# Linear Algebra - Inner Product Proof

The question is:

Show that the following defines an inner product on $\mathbb{R}^2$: $$\langle ( x_1 , y_1 ) , ( x_2 , y_2) \rangle = x_1 y_1 + 2 x_1 y_2 + 2 x_2 y_1 + 8 x_2 y_2$$

Tried to show that $\langle ( x_2 , y_2) , ( x_1 , y_1) \rangle$ is the same thing but came at a roadblock (it isn't).

It's a past exam, and the solution says that the inner product is defined by $X_1^T A X_2$ where $$A = \begin{pmatrix} 1&2\\2&8 \end{pmatrix}$$

The exam is in two days, so any help would be greatly appreciated.

• Well the thing you say isn't the same thing is, and the matrix formula gives you the same formula as the original. So what actually is the question? Jun 3, 2017 at 13:41
• Here's a tutorial in MathJax Jun 3, 2017 at 13:42

HINT: You have to show, for arbitrary $x,y,z$

$\langle x,y \rangle = \langle y,x \rangle$

$\langle ax+by,z \rangle = a\langle x,z \rangle + b\langle y,z \rangle$

(linearity in the second component follows by symmetry)

$\langle x,x \rangle > 0$ if $x \neq 0$

The first two things is calculating and see if it works out. For the last thing, you have to show that the expression is always greater than zero for every non zero vector. Since you know you can rewrite the inner product as:

$$X_1^TAX_2$$

it suffices to show that $$X^TAX >0$$

which is the same thing as saying that $A$ is positive definite. For this, you can for example use this post:

Proving that a symmetric matrix is positive definite iff all eigenvalues are positive

(Or even more simple, write the expression as a square)

Bilinearity is easy, particularly if you use the matrix form.

For the final condition, consider $$\langle(x,y),(x,y)\rangle=x^2+4xy+8y^2=x^2+4xy+4y^2+4y^2= (x+2y)^2+4y^2\ge0$$ When is such an expression equal to $0$?

Hint:

Definition: $\langle\alpha,\beta\rangle$ is an inner product if:

1. $\langle\alpha+\beta,\gamma\rangle=\langle\alpha,\gamma\rangle+\langle\beta,\gamma\rangle$
2. $\langle c\alpha,\beta\rangle=c\langle\alpha,\beta\rangle$; where c is scalar.
3. $\langle\alpha,\beta\rangle=\langle\beta,\alpha\rangle$
4. $\langle\alpha,\alpha\rangle\gt0 ; \alpha\neq0$ So, if you prove that $\langle(x_1,y_1),(x_2,y_2)\rangle$ satisfies the 4 conditions of definotion, you've already proved that it's an inner product.
• Scalar missing in 2. $\langle c \alpha, \beta \rangle = c \langle \alpha, \beta \rangle$. Jun 3, 2017 at 16:50
• thank you, I will edit it right now @GilbertoLópez
– user192508
Jun 3, 2017 at 19:49

$A$ is positive-definite (both eigenvalues are strictly positive) and symmetric.

Since it is symmetric: $\langle X_1,X_2\rangle_A = X_1^TA\,X_2 = (A\,X_1)^TX_2 = X_2^T(A\,X_1) = X_2^TA\,X_1 = \langle X_2,X_1\rangle_A$

Since it is positive-definite: $\langle X,X\rangle_A \geq 0$ (it is zero iff $X=0$)

Therefore $\langle X_1,X_2\rangle_A$ defines an inner product!

$\langle (x_1,y_1),(x_2,y_2)\rangle_A = x_1\,x_2 + 2\,x_1\,y_2 + 2\,y_1\,x_2 + 8\,y_1\,y_2$