# Complex inner product space.

#### hsazerty2

Joined Sep 25, 2015
28
Hi there, I encountered the above problem in Schaum’s Outlines of Linear Algebra 6th Ed (2017, McGraw-Hill) Chapter 7 - Inner Product Spaces, Orthogonality.
Using some particular values for u and v, I proved that a and d must be real positive, and b is the conjugate of c. The solution indicates that a.d-b.c must also be positive, but i can't figure that out.

#### Papabravo

Joined Feb 24, 2006
18,806
Hi there, I encountered the above problem in Schaum’s Outlines of Linear Algebra 6th Ed (2017, McGraw-Hill) Chapter 7 - Inner Product Spaces, Orthogonality.
Using some particular values for u and v, I proved that a and d must be real positive, and b is the conjugate of c. The solution indicates that a.d-b.c must also be positive, but i can't figure that out.

You can't prove something by using particular values. That is not the way it works. There is no particular reason why should be able to assert that a and d must be real, when it is given that they are members of ℂ. You use the definition to show that all of the inner product axioms are satisfied. Those axioms are Conjugate Symmetry, Linearity, and Positive Definite.

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#### bogosort

Joined Sep 24, 2011
694
Hi there, I encountered the above problem in Schaum’s Outlines of Linear Algebra 6th Ed (2017, McGraw-Hill) Chapter 7 - Inner Product Spaces, Orthogonality.
Using some particular values for u and v, I proved that a and d must be real positive, and b is the conjugate of c. The solution indicates that a.d-b.c must also be positive, but i can't figure that out.

Hi, good job on finding that $$a$$ and $$d$$ must be positive, and that $$b$$ and $$c$$ must be conjugate pairs. This follows from the condition that $$f$$ is positive-definite, i.e., $f(u, u) > 0$ for any non-zero $$u$$. That is, $az_1 \bar{z_1} + bz_1 \bar{z_2} + cz_2 \bar{z_1} + dz_2 \bar{z_2} > 0$ By definition, the conjugate pairs are positive definite, e.g., $z_1 \bar{z_1} = (x+iy)(x-iy) = x^2 + y^2 = |z_1|^2$ which is always a non-negative real number. Consequently, $$a$$ and $$d$$ must be real and positive.

Next, we require that $bz_1 \bar{z_2} + cz_2 \bar{z_1} \in \mathbb{R}$ Otherwise, the sum of a non-real complex number with a real number is complex and cannot be ordered with 0. Fortunately, we can guarantee that this sum is real by forcing $$b$$ and $$c$$ to be complex conjugates. The most straightforward way I know how to prove this is using polar notation: \begin{align} z \in \mathbb{C} &= re^{i \theta} = r(\cos \theta + i \sin \theta) \\ \bar{z} \in \mathbb{C} &= re^{i \theta} = r(\cos \theta - i \sin \theta) \end{align} where $$r, \theta \in \mathbb{R}$$.

Let \begin{align} z_1 &= r_1 e^{i \theta_1} \\ z_2 &= r_2 e^{i \theta_2} \\ b &= r_3 e^{i \theta_3} \end{align} Then, \begin{align} bz_1 \bar{z_2} + cz_2 \bar{z_1} &= r_3 r_1 r_2 e^{i(\theta_ + \theta_1 - \theta_2)} + r_3 r_2 r_1 e^{(-\theta_3 + \theta_2 - \theta_1)} \\ &= k \left[ e^{i(\theta_1 - \theta_2 + \theta_3)} + e^{-i(\theta_1 - \theta_2 + \theta_3)} \right] \\ &= k \left[ \cos(\theta_1 - \theta _2 + \theta_3) + i \sin(\theta_1 - \theta_2 + \theta_3) + \\ \cos(\theta_1 - \theta _2 + \theta_3) - i \sin(\theta_1 - \theta_2 + \theta_3) \right] \\ &= 2k \cos(\theta_1 - \theta_2 + \theta_3) \end{align} which is, of course, a real number.

As for why $$ad - bc > 0$$ is a condition of the solution, it may help to know about positive-definite matrices. Given any non-zero vector $$v$$ over ℂ, a square matrix $$M$$ over ℂ is positive-definite iff $v^* M v > 0$ where $$v^*$$ indicates the conjugate transpose of $$v$$.

Such matrices are necessarily Hermitian (equal to their own conjugate transpose) and are useful because they don't change the sign of the vectors they act upon. Now, consider how we might construct the sum in $f(u,u) = az_1 \bar{z_1} + bz_1 \bar{z_2} + cz_2 \bar{z_1} + dz_2 \bar{z_2}$ Notice that this looks like a linear equation with coefficients a,b,c,d. Suppose that we define a matrix of coefficients $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and a row vector $u = \begin{bmatrix} z_1 & z_2 \end{bmatrix}$ Then, the vector's conjugate transpose $$u^*$$ is a column vector of complex conjugates. Multiplying this by the matrix gives us $Mu^* = \begin{bmatrix} a \bar{z_1} + b \bar{z_2} \\ c \bar{z_1} + d \bar{z_2} \end{bmatrix}$ Finally, multiplying our row vector by this quantity gives us the desired sum: $uMu^* = az_1 \bar{z_1} + bz_1 \bar{z_2} + cz_2 \bar{z_1} + dz_2 \bar{z_2}$ Therefore, our inner product $$f$$ is positive-definite iff $uMu^* > 0$ Notice that this has the same form of the definition of a positive-definite matrix. One of the properties of a positive-definite matrix is that it's determinant is always positive, which is an intuitive result when we think about how a determinant of $$-1$$ characterizes a reflection (such a matrix takes vectors to their opposite signs).

Of course, the determinant of $$M$$ is precisely $$ad - bc$$, and so the condition $ad - bc > 0$ tells us that the matrix of coefficients must be positive definite. And since $$M$$ is positive-definite, then $$f$$ must be positive-definite, and so the assertion is proved.

Note that, by the definition of Hermitian matrices, this proof automatically implies that $$a$$ and $$d$$ are real, and that $$b$$ and $$c$$ are complex-conjugates. So, the simplest strategy (I think) would have been to start with a,b,c,d as a matrix of coefficients and then show that it must be Hermitian and positive-definite.

Hope that helped.

Big thanks for your detailled answer. 