interpretation of the semidefinite matrix. By definition $m\times m$ symmetric matrix $A$ is positive semidefinite iff for any $m\times 1$ vector $z$ $$z'Az \geq 0.$$

Let $A \equiv \Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right) - \Sigma_z(h|\Omega_t).$ So what is the intuition that if $x'Ax \geq 0$, then I can conclude that $x_t$ helps in predicting $z_t$?

Edit: Just to clarify, the question is more about the concept that the variance of one random vector $x$ (denote it by $\Omega_x$) is greater than for the other random vector $y$ (denote it by $\Omega_y$) if the difference between their covariance matrices, $\Omega_x - \Omega_y$ is psd. As one more example, consider GLS and OLS estimators in a linear regression setting, $\hat\beta_{GLS}$ and $\hat\beta_{OLS}$ respectively. We say that the GLS estimator is more efficient if $B \equiv \Omega_{OLS} - \Omega_{GLS}$ is psd matrix. That is, for any vector $z$ it holds that $$z'Bz \geq 0.$$ In other words, covariance matrix $\Omega_{GLS}$ is smaller in a matrix sense than $\Omega_{OLS}$. What I am looking for is an explanation/intuition of why the case when the matrix $\Omega_{OLS} - \Omega_{GLS}$ is psd implies that the GLS estimator is more efficient then the OLS? Maybe another formulation: what does it mean for one matrix to be smaller then the other?

Suppose that we have vector $y_t = (z_t, x_t)'$ for $t = 1, 2, \dots, T$. Let $\Omega_t$ be the information set available at time $t$ and $z_t(h|\Omega_t)$ be optimal $h$-step predictor. $\Sigma_z(h|\Omega_t)$ denotes forecast MSE matrix.

We say that $x_t$ helps to predict $z_t$ (Granger Causality) if $$\Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right) - \Sigma_z(h|\Omega_t)$$ for at least one $h = 1,2, \dots$ is positive semidefinite and $$\Sigma_z(h|\Omega_t) \neq \Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right).$$ The definition is clear for me. $x_t$ helps predicting $z_t$ if the covariance matrix of the prediction without using $x_t$ is bigger then when we use $x_t$ in a matrix sense. My confusion is more about interpretation of the semidefinite matrix. By definition $m\times m$ matrix $A$ is psd iff for any $m\times 1$ vector $x$ $$x'Ax \geq 0.$$

Let $A \equiv \Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right) - \Sigma_z(h|\Omega_t).$ So what is the intuition that if $x'Ax \geq 0$, then I can conclude that $x_t$ helps in predicting $z_t$?

Suppose that we have vector $y_t = (z_t, x_t)'$ for $t = 1, 2, \dots, T$. Let $\Omega_t$ be the information set available at time $t$ and $z_t(h|\Omega_t)$ be optimal $h$-step predictor. $\Sigma_z(h|\Omega_t)$ denotes forecast MSE matrix.

We say that $x_t$ helps to predict $z_t$ (Granger Causality) if $$\Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right) - \Sigma_z(h|\Omega_t)$$ for at least one $h = 1,2, \dots$ is positive semidefinite and $$\Sigma_z(h|\Omega_t) \neq \Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right).$$ The definition is clear for me. $x_t$ helps predicting $z_t$ if the covariance matrix of the prediction without using $x_t$ is bigger then when we use $x_t$ in a matrix sense. 

My confusion is more about interpretation of the semidefinite matrix. By definition $m\times m$ matrix $A$ is positive semidefinite iff for any $m\times 1$ vector $x$ $$x'Ax \geq 0.$$

Let $A \equiv \Sigma_z\left(h|\Omega_t\setminus\{x_s|s\leq t\}\right) - \Sigma_z(h|\Omega_t).$ So what is the intuition that if $x'Ax \geq 0$, then I can conclude that $x_t$ helps in predicting $z_t$?