Orthogonal or Symmetric Matrix?

Algebra Level pending

Given a column matrix $u$ such that

$u^Tu = 1$

Here $u \in {\rm I\!R}^{n\times1}$ . Consider the matrix:

$H = I_n - 2uu^T$

Here, $I_n$ is the identity matrix of size $n \times n$ . Is the matrix $H$ orthogonal, or symmetric, or both symmetric and orthogonal or neither of the two? Justify your answer.

Look up:

Orthogonal Matrix

Symmetric Matrix

None Orthogonal Both Symmetric

2 solutions

Karan Chatrath
Aug 28, 2019

Given:

$H = I_n - 2uu^T$ $H^T = \left(I_n - 2uu^T\right)^T \implies = H^T = I_n^T - 2(uu^T)^T$ $H^T = I_n^T - 2(uu^T)^T = I_n - 2(u^T)^Tu^T = I_n - 2uu^T = H$

Therefore:

$\boxed{H = H^T}$

$H$ is a symmetric matrix.

To check for orthogonality, the following must be true:

$HH^T = I_n$

$HH^T = (I_n - 2uu^T)(I_n - 2uu^T) = I_n - 4uu^T + 4uu^Tuu^T = I_n - 4uu^T + 4u(u^Tu)u^T$

Recognizing that $u^Tu =1$ :

$HH^T= I_n - 4uu^T + 4u(u^Tu)u^T = I_n - 4uu^T + 4uu^T = I_n$

Therefore:

$\boxed{HH^T=I_n}$

$H$ is orthogonal.

Mark Hennings
Aug 28, 2019

Note that $Hu=-u$ , while $Hv=v$ for any $v$ that is orthogonal to $u$ . Thus $H$ represents reflection in the hyperplane $r \cdot u = 0$ , and is therefore both symmetric and orthogonal.

Nice catch. The inspiration for this problem came while I was reading about Householder reflections.

Karan Chatrath - 1 year, 9 months ago

Orthogonal or Symmetric Matrix?

2 solutions

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