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Given a 3x3 symmetric matrix $A$ =\begin{pmatrix} 1 & -1 & -1 \\ -1 & 1 & 1 \\ -1 &1& 1 \end{pmatrix}
Find the orthogonal basis corresponding to the above matrix $A$

The eigen values for the above are coming out to be repeated i.e. 0 , 0 , 3
So for the symmetric matrix $A$=$PD$$P^{T}$ where $P$ is the eigen vectors corresponding to eigen values in $D$ and from $P$ matrix we get the orthonormal basis. But here the eigen vectors corresponding to the eigen value 0 would not be unique so the matrix $P$ formed will have linearly dependent columns.

Can the above approach be used if the eigen values are repeated and eigen vectors are not distinct.

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2 Answers 2

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No, the matrix $P$ will not have linearly dependent columns. If you compute the eigenvectors corresponding to the eigenvalue $0$; you will see that they are the linear combinations of $(1,1,0)$ and $(1,0,1)$. Besides, $(-1,1,1)$ is an eigenvector corresponding to the eigenvalue $3$. Can you take it from here?

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  • $\begingroup$ for eigen value 0 if i get eigen vector as (1,1,0) should it be the same for other column as 0 is repeated eigen value or another combination of eigen vector corresponding to it e.g. (1,0,1) to be taken? $\endgroup$
    – ten do
    Commented Apr 23, 2019 at 21:34
  • $\begingroup$ No, it cannot be the same, since you want to get an invertible matrix. $\endgroup$ Commented Apr 23, 2019 at 21:36
  • $\begingroup$ Ok got it thanks $\endgroup$
    – ten do
    Commented Apr 23, 2019 at 21:38
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You are in fact free to choose the null space $P$ columns as $0$ if you want to if your only goal is to write:

$$A = PDP^T$$

You can even choose to omit them completely. Then you will get something called a compressed Singular Value Decomposition.

(But it would not work if that subspace had another eigenvalue than $0$.)

Now your only goal was not to write $A$ factored in this way but to find an ON basis. Then you will need to search for vectors spanning this null space anyways no matter if you choose to put them in $P$ or not.

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  • $\begingroup$ Ok got it thanks $\endgroup$
    – ten do
    Commented Apr 23, 2019 at 21:38

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