Question

See attachment, at the end of the page, I don't understand the stuff how to get v1, v2. anyone explains me, please

Answer 1

v1 looks like the result of normalizing u1, how about v2? if using Gram-Schmidt, for them, how to apply, they didn't work

Answer 2

\[\text{Let }v_1=u_1\Rightarrow \text{unit vector }=\frac{v_1}{||v_1||}\\
v_1=\left[\begin{matrix}-1\\1\\0\end{matrix}\right]\Rightarrow\text{unit vector}=\frac{1}{\sqrt{(-1)^2+1^2+0^2}}\left[\begin{matrix}-1\\1\\0\end{matrix}\right]=\left[\begin{matrix}-\frac{1}{\sqrt2}\\\frac{1}{\sqrt2}\\0\end{matrix}\right]\\
\begin{align*}v_2&=u_2-\frac{\langle v_1,u_2\rangle}{\langle v_1,v_1\rangle}v_1\\
&=u_2-\frac{v_1\cdot u_2}{ v_1\cdot v_1}v_1\\\\
&=\left[\begin{matrix}-1\\0\\1\end{matrix}\right]-\frac{(-1,1,0)\cdot (-1,0,1)}{ (-1,1,0)\cdot (-1,1,0)}\left[\begin{matrix}-1\\1\\0\end{matrix}\right]\\\\
&=\left[\begin{matrix}-1\\0\\1\end{matrix}\right]-\frac{1}{2}\left[\begin{matrix}-1\\1\\0\end{matrix}\right]\\\\
&=\left[\begin{matrix}-1\\0\\1\end{matrix}\right]+\left[\begin{matrix}\frac{1}{2}\\-\frac{1}{2}\\0\end{matrix}\right]\\\\
&=\left[\begin{matrix}-\frac{1}{2}\\-\frac{1}{2}\\1\end{matrix}\right]\\&\Rightarrow \text{unit vector}=\frac{1}{\sqrt{(-1/2)^2+(-1/2)^2+1^2}}\left[\begin{matrix}-\frac{1}{2}\\-\frac{1}{2}\\1\end{matrix}\right]=\sqrt{\frac{2}{3}}\left[\begin{matrix}-\frac{1}{2}\\-\frac{1}{2}\\1\end{matrix}\right]\\&=\left[\begin{matrix}-\frac{1}{\sqrt6}\\-\frac{1}{\sqrt6}\\\frac{2}{\sqrt6}\end{matrix}\right]
\end{align*}\]

Answer 3

i know you are not here, but I still want to make question, when you are online, please answer my question: like what the problem states, the eigenvalue 2 has 2 eigenvectors u1, u2 and we use Gram-Schmidt to convert them into orthonormal eigenvectors v1, v2. We still have another eigenvalue, it's 8. to that eigenvalue, we just normalize it to get v3 like we do with v1, is it right?
If we have 3 eigenvalues and 3 eigenvectors corresponding to them, we don't apply Gram-Schmidt, right?
I am waiting for your answer. Thanks.

Answer 4

@Hoa, I'm pretty sure you would only apply G-S to the eigenvalue that has more than one corresponding eigenvector. For \(\lambda=8\), I believe you would just normalise the eigenvector (without apply G-S).