similarity transform mapping diagonal matrix of complex conjugates, to real matrix

Question

I know there is some unitary matrix $T$ that maps a diagonal matrix $A = \begin{bmatrix}a+bj & 0 \cr 0 & a-bj \end{bmatrix}$ to a real matrix, namely

$$T^{-1}AT = \begin{bmatrix}a & b \cr -b & a \end{bmatrix}$$

but I am having trouble figuring how to construct $T$. At first I thought it was just a Givens rotation, that can't be right because Givens rotations have real entries and that wouldn't eliminate the non-real components.

How can I compute $T$?

How do you know it is unitary if the matrix it diagonalizes is not symmetric? (e.g. I know that it is true for real, symmetric matrices: https://math.stackexchange.com/questions/82467/eigenvectors-of-real-symmetric-matrices-are-orthogonal) — makansij, Sep 01 '18 at 19:37
I'm sorry, I can't answer your question. I don't remember, it was two-and-a-half years ago. — Jason S, Sep 02 '18 at 02:33

Ben Grossmann · Accepted Answer · 2016-01-03T19:52:40.337

2

Let $B$ denote $\pmatrix{a&b\\-b&a}$. Note that $T^{-1} = T^*$ diagonalizes $B$, so its columns are eigenvectors of $B$.

In particular, calculate the first eigenvector $v$ using $$ \pmatrix{-jb & b\\-b & -jb} v = 0 $$ And see that $v = (1,j)$ (which we normalize to $(1/\sqrt 2)(1,j)$) works.

Similarly, find $v = (1,-j)$ for the other eigenvector.

So, we can take $$ T^* = \frac{1}{\sqrt 2}\pmatrix{1&1\\j&-j} $$ Or, multiplying the second column by $j$, we get the slightly neater looking answer $$ T^* = \frac{1}{\sqrt 2}\pmatrix{1&j\\j&1} \implies T = \frac{1}{\sqrt 2}\pmatrix{1&-j\\-j&1} $$ You'll notice that this choice of matrix does not depend on $a$ or $b$.

edited Jan 03 '16 at 19:52

answered Jan 03 '16 at 19:42

Ben Grossmann

225,327

Oh. Huh, that seems too easy. Is there a way to express it directly in terms of $a$ and $b$? I figure it would help to express them in polar form as $a=r \cos \theta$ and $b = r \sin \theta$. – Jason S Jan 03 '16 at 19:44
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Here's a tricky bit: the same unitary $T$ can be used independently of the values of $a$ and $b$ – Ben Grossmann Jan 03 '16 at 19:45
really?! weird. – Jason S Jan 03 '16 at 19:45
See my latest edit – Ben Grossmann Jan 03 '16 at 19:54
excellent -- thanks! I would have thought discussions of diagonalization and block diagonalization and similarity transforms would include this, but I didn't see any mention of it. – Jason S Jan 03 '16 at 19:54
Which textbook are you using? – Ben Grossmann Jan 03 '16 at 19:57
well, I've been searching the web, but I have Strang's Intro to Linear Algebra. – Jason S Jan 03 '16 at 20:03
Interesting that this isn't mentioned there, then. I believe that I saw this in Lay's text, but I don't remember at this point. It should be mentioned in any basis lin-alg/diff eq course/textbook. – Ben Grossmann Jan 03 '16 at 20:11
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where does the matrix in your sentence "In particular, calculate the first eigenvector" come from? Is that $A-\lambda I$ for $\lambda = a+bj$? – Jason S Jan 03 '16 at 20:19
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Yes, exactly. ${}{}{}$ – Ben Grossmann Jan 03 '16 at 20:20

similarity transform mapping diagonal matrix of complex conjugates, to real matrix

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