Let A be a matrix with real entries. A is positive semidefinite if the dot product of Ax and x is nonnegative for any vector with real components. $\langle Ax, x \rangle \geq 0$
Let A = (1,2)(-2,1) Then Ax = $(x_1+2x_2)(-2x_1+x_2)$, Finally $\langle Ax, x \rangle$ =$(x_1^2+2x_2^2-2x_1^2+x_2^2)$
but the text I am reading just argues: $\langle Ax, x \rangle$ = $x_1^2+x_2^2 \geq 0$ implying that A is positive semidefinite.
The question is : What happened to the $2x_2^2-2x_1^2$ part? why can we just claim that $x_1^2+x_2^2 \geq 0$ ,which may be true. but still we have one negative in our equation, right?
This definition of positive semidefinite is wrong. Let $ v=[x_1,x_2] $. A 2x2 matrix is positive semidefinite if $ vAv^T \ge 0 $ for all v.Note that A can be replaced by $ (A+A^T)/2 $ so there is no loss of generality in assuming that A is symmetric. Note also that some authors include the additional requirement that A not be positive definite in their definition of positive semidefiniteness. Note that the definition can be generalised to vectors with n components and nxn (symmetric) matrices. An old book by Ferrar "Algebra: a text-book of determinants, matrices and algebraic forms" is an excellent reference. A 2x2 symmetric matrix is positive semidefinite (without imposing the condition that it not be positive definite) iff its diagonal elements and its determinant are all non-negative.