If you are working with complex numbers in linear algebra then you will definitely need to know about
something called the her mission transpose which is also sometimes called the conjugate transpose or
the complex conjugate transpose.
And to understand the her mission transpose you will have to know about an operation that applies to
complex numbers which is called the conjugate or the complex conjugate.
So in this video I'm going to show you how to compute the conjugate of a complex number.
I'll show you how to compute the commission transpose and I will briefly explain why her mission transpose
is necessary for vectors that contain complex numbers the complex conjugate is a really simple operation
to apply.
All you have to do is flip the sign of the imaginary part of the complex number without changing the
real part of the complex number.
So that's the algebraic perspective on complex conjugate.
So this complex number turns into this complex conjugate and this complex number if there already is
a negative sign in the imaginary component turns into this complex number where now the imaginary component
is positive again because the idea is not to just make sure the imaginary component is negative but
instead to flip the sign.
You can also see here that the real part doesn't change.
So this is plus a and this is plus a.
And notice also that with the complex conjugate you're not changing the magnitude of the imaginary part
you're only changing its sine.
There is also a geometric interpretation of the complex conjugate.
If you plot a complex number on the complex plane then the complex conjugate of this number represented
as this coordinate becomes this number or this coordinate down here.
So essentially you are flipping the number the complex number across the real axis.
So the real component doesn't change and the magnitude of the imaginary component doesn't change only
the sign of the imaginary component changes which means flipping the complex number across the real
axis notice also that if the complex number is sitting on the real number line then the complex conjugate
actually does nothing whatsoever.
You can also see that algebraic Lee because if the imaginary part is 0 then plus 0 is the same thing
as minus zero.
Okay so that's the complex conjugate now for the conjugate transpose which is also sometimes called
the her mission transpose the her mission transpose is indicated using either an h a capital H or an
asterisk in place of the T for the normal transpose.
So the her mission transpose is just like the regular transpose but any complex numbers get their imaginary
components flipped in sine.
So here you see a column vector of four dimensional column vector.
It contains complex numbers of these elements have real parts and imaginary parts.
You can pause the video and compute the her mission transpose of this vector.
Here it is.
You can see that these signs of the imaginary components are flipped and nothing happens to the real
part of the vector elements.
So this is how you compute the her mission transpose.
The question is why do we need such a thing.
Why can't you just use the regular transpose.
And why is the mission transpose important in linear algebra and in geometry.
To understand why we need the permission transpose.
Let's consider an example of this vector in the complex plane.
So this is the real axis.
This is the imaginary axis.
And here is the coordinate represented by this complex number no Z 3 and 4 AI.
So here's a question for you.
What is the length of this line.
What is the distance of this point away from the origin of the complex plane.
If you remember your Pythagorean triplets then you will recognize this as a three four five right triangle.
So this is three units over on the real axis.
Four units up on the imaginary axis so therefore the partners must be five units away from the origin.
OK so we know that this has to be the right answer.
Now we should be able to confirm this algebraic CLI by computing the magnitude of this vector.
Now in this case I'm going to conceptualize this complex number as just a single number.
So there'll be a vector that has only one element.
So the length of this line squared is this complex number transposed.
And then times itself which of course is this number of squared.
Okay so here you see all the math.
So we take 3 for AI as the complex numbers Z.
Multiply it or transpose it and multiply it by itself.
And then we just have to work through all the multiplication.
So we get three times three is nine three times for AI is 12 VI for AI times three is also twelve AI
and then four AI times for AI is sixteen AI squared.
So I squared is minus one.
So this turns into minus sixteen.
And then we can group these terms and discover that the length of this line is apparently minus seven
plus twenty four AI.
So I think that's not really a sensible length squared.
Now let's try this again using the her mission and transpose.
So I'm going to repeat this operation but first flip the sign of the imaginary component.
And actually I'm here flipping the sign of the second one was I wrote that I'm computing her mission
transpose of the first one but in this special case that doesn't matter because these are both the same
number.
Okay so here we get three times three is nine three times minus four AI is minus 12 AI and then this
for AI times three is plus twelve AI and then we get for AI times minus for AI and that gives us minus
sixteen AI squared and now these two terms cancel this again becomes plus sixteen because the AIS squared
is minus one and ultimately that gives us twenty five which is indeed the length of this line squared.
Now I've computed this here using or treating this complex number as a one element vector you can also
repeat the procedure by thinking about this as a two element vector where the first element is the real
part and the second element is the imaginary part I encourage you to work through the math and compute
again the length of this line by considering this to be a two element vector and you will discover the
same concept which is that the regular transpose gives a nonsensical and obviously geometrically incorrect
answer whereas computing the dot of this vector with itself replacing the normal transpose with the
commission transpose gives you the correct answer.
Now I'm going to switch to matlab and show you how to implement these two forms of the transpose so
here I'm creating a complex number amusingly function complex.
The real part is the first input and the imaginary part is the second input.
And notice here that I'm just inputting the magnitude of the imaginary number.
I don't need to include the I hear the explicitly square root of minus one.
Here use the norm function to compute the magnitude of this complex number.
So that's 5 that basically just confirms what we already discovered in the slides.
And what you should also know from the Pythagorean triplets the three four five triplets.
OK.
So there's a function transpose and let's see what that does so this function transpose preserved the
sign of this complex numbers to the imaginary part is still positive.
Okay.
And let's see what happens when we multiply said transposed by z.
Here we get some weird number that's obviously not what we expect the magnitude the length of this complex
number to be.
So now it turns out that the mission transpose is actually given by a single apostrophe and that's the
symbol the single apostrophe.
That's the character that you've been using so far to compute what you thought was the regular transpose.
In fact you were secretly computing the her mission transposed but normally so if you have just real
value vectors or real valued matrices the mission transpose is the same thing as the regular say non
termination transpose because there are no imaginary components to the real value numbers.
Okay.
But now I can show you
Z transpose with an apostrophe gives us the complex number three minus four eyes.
So the imaginary part has been flipped and now multiplying out.
This operation gives us twenty five.
And again the square root of that is going to be the norm of the magnitude of that vector.
Now if you put a dot a period before the transpose then Matlab is going to interpret that as computing
the regular transpose.
So not the mission transposed so this is still the wrong answer.
Okay.
Here again I'm just showing you basically the same thing that I illustrated here.
But in a slightly larger vector.
So this is going to be a one two three four element or four dimensional vector.
It's gonna be a row vector.
So here's the first element is three and this has zero.
I because the entire vector is complex because some of the elements are complex but this particular
number.
So I'd only specify a real value.
No Matlab converted this into a complex number and it's a complex number that has zero imaginary part.
Okay.
So then basically with the apostrophe that's going to flip the signs of the imaginary components so
you can see this.
So the imaginary components have sine positive positive positive negative.
And here this is going backwards positive negative negative.
Now this might seem incorrect because the sign here is positive and the sign here is also positive.
But this is zero.
So positive zero is the same thing as negative zero.
Okay.
And then with a dot asterisk that is not going to flip the signs and with the function transpose that
will also not flip the signs in this video.
I told you about the complex conjugate I showed you how to compute the her mission transpose and I used
the concept of length in geometry and linear algebra to explain why the Romanian transpose is necessary
for computing the dot product of complex value vectors.
And finally I showed you how to implement both the her mission and the regular transpose in Matlab.