Prove $I-YY^\dagger$ is a projection operator for $N(Y)$

Question

Let $Y\in \mathbb{R}^{m*m}$ be a symmetric matrix and how to prove $I-YY^\dagger$ is a projection operator for $N(Y)$? I know $I-Y^\dagger Y$ is a projection operator for $N(Y)$.

Thanks in advance!

Does $N(Y) = \ker Y = { y \mid Yy = 0}$? – Robert Lewis Feb 01 '19 at 08:18 — Robert Lewis, Feb 01 '19 at 08:18

score 0 · Answer 1 · answered Feb 01 '19 at 08:55

0

How can you rewrite $I-YY^\dagger$ if you know Y is symmetric and you already know that $I-Y^\dagger Y$ is a projection operator?

answered Feb 01 '19 at 08:55

GenericUsrnme

395

score 0 · Answer 2 · answered Feb 02 '19 at 05:34

0

Since $Y$ is real symmetric,

$Y^\dagger = Y^T = Y; \tag 1$

thus

$Y^\dagger Y = YY^\dagger = Y^2, \tag 2$

whence

$I - Y^\dagger Y = I - YY^\dagger = I - Y^2; \tag 3$

it therefore follows that $I - YY^\dagger$ is a projection onto $N(Y)$ if and only if $I - Y^\dagger Y$ is . . .

answered Feb 02 '19 at 05:34

Robert Lewis

71,180

why is $Y^\dagger = Y^T$ true? – Cris Feb 03 '19 at 08:44
@Jingyang: Because $Y$ is real and symmetric. – Robert Lewis Feb 03 '19 at 12:02

Prove $I-YY^\dagger$ is a projection operator for $N(Y)$

2 Answers2