Inner product space computation

Question

If $x = (x_1,x_2)$ and $y = (y_1,y_2)$ show that $\langle x,y\rangle = \begin{bmatrix}x_1 & x_2\end{bmatrix}\begin{bmatrix}2 & -1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}y_1 \\ y_2\end{bmatrix}$ defines an inner product on $\mathbb{R}^2$.

Is there any hints on this one? All I'm thinking is to compute a determinant, but what good is that?

Answer 1

Use the definition of an inner product and check whether your function satisfies all the properties. Note that in general, as kahen pointed out in the comment, $\langle \mathbf{x}, \mathbf{y}\rangle = \mathbf{y}^*A\mathbf{x}$ defines an inner product on $\mathbb{C}^n$ iff $A$ is a positive-definite Hermitian $n \times n$ matrix.

Answer 2

An inner product has to have all of the following properties:

Positivity, $\langle v, v \rangle \ge 0$ for all $v \in V$.

Definiteness, $\langle v, v \rangle = 0$ if and only if $v = 0$.

Linearity in the first slot, $\langle \alpha(u+v), w \rangle = \alpha \langle u, w \rangle + \alpha \langle v, w \rangle$.

Conjugate symmetry, $\langle u, v \rangle = \overline{\langle v, u \rangle}$. Since the vector space is real, any scalar is equal to its conjugate, so you have to show that $\langle u, v \rangle = \langle v, u \rangle$.

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To prove that your inner product has all of these properties, you can just calculate the inner product symbolically for appropriate arbitrary vectors and then show that the result has the desired property. For example, to show positivity simply calculate $\langle (x_1, x_2),(x_1, x_2) \rangle $ and show that the resulting expression must always be positive. Similar arguments should work for the rest.

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Concretely, here is the argument for the first two parts. Positivity and definiteness concern an inner product of a vector with itself; so we take an arbitrary vector $x = (x_1, x_2)$ and compute its inner product with itself. $$ \langle (x_1, x_2), (x_1, x_2) \rangle = \begin{bmatrix}x_1 & x_2\end{bmatrix} \begin{bmatrix}2&-1\\1&1\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}2x_1 + x_2 & -x_1 + x_2\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} $$ $$ = 2x_1^2 + x_1 x_2 - x_1 x_2 + x_2^2 = 2x_1^2 + x_2^2 $$

Since any real number squared is non-negative, we can see that the inner product of a vector with itself is never negative, proving positivity. Additionally, we know that if the sum of a sequence of non-negative terms is zero, then each term must be zero. From our expression for the inner product, this implies that if $\langle v, v \rangle = 0$ then $v = 0$. Plugging in $x_1 = 0, x_2 = 0$ shows that the converse is also true, so this proves definiteness.

For linearity, compute $\langle (\alpha x_1, \alpha x_2), (y_1, y_2) \rangle$ and $\alpha \langle (x_1, x_2), (y_1, y_2) \rangle$; you will find that the results are the same. Then compare $\langle (x_1 + x_3, x_2 + x_4), (y_1, y_2) \rangle$ and $\langle (x_1, x_2), (y_1, y_2) \rangle + \langle (x_3, x_4), (y_1, y_2) \rangle$; the results will be the same.

For conjugate symmetry (or in this case, symmetry) compute $\langle (x_1, x_2), (y_1, y_2) \rangle$ and $\langle (y_1, y_2), (x_1, x_2) \rangle$; again you should find that the results are the same.