# simplifying complex inner product by factoring out complex constants

From "Linear Algebra Demystified", David McMahon, 2006, problem 2, page 132 and page 235:

$$<\underline{u},\underline{v}> = 2i$$ $$<\underline{u},\underline{w}> = 1 + 9i$$

$$y = (2\ <3i\underline{u},\ \ \underline{v}>\ ) - <\underline{u},\ \ i \underline{w}>$$

Find value of y:

$$y = (2(-3i)<\underline{u},\ \ \underline{v}>\ )\ - (i<\underline{u},\ \ \underline{w}>) \\$$

$$y = ((-6i)<\underline{u},\ \ \underline{v}>\ )\ - (i<\underline{u},\ \ \underline{w}>) \\$$

$$y = ((-6i)(2i) - (i)(1+9i)$$

$$y = -12i^2 - i-9i^2)$$

$$y = -21i^2 - i$$

$$y = 21 - i$$

However, book says the answer is:

$$-3 -i$$

How did they get that? are they wrong or right? just curious because complex inner product has some strange factoring rules for pulling out constants, being first argument anti-linear, and second argument linear.

If i understand correctly, when pulling the complex constant out of the first argument of dot product then i need to conjugate the imaginary term, if i pull the complex constant out of the second argument of dot product, then i don't need to conjugate it.

and as usual: $$i^2 = \sqrt{-1} \sqrt{-1} = -1$$

Actually, i'm a little bit confused about this part. The book say that complex vectors have the following inner space properties:

$$ = a^* <\underline{u}, \underline{w}>$$

$$<\underline{u}, b\underline{w}> = b <\underline{u}, \underline{w}>$$

And they had a proof in the book that proves it this way...but when i look on the internet, they say its this, instead:

$$ = a <\underline{u}, \underline{w}>$$

$$<\underline{u}, b\underline{w}> = b^* <\underline{u}, \underline{w}>$$

(my opinion, is that mathworld wrote it wrong then wikipedia copied it, and then everybody else believed them)

• mathworld.wolfram.com/HermitianInnerProduct.html – DiscreteMath Feb 26 at 4:04
• en.wikipedia.org/wiki/Hilbert_space – DiscreteMath Feb 26 at 4:09
• turns out that on 2/26/2019, both mathword and wikipedia are printing the equations wrong. I complained to mathworld using their complaint form. – DiscreteMath Feb 26 at 12:48
• When pulling out a scalar, whether the scalar in the first argument or the second one should be conjugated is just a convention. Some choose the first one and some choose the second one. As long as one is consistent (in your case, the book author is not), this shouldn't be a problem. – user1551 Feb 26 at 13:34

$$2<3iu,v>-=-6i\cdot2i-i(1+9i)=12-i+9=21-i,$$ which says that the book answer is wrong and you are right.
• but don't you need to apply complex-conjugate to remove complex constant from LHS of dot product? $2<3iu,v>=2(-3i)<u,v> = -6i<u,v>= (-6i)(2i)=-12i^2=12$ – DiscreteMath Feb 26 at 3:57
• @DiscreteMath Yes, I think you are right. I fixed. Maybe in the given $<u,v>=-2i$? – Michael Rozenberg Feb 26 at 6:03