Condition for being positive definite matrix A Hermitian matrix is positive definite if the scalar $z^{\dagger} M z$ is strictly positive for every nonzero $z$.
But what does expression $z^{\dagger}Mz$ mean? I am not getting any intuition here. The spectral decomposition has a similar type of expression. Is there any relation with that?  
 A: 
An $n\times n$ complex matrix $A$ is called positive definite if
  for all nonzero complex vectors $x$ in $\mathbb C^n,$ where 
  $x^\dagger$ denotes the conjugate transpose of the vector $x$ fulfill the condition:
$$\Re\left[x^\dagger Ax \right]>0$$     

Generally, this concept is applied to symmetric matrices.
For example, given the matrix
$$A = \begin{bmatrix} 
3+3i & -5+0i & 2-9i \\
-5+0i & 6+2i & 7-1i \\
2-9i & 7-1i & -1+0i
\end{bmatrix}$$
and an example of a complex vector
$$x = \begin{bmatrix} 9-1i \\0-3i \\0-1i\end{bmatrix}$$
with conjugate transpose
$$x^\dagger = \begin{bmatrix} 9+1i &0+3i &0+1i\end{bmatrix}$$
we get
$$x^\dagger A x = 315+240i$$
with $\Re(x^\dagger A x) = 315 >0.$
In the case of a Hermitian matrix $x^\dagger A x$ will be real. For example,
$$A =\begin{bmatrix}
3 & -5-3i & 2-1i\\
-5+3i & 6 & 7-1i\\
2+1i & 7+1i & -1
\end{bmatrix}$$
$$x^\dagger A x = 135$$
The intuition is that of a quadratic operation, which renders itself to finding a global minimum when the result is consistently positive (picture a bowl).
From Prof. Strang's lectures:
$$x^\top A x = \begin{bmatrix}x_1&x_2\end{bmatrix}\begin{bmatrix}a&b\\b&c\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=ax_1^2 + 2 b x_1 x_2 +cx_2^2$$
The idea is that $x_1$ and $x_2$ are variables. Both $x_1^2$ and $x_2^2$ are naturally positive (or zero); therefore, the term $2bx_1x_2$ determines the PSD-ness of the matrix.
The most widespread application is in ordinary least squares (OLS) parameter estimation, where given a model matrix $X$ and dependent variable realization $y,$ a vector of parameters or coefficients $\hat \beta$ are calculated in closed form to minimize the sum of square residuals:
$$\begin{align}
e^\top e &= (y - X\hat \beta)^\top(y - X\hat\beta)\\
&=y^\top y
- y \hat\beta^\top X^\top y + \color{red}{\hat \beta^\top X^\top X\hat \beta}
\end{align}$$
by differentiating:
$$\frac{\partial e^\top e}{\partial \hat \beta}=-2 X^\top y + 2 X^\top X \hat \beta =0$$
$\color{red}{ X^\top X}$ is a Gramian matrix, and hence, positive definite.
Also, a function $f : \mathbb R^n \to \mathbb R$ of the form $f(x) = x^\top Ax = \sum_{i,j=1}^n A_{ij} x_ix_j$ is called a quadratic form.
