# Thread:

1. ## T is a projection and normal, then T is orthogonal projection

Question: Let $T$ be a normal operator on a finite-dimensional inner product space. Prove that if $T$ is a projection, then $T$ is also an orthogonal projection.

Attempt: Since $T$ is normal, $TT^* = T^*T$ and since $T$ is a projection, then $T=T^2$.

Since $T$ is an orthogonal projection if and only if $T^2 = T = T^*$, we already have the left equality by that fact that $T$ is a projection. We only need to show that $T=T^*$ and we'll be done.

Somehow, I figure that I can do this by showing that $<T(x),y>=<x,T(y)>$ but how?

2. Since $T$ is normal, it is unitarily equivalent to a diagonal matrix. Since $T=T^2$, its eigenvalues can be only $\pm 1$. Thus $T=UDU^*$ and $T^*=U^{**}D^*U^*=UDU^*=T$.
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "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": 0.9511876106262207, "perplexity_flag": "head"}