## Adjoints: A Detailed Example of the Conjugate Transpose Property.

### July 24, 2010

My next post is going to be this nice proof of the fact that if we have $M(T)$ for some linear map $T: V\rightarrow V$ for $V$ is a nontrivial finite vector space, then $M(T^{\ast})$ is really easy to find: it’s simply the conjugate transpose of $M(T)$.  This is exactly what it sounds like: we find the transpose of $M(T)$ (by switching $a_{i,j}$ with $a_{j,i}$ for all $i$ and $j$) and then replacing every element in the matrix with its conjugate.  I guess the notation should be $(M(\overline{T}))^{t} = M(T^{\ast})$.  Which is pretty complicated looking, but it’s really not hard to do at all.  In fact, the point of this post is to give a detailed example.  Actually, let’s do two examples, and you’re going to find that half of the time this is much easier to do than what I’ve said above.

## Example 1: A Real Space.

Let’s let our vector space be ${\mathbb R}^{3}$ with the standard basis, and let’s let our linear map be $T:{\mathbb R}^{3}\rightarrow {\mathbb R}^{3}$ defined by $T((x,y,z)) = (2x + y, -y, x + z)$.  So, what’s the matrix for this?  Well, let’s write out where the basis elements go.

$T((1,0,0)) = (2, 0, 1)$

$T((0,1,0)) = (1, -1, 0)$

$T((0,0,1)) = (0, 0, 1)$

and so our matrix is

$M(T) = \left(\begin{array}{ccc} 2 & 1 & 0 \\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{array}\right)$

which is a pretty cute matrix, all things considered.  Now, let’s do this the hard way: what is $T^{\ast}$?  Let’s figure this crap out.  Fix $(a,b,c)\in {\mathbb R}^{3}$, and then

$\langle T((x,y,z)), (a,b,c)\rangle = \langle (2x + y, -y, x + z), (a,b,c)\rangle$

$= (2x + y)a + (-y)b + (x + z)c = 2xa + ya - yb + xc + zc$

$= x(2a + c) + y(a-b) + zc = \langle (x,y,z), (2a+c, a-b, c)\rangle$

$= \langle (x,y,z), T^{\ast}((a,b,c))\rangle$.

So we have to define our $T^{\ast}(x,y,z) = (2x + z, x - y, z)$.

Let’s see if this jives with our crazy conjugate transpose idea.

What was our matrix again?  Oh, right.

$M(T) = \left(\begin{array}{ccc} 2 & 1 & 0 \\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{array}\right)$

Now what’s the conjugate transpose?

$M(T^{\ast}) = \left(\begin{array}{ccc} 2 & 0 & 1 \\ 1 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right)$

So what should we have our transpose do?  Let’s tease this out.  This matrix implies:

$T^{\ast}((1,0,0)) = (2, 1, 0)$

$T^{\ast}((0,1,0)) = (0,-1,0)$

$T^{\ast}((0,0,1)) = (1,0,1)$

which means that, in general,

$T^{\ast}((x,y,z)) = (2x + z, x-y, x + z)$

and, what!  Look at that!  It’s the same thing we got from before!  Is it magic?  Sort of.  But note that we didn’t really do any conjugating in this example, since, you know.  Real stuff doesn’t change when you conjugate it.

Maybe we should do an example where you need to conjugate things then.  Complex stuff.  Yeah.  Good idea.

## Example 2: Complex Stuff.

I just want to note, before doing this, that it was extremely difficult finding an actual example of this being done the “long way.”  In fact, I found no examples of this being done the long way, so I needed to kind of “invent” my own way of doing this.  Therefore, as a caveat, note that my method may differ from those which you have seen before.  But it may not.  I don’t know.

We will be using the vector space ${\mathbb C}^{2}$ and the linear map $T:{\mathbb C}^{2}\rightarrow {\mathbb C}^{2}$ defined by $T((z_{1}, z_{2})) = (iz_{2}, z_{1} + z_{2})$.  Okay?  Okay.

First, because this was really weird to do, let’s talk about $M(T)$.  Our standard basis for the complex vector space ${\mathbb C}^{2}$ is the same as the standard basis for ${\mathbb R}^{4}$; explicitly, it is $\{(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)\}$.

Look at those basis elements for a second.  What do they correspond to?  Well, if we consider a point $(z_{1}, z_{2})$ then $(1,0,0,0)$ corresponds to the real part of $z_{1}$ and, similarly, $(0,1,0,0)$ corresponds to the imaginary part of $z_{1}$.  The other two correspond to the real and imaginary parts, respectively, of $z_{2}$.  So, let’s start doing that thing with the basis elements acted on by the linear map so that we can make $M(T)$.  Note that if we have $(z_{1}, z_{2})$, I will call the spot occupied by $z_{1}$ the “first complex coordinate” and I’ll call the spot occupied by $z_{2}$ the “second complex coordinate.”  Notice that in the first complex coordinate, there is the “real part” and the “imaginary part,” and similarly for the second complex coordinate.

$T((1,0,0,0)) = (0,0,1,0)$

Note that this says, “where does the real part of $z_{1}$ go?  It goes to the second complex coordinate in the real spot, as it is, itself, real.”

$T((0,1,0,0)) = (0,0,0,1)$

This says, “where does the imaginary part of $z_{1}$ go?  It goes to the second complex coordinate in the imaginary spot, as it is, itself, imaginary.”

$T((0,0,1,0)) = (0,1,1,0)$

Here it gets a little tricky.  Where does the real part of the second complex coordinate go?  Well, the second complex coordinate goes into the first complex coordinate, but it is multiplied by $i$ and so it becomes imaginary.  Therefore, we put the real part of the second complex coordinate into the imaginary part of the first complex coordinate.  In addition to this, it also goes to the second complex coordinate in the real part, since we’re not multiplying it by $i$ in the second coplex coordinate.

$T((0,0,0,1)) = (-1,0,0,1)$

This is the last part.  Where does the imaginary part of the second complex coordinate go?  Well, it goes into the first complex coordinate, but since we must multiply it by $i$ it become real but we must multiply it by $-1$ since we have that $i^{2} = -1$.  Similarly, it must go into the second complex coordinate, in the imaginary spot.

If you’ve followed all that, congratulations.  It’s not all that easy, but after doing it for a bit, it gets really intuitive.  Anyhow, this means that

$M(T) = \left(\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1\end{array}\right)$

which is a nice compact way to look at this.

Now, we do this the long way.  This way becomes really irritating really fast, but in this case it isn’t too bad.  Let’s do it.  Fix $(w_{1},w_{2})\in {\mathbb C}^{2}$.  Also recall that complex conjugation is defined by $\langle (z_{1},z_{2}),(w_{1},w_{2})\rangle = z_{1}\overline{w_{1}} + z_{2}\overline{w_{2}}$.

$\langle T((z_{1}, z_{2})), (w_{1},w_{2})\rangle = \langle (iz_{2}, z_{1} + z_{2}), (w_{1},w_{2})\rangle$

$= iz_{2}\overline{w_{1}} + z_{1}\overline{w_{2}} + z_{2}\overline{w_{2}}$

$= z_{1}(\overline{w_{2})} + z_{2}(\overline{w_{1}}i + \overline{w_{2}})$

and since we have

$= \langle (z_{1}, z_{2}), T^{\ast}(w_{1},w_{2})\rangle$

this implies that

$T^{\ast}(w_{1},w_{2}) = (\overline{\overline{w_{2}}}, \overline{\overline{w_{1}}i + \overline{w_{2}}})$

$= (w_{2}, -w_{1}i + w_{2})$

Whew.

(Quick Note!) You might be a little stumped about the last part of the second-to-last line and how we got rid of that big nasty bar over all those other bars.  Note that it follows easily if we quickly prove these two quick lemmas:

Quick Lemma 1: If we have $z, w\in {\mathbb C}$, then $\overline{z + w} = \overline{z} + \overline{w}$.

Quick Proof. Write $z = a + bi$ and $w = c + di$ with $a,b,c,d\in {\mathbb R}$ as we usu’ do.  Then we have

$\overline{z + w} = \overline{(a + c) + i(b + d)}$

$= (a + c) - i(b + d) = (a - ib) + (c - id)$

$= \overline{z} + \overline{w}$$\Box$

Quick Lemma 2:  If we have $z, w\in {\mathbb C}$, then $(\overline{zw}) = (\overline{z})(\overline{w})$.

Quick Proof. Write $z$ and $w$ as above.  Then we have

$(\overline{zw}) = \overline{(a + bi)(c + di)}$

$\overline{(ac - bd) + (ad + bc)i} = (ac - bd) - (ad + bc)i$

$= (a - bi)(c-di) = (\overline{z})(\overline{w})$$\Box$

(End of Quick Note!)

Okay, so, let’s set up the matrix for $T^{\ast}$ above.  We do the exact same thing as we did for the one above.  Exact same thing.  We find that we get the matrix (and do this yourself, it’s very magical!):

$M(T^{\ast}) = \left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ -1 & 0 & 0 & 1 \end{array}\right)$

Now, look at the original matrix and compute the transpose conjugate (and, since we never used any complex variables in the matrix, this is just the transpose), and it should match up to the one we just got!  Magical.

Let’s just note here that the longer process was tedious even for ${\mathbb C}^{2}$ which is relatively “small” compared to some other spaces.  I’m sure there’s a better way to do “the long way” of finding the adjoint, and when I find it, I’ll tell you.  In the meantime, use the conjugate transpose theorem!

Last, note that this entire time we’ve been using orthonormal bases (quietly, but we’ve been doing it) and, in fact, this conjugate transpose theorem (which we will prove next post) only works when we have an orthonormal bases.  So don’t just think we can apply this any-old-which-way.  I want to emphasize, again, that the complex example is probably not done in the most efficient way, but, again, I could not find an example of it done elsewhere.  If you find one, please comment and let me know!