# Show the matrix product $U = (I+T+iS)(I-T-iS)^{-1}$ is unitary

$S$ is real symmetric. $T$ is real skew-symmetric. I have shown that $T\pm iS$ is skew-Hermitian. I am further asked to show that $U = (I+T+iS)(I-T-iS)^{-1}$ is unitary.

Denoting by $^\dagger$ the conjugate transpose, I have that
$U^\dagger = [(I+T+iS)(I-T-iS)^{-1}]^\dagger$
$= ((I-T-iS)^\dagger)^{-1}(I+T+iS)^\dagger$
$= (I+T+iS)^{-1}(I-T-iS)$

But then the products $UU^\dagger$ and $U^\dagger U$ have the factors in the wrong order to cancel out to make identity. I must be missing something obvious here; I'd appreciate a steer!

-

For any $A$, $(I+A)$ commutes with $(I-A)$, and hence with $(I-A)^{-1}$ if it exists. – Chris Eagle Jan 5 '11 at 19:25
Thanks Chris, that makes it easy. I don't think I've seen that simple (but useful-looking) result until now! I have proved it using index notation; is there a nicer way? – user5426 Jan 5 '11 at 19:50
If $A$ commutes with $B$, then $AB=BA$. If further $B$ is invertible, then $B^{-1}(AB)B^{-1}=B^{-1}(BA)B^{-1}$, and hence $B^{-1}A=AB^{-1}$. – Chris Eagle Jan 5 '11 at 19:58
Sorted, thanks very much indeed. – user5426 Jan 6 '11 at 1:09
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9416186809539795, "perplexity_flag": "head"}