Let $A = \begin{bmatrix}a & b\\c & d\end{bmatrix}$, where $a,b,c,d$ are real numbers.
For $x,y \in \Bbb{R}^2$ , let $f_A(x,y) = y^T A x$ be an inner product on $\Bbb{R}^2$.
Show that $b = c$ , $a > 0$ , $d > 0$ , and $ad − bc > 0$.
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SacredMechanic
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3Does this answer your question? Show that a 2x2 matrix A is symmetric positive definite if and only if A is symmetric, trace(A) > 0 and det(A) > 0 – Kandinskij Dec 12 '20 at 11:20
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If I'm right, you want to know the conditions for it to be a inner product, right? If it is, you need to show the following: Use your muscles to "math this out"!
- Linearity:
In this case, prove that $$ f_{A}(ax,y) = a f_{A}(x,y), a \in \Bbb{R} $$ and $$ f_{A}(x+y,z) = f_{A}(x,z) + f_{A}(y,z). $$
- Symmetry
Show that $$ f_{A}(x,y)=f_{A}(y,x). $$
- "Positive-definitedness"
Just show that, if $x \neq 0$, then $$ f_{A}(x,x) > 0. $$
While you prove this, you will be able to achieve the conditions that you want. To make it easiear, take $x = (x_{1},x_{2})$ and $y = (y_{1},y_{2})$.