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Everything below is correct. But I got a question at the bottom (yellow).

Given is matrix $A=\begin{pmatrix} 11 & 0 & -6\\ 0 & 5 & -6\\ -6 & -6 & -2 \end{pmatrix}$ which is diagonalizable.

The eigenvalues are

$$\lambda_1= -7\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \lambda_2= 7\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \lambda_3=14$$

The eigenspaces are

$E_{A}(\lambda_1) = \left\{ \begin{pmatrix} \frac{z}{3}\\ \frac{z}{2}\\ z \end{pmatrix} \mid z \in \mathbb{R}\right\}, E_{A}(\lambda_2) = \left\{ \begin{pmatrix} \frac{3}{2}z\\ -3z\\ z \end{pmatrix} \mid z \in \mathbb{R}\right\}$

$E_{A}(\lambda_3) = \left\{ \begin{pmatrix} -2z\\ -\frac{2}{3}z\\ z \end{pmatrix} \mid z \in \mathbb{R}\right\}$

Let's say I was supposed to to determine an orthonormal basis $(v_1,v_2,v_3) \in \mathbb{R}^{3}$ of the eigenvectors of $A$, can I just write the matrix in its diagonalized form and continue there determining an orthonormal basis? It should be possible iff each column is an eigenvector of the eigenspace, right? Is it?

So matrix A diagonalized is $A_D= \begin{pmatrix} -7 & 0 & 0\\ 0 & 7 & 0\\ 0 & 0 & 14 \end{pmatrix}$

Now can I say that $v_1= \begin{pmatrix} -7\\ 0\\ 0 \end{pmatrix}, v_2= \begin{pmatrix} 0\\ 7\\ 0 \end{pmatrix}, v_3= \begin{pmatrix} 0\\ 0\\ 14 \end{pmatrix}$ and determine the orthonormal basis with these? Sorry for making this question so long, I didn't know how to keep it short. I hope you can help me anyway.

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    $\begingroup$ Minor correction: it's "orthonormal basis" (two separate words). $\endgroup$ – Alex M. Mar 21 '17 at 17:43
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    $\begingroup$ @AlexM. When I saw that mistake I immediately wondered if the OP was German, and well, sure enough. :) Also, OP: what do you mean when you refer to "an orthonormal basis of the eigenvectors"? $\endgroup$ – Jack M Mar 21 '17 at 17:48
  • $\begingroup$ @jackm I'm supposed to determine the orthonormal basis of $(v_1,v_2,v_3)$. $v_1$ is an eigenvector of the first eigenspace, $v_2$ is the eigenvector of the second eigenspace, etc. $\endgroup$ – cnmesr Mar 21 '17 at 17:52
  • $\begingroup$ @jackm If you want I can upload the PDF (in German). It will surely explain things better than me :) $\endgroup$ – cnmesr Mar 21 '17 at 17:55
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But the columns of $A_D$ are NOT eigenvectors of $A$. They rather describe the action of $A$ on those eigenvectors. The eigenvectors are essentially listed in those eigenspaces that you already found. For example, from $E_A(\lambda_1)$ we can see that an eigenvector corresponding to $\lambda_1=-7$ is $$\mathbf{v}_1=\begin{pmatrix}1/3\\1/2\\1\end{pmatrix}$$ or any vector parallel to it, i.e. any nonzero multiple of it. In fact, if you need an orthonormal basis, you'll have to renormalize this vector to get its multiple whose length (norm) is $1$. Similarly for the other two.

Good news, though: you don't need to worry about the vectors being orthogonal to each other, because eigenvectors corresponding to different eigenvalues are automatically orthogonal.

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  • $\begingroup$ "you don't need to worry about the vectors being orthogonal to each other".. however I still have to use some annoying method like Gram-Schmidt to determine the orthonormal basis of $(v_1,v_2,v_3)$? $\endgroup$ – cnmesr Mar 21 '17 at 17:49
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    $\begingroup$ @cnmesr: Generally speaking -- yes. In this case -- no, you don't. You only need to renormalize each vector to make them unit (of norm $1$), and that's it. Because all three eigenvalues are distinct, all three eigenvectors are mutually orthogonal -- you can go ahead and check that manually. Even if you apply Gram-Schmidt now, it will result in converting the vectors into unit vectors, but wouldn't change anything else. So in this case you don't need to waste your time applying it -- you've already got orthogonality for free. $\endgroup$ – zipirovich Mar 21 '17 at 17:52
  • $\begingroup$ So in this case I don't need to use Gram-Schmidt because the vectors are eigenvectors? $\endgroup$ – cnmesr Mar 21 '17 at 17:54
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    $\begingroup$ Because the vectors are eigenvectors and because the matrix is symmetric. For a general matrix, the eigenvectors might not be orthogonal (and if they're not, there's nothing you can do about it, because applying Gram-Schmidt will make them stop being eigenvectors!) $\endgroup$ – Misha Lavrov Mar 21 '17 at 17:57
  • $\begingroup$ @Misha Yes, and of course the other thing that can happen in general is that the eigenspaces do not "combine" to fill the whole space. As an example, the matrix $\begin{pmatrix} 3 & 0 & 0\\ -1 & 7 & -4\\ -1 & 2 & 1 \end{pmatrix}$ has eigenvalues $3$, $3$, and $5$. Two of them equal. The eigenspace for $3$ is just a line (1D), not a plane (2D). And the eigenspace for $5$ is another line which is not perpendicular to the first line. Wolfram Alpha query $\endgroup$ – Jeppe Stig Nielsen Mar 22 '17 at 12:20
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So, you want to find a basis of the eigenvectors but the $v_i$'s you listed correspond to the diagonalized matrix and thus correspond to the eigenvalues.

How do we find a basis for each eigenspace? Well consider $E_A(\lambda_1)$. Every vector in this space is a multiple of $(1/3, 1/2, 1)^T$ (this is easy to check), and thus $(1/3, 1/2, 1)^T$ is a basis for the eigenspace $E_A(\lambda_1)$. To make this normalized, we want the length of the vector to be exactly $1$. We can do this by calculating the norm of the vector and dividing each element by the norm. So for this basis, the length is $\sqrt{(1/3)^2 + (1/2)^2 + 1^2}$. Dividing every element in the vector by $\sqrt{(1/3)^2 + (1/2)^2 + 1^2}$ gives us the normalized vector $(2/7, 3/7, 6/7)^T$ (after some algebra and reducing fractions).

We can continue doing this for each eigenspace. We should get three normalized vectors.

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  • $\begingroup$ "We should get three normalized vectors" And after I did that, I got the orthonormal basis, right? $\endgroup$ – cnmesr Mar 21 '17 at 18:01
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    $\begingroup$ Yes, if you have orthonormal eigenvectors $v_1, v_2, v_3$, this forms the orthonormal basis of the eigenvectors. Quick sanity check: is this set of vectors in $\mathbb R^3$ as desired? Yes. Is this set of vectors linearly independent? Yes, since eigenvectors corresponding to distinct eigenvalues are independent. $\endgroup$ – hausdork Mar 21 '17 at 18:15
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Note that $A$ is a symmetric matrix, so it can be diagonalized by an orthogonal matrix ( see the properties of symmetric matrices here).

The matrix that diagonalize $A$ is not $A_D$ but the matrix $M$ in $A=MA_DM^{-1}$, that is the matrix that has as columns the eigenvectors of the corresponding eigenvalues in the diagonal of $A_D$.

In fact the eigenspaces that you have found are orthogonal as you can easily verify. This means that for any value of $z \ne 0$ that you can chose, you have three orthogonal vectors and, if you want, you can normalize these vectors to obtain an orthonormal basis.

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The answer is no. If you want the eigenvectors, take them from the eigenspaces; they're the spans of each of the eigenvectors.

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The matrix $A_D$ contains the eigenvalues along the diagonal entries; the columns are not the eigenvectors. You already have the eigenspaces, which are just the spans of the eigenvectors.

For example, just take $z = 1$. It is usually best pick a basis of orthonormal eigenvectors, so just normalize each eigenvector once you have your basis.

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