# Eigenvalues, orthonormal eigenvectors

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?

• I think you made a typo in $A$. – Gonzalo Benavides Jun 22 '18 at 18:17
• Yup lol corrected it! – bladiebla Jun 22 '18 at 18:18
• Why using Gram-Schmidt? You just normalized them, since they were already orthogonal. Moreover, if you orthogonalize vectors, in general, they will not be eigenvectors anymore. Hence, that is a bad ansatz. – Mundron Schmidt Jun 22 '18 at 18:22

## 2 Answers

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.

• Ah yeah makes sense, still studying the topic so didn't think about that. Thanks! – bladiebla Jun 22 '18 at 18:45

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 $i=j$