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I have an exercise where I found the two correct eigenvalues $\lambda_{1} = 0$ and $\lambda_{2} = 6$.

The algebraic multiplicity of $\lambda_{2}$ is 2. Now I try to find $E_{\lambda_{2}}$ with the following matrix. I've already done the elementary operations needed to solve the system.

$(A-6I | 0) = \begin{pmatrix} 1 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$

Now we get the following system of linear equations.

$x+y-2z = 0$

$y = y$ (first free variable)

$z = z$ (second free variable)

This basically implies that $x = -y +2z$

So, to get my eigenvectors of my basis I do: $\begin{pmatrix} x \\y\\z \end{pmatrix} = \begin{pmatrix}-1\\1\\0 \end{pmatrix}y + \begin{pmatrix} 2\\0\\1 \end{pmatrix}z$

therefore $E_{\lambda_{2}} = \bigg\{ \begin{pmatrix}-1\\1\\0 \end{pmatrix} , \begin{pmatrix} 2\\0\\1 \end{pmatrix} \bigg\}$.

However, this not what the solution in my textbook is even though we had the same exact matrix.

They have as solution that $E_{\lambda_{2}} = \bigg\{ \begin{pmatrix}1\\-1\\0 \end{pmatrix} , \begin{pmatrix} 1\\1\\1 \end{pmatrix} \bigg\}$.

I know that they are different vectors which you can take by multiplying the vectors found in the linear equation $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = ()y + ()z$ by some scalar.

For example I could do $-1\begin{pmatrix} -1\\1\\0 \end{pmatrix}$ = $\begin{pmatrix} 1\\-1\\0 \end{pmatrix}$ which gives me the first vector they had in their solution.

However, I don't get how they got $\begin{pmatrix} 1\\1\\1 \end{pmatrix}$ as their second vector of the basis. Where did I went wrong?

While writing the question I realized that in the end both my solution and theirs are correct however their solution has the advantage of proposing two orthogonal vectors that are orthogonal with the basis of $E_{\lambda_{1}} = \bigg\{\begin{pmatrix} 1\\1\\-2 \end{pmatrix}\bigg\} $ without resorting to doing Gram-Schmidt afterwards. So now, I want to know what was the method used to get those orthogonal vectors easily is there a process or was it just trial and error?

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  • $\begingroup$ since you can choose 2 eigenvectors, you get the first by setting y=-1 and z=0 like you suggested yourself. you get the other one by setting y=z=1 $\endgroup$ Commented Apr 21, 2019 at 19:42
  • $\begingroup$ why do you do y=z=1? $\endgroup$ Commented Apr 21, 2019 at 21:03
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    $\begingroup$ could you be a little more specific? of course you can choose whatever you want for y and z, but in this case it's easy to see that we get orthogonal vectors if we choose them this way. why? well the first vector has a 1, a -1 and a zero. so the third coordinate is of no interest. we only want the first and second coordinate to be the same, so they cancel out if we take the scalar product. $\endgroup$ Commented Apr 21, 2019 at 21:58

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If you just want a vector that satisfies $x=-y+2z$ and is orthogonal to $\begin{pmatrix}-1\\1\\0\end{pmatrix}$, then solve $x=-y+2z$ and (from the dot product being $0$) $-x+y=0$. You'll find that $x=y=z$ and then might as well pick $\begin{pmatrix}1\\1\\1\end{pmatrix}$. As you mentioned, if you wanted an orthonormal basis, Gram-Schmidt is a process you could follow.

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There is nothing wrong with the basis vectors that you have found. As you know basis for a vector space is not unique even if you orthogonalize your basis your solution may be different from the book's answer. You may use Gram-Schmidt process for togonalization of your basis.

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Applying Gram-Schmidt without normalization to your two vectors gets you $(1,1,1)^T$. Not having seen your original matrix, having this as an eigenvector means that the rows all sum to zero, so you might’ve found this one without going through any of the other work.

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