As Algebraic Pavel stated correctly:
Every principal submatrix of a positive (semi)definite matrix is positive (semi)definite.
Here some elaborate statements:
If $A\in\mathbb{R}^{n\times n}$ is positive definite, then all of its principal submatrices $a_{1:m,1:m}$ ($m=1,\dots, n$) are positive definite. If $A$ is positive semi-definite, then all of its principal submatrices $a_{1:m,1:m}$ ($m=1,\dots, n$) are positive semi-definite.
This also works for negative (semi)-definite matrices, by simply multiplying the matrix by -1, i.e.
If $-A\in\mathbb{R}^{n\times n}$ is positive definite, then all of its
principal submatrices $-a_{1:m,1:m}$ ($m=1,\dots, n$) are positive
definite. If $-A$ is positive semidefinite, then all of its principal
submatrices $-a_{1:m,1:m}$ ($m=1,\dots, n$) are positive semidefinite.
For a reference, see Observation 7.1.2 from Matrix Analysis (Horn, Johnson), 2nd edition.
See also Sylvester's criterion, which is a similar statement regarding principle minors.
You have done everything correct except for the small mistake in the last step. While normalising the vector, you have multiplied by the norm instead of dividing it and hence you are getting different values. For the first column, instead of $\frac{\sqrt{14}}{2}$ the term should be $\frac{2}{\sqrt{14}}$. And similarly for the other column. That should fix the error.
Best Answer
one understanding of positive definitive is to have all eigenvalue positive,or whole determinant and also minor determinants should be greater then $0$,for example let us consider
$(1-a^2)>0$
which means that from negative infinity to $-1$ or from $1$ to infinity,also
$(a^2-a)>0$
or
$a(a-1)>0$
or from negative infinity to $0$ or $1$ to infinity
as determinant we have
ans must be positive,or
we have
$(1-a)>0$ means that $1>a$
$2*a+1>0$
means that $a>-0.5$ ,so we have
$1>a$ and $a>-0.5$
about unitary matrix
Finding a unitary matrix that diagonalizes a given matrix