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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?

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  • $\begingroup$ Could you format your question with $\LaTeX$ to make it more clear what you're saying? $\endgroup$
    – smackcrane
    Commented Nov 5, 2011 at 19:57

2 Answers 2

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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.

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  • $\begingroup$ In fact more is true: Positive definite matrix [Wikipedia] $\endgroup$
    – kahen
    Commented Nov 5, 2011 at 20:05
  • $\begingroup$ @matteo: How many properties do I need to satisfy? Can you list them? $\endgroup$ Commented Nov 5, 2011 at 20:15
  • $\begingroup$ Thanks @kahen! I added your input to my answer. $\endgroup$
    – matteo
    Commented Nov 5, 2011 at 20:23
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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 $\langle u, v \rangle = \langle v, u \rangle$.


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.


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.

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  • $\begingroup$ This makes sense, but when I try it it does not. Can you show the other arguments (organized) so I can see it clearly? Thanks $\endgroup$ Commented Nov 5, 2011 at 21:03
  • $\begingroup$ For the linearity: <α(u+v),(u+v)> = [(αu + αv u+v][ 2 -1 $\endgroup$ Commented Nov 6, 2011 at 4:37
  • $\begingroup$ and then I get: [2αu + αv - u + v]* column vector of [x1 x2] = 2αux1 + αvx1 -ux2 -vx2 = α(2ux1 + x1) - x2(u-v) $\endgroup$ Commented Nov 6, 2011 at 4:39
  • $\begingroup$ I don't see how the conjugate symmetry argument follows in this example $\endgroup$ Commented Nov 6, 2011 at 4:39

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