Inner products

Algebra Level 4

Which of the following are inner products ?

I. On R 2 , {\mathbb R}^2, with vectors x , y x,y written as 2 × 1 2\times 1 column vectors, define ⟨ x , y ⟩ = x T A y , \langle x,y \rangle = x^T A y, where A = ( 0 − 1 2 1 ) . A = \begin{pmatrix} 0&-1 \\ 2&1 \end{pmatrix}.

II. Same as in I, but A = ( 1 2 2 1 ) . A = \begin{pmatrix} 1&2 \\ 2&1 \end{pmatrix}.

III. The Minkowski product : on R 4 , {\mathbb R}^4, with vectors v = ( x 1 , y 1 , z 1 , t 1 ) {\bf v} = (x_1,y_1,z_1,t_1) and w = ( x 2 , y 2 , z 2 , t 2 ) , {\bf w} = (x_2,y_2,z_2,t_2), write ⟨ v , w ⟩ = x 1 x 2 + y 1 y 2 + z 1 z 2 − t 1 t 2 . \langle {\bf v},{\bf w} \rangle = x_1x_2+y_1y_2+z_1z_2-t_1t_2.

I only I and II I and III II only II and III None of them

Patrick Corn by Patrick Corn , 44, USA

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2 solutions

Otto Bretscher
May 25, 2016

None of them is positive definite. Consider, in this order, the vectors, e 1 , e 1 − e 2 e_1, e_1-e_2 and e 4 e_4 .

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but in first sentence det A = 2. why this sentence is false?

Gabriel Cabral - 3 years ago

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the second messed me up too. you need to calculate the eigenvalues and you'll see that you can't have <x,x> >= 0 forall x

Giuseppe C - 6 months, 3 weeks ago

In the first case let us look at the expanded form of x T A y \displaystyle x^{T}Ay for column vector ( x 1 x 2 ) \displaystyle \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} and column vector ( y 1 y 2 ) \displaystyle \begin{pmatrix} y_{1}\\ y_{2} \end{pmatrix} both belonging to R 2 \mathbb{R^{2}}

We get − x 1 y 2 + 2 x 2 y 1 + y 2 x 2 \displaystyle -x_{1}y_{2}+2x_{2}y_{1} + y_{2}x_{2}

Take the vector x x in R 2 \mathbb{R^{2}} to be ( 1 0 ) \displaystyle \begin{pmatrix} 1 \\ 0 \end{pmatrix}

Then ⟨ x , x ⟩ = 0 \langle x,x\rangle=0 but x x is not the null vector which contradicts the positivity criteria for inner products.

For the 2nd case we can simply examine the eigen values of A A to determine the positive definiteness of ⟨ x , x ⟩ \langle x,x\rangle . It has eigen values − 1 -1 and 3 3 . Hence it is indefinite and hence it again violates the positivity criteria.

For the third case just consider the vector x x to be ( 1 , 0 , 0 , 1 ) (1,0,0,1) in R 4 \mathbb{R^{4}} . Then again ⟨ x , x ⟩ = 0 \langle x,x\rangle =0 . But x x is not the null vector. Hence all three of them cannot be an inner product.

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