Velvet Star Monitor

Standout celebrity highlights with iconic style.

news

Eigenvalues, orthonormal eigenvectors

Writer Andrew Mclaughlin
$\begingroup$

While studying Linear Algebra, I encountered the following exercise:

Let

$$A = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$$

Write $A$ as a sum $$\lambda_{1} u_{1}{u_{1}}^T + \lambda_{2} u_{2}{u_{2}}^T$$ where $\lambda_1$ and $\lambda_2$ are eigenvalues and $u_1$ and $u_2$ are orthonormal eigenvectors.

So what I did is, I computed the eigenvalues: $1$ and $-1$. Then I computed the eigenvectors: \begin{bmatrix} 1\\1 \end{bmatrix}

and \begin{bmatrix} -1\\1 \end{bmatrix}.

But, these vectors are not an orthonormal set, thus I used Gram-Schmidt to obtain an orthonormal set of eigenvectors. This gave me: $u_1 = 1/\sqrt{2}\left( \begin{smallmatrix} 1\\ 1 \end{smallmatrix} \right)$ and $u_2 = 1/\sqrt{2}\left( \begin{smallmatrix} -1\\ 1 \end{smallmatrix} \right)$.

This outcome seems okay, however, I am wondering whether there is a faster method to do this?

$\endgroup$ 3

2 Answers

$\begingroup$

The vectors $(1,1)^T$ and $(-1,1)^T$ are orthogonal, so you just had to normalize them (divide them by their norm) to get an orthonormal set.

$\endgroup$ 1 $\begingroup$

Recall that a set of vectors is defined orthogonal when

  • $u_i \cdot u_j=0$ for $i\neq j$

and it is defined orthonormal when also

  • $u_i \cdot u_j=1$ for $i=j$
$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy