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To keep things simple, assume a matrix $A$ with the following properties:

  • $A$ is real and symmetric
  • $A$ is of full rank and the columns of $A \in \mathbb{R}^n$

What can be stated about the Eigen Value Decomposition (EVD) of such a matrix?

I am aware of a few things such as:

  • The eigenvectors of $A$ must be orthogonal (at least those that correspond to distinct eigenvalues)
  • Since $A$ is of full rank, $0$ is not an eigenvalue of $A$

But what else can be stated? Or better yet, under what conditions can the following be stated?

  • If the EVD of $A$ produces diagonalization $A = V \Lambda V^T$, where $\Lambda$ is the diagonal matrix of eigenvalues and $V$ is a matrix whose columns are the eigenvectors of $A$, when is $V$ a unitary matrix i.e. when is $V^T V = VV^T = I$ where $I$ is the identity or one can say when does the columns of $V$ yield an orthonormal basis for $\mathbb{R}^n$?
Prithvidiamond
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  • In the real case, symmetric matrices are precisely the matrices that have an eigendecomposition with orthonormal eigenvectors, i.e. $V^TV = I$. Such a matrix is invertible if and only if every eigenvalue is nonzero. – Kakashi Oct 05 '23 at 17:32
  • @Kakashi So the EVD always yields the orthonormal eigenvectors? What about the case of multiplicities? – Prithvidiamond Oct 05 '23 at 18:06
  • Yeah if $u$ and $v$ are eigenvectors corresponding to different eigenvalues, then $(u, v) = 0$. This is because $\lambda_1(u, v) = (Au, v) = (u, Av) = (u, v)\lambda_2$. – Kakashi Oct 05 '23 at 18:09
  • Okay, I was aware of that from this post here. But what happens when the eigenvalues are not distinct? Also, can you link me to a proof for why $V^T V = I$ (I understand orthogonality when the eigen values are distinct but why must the EVD necessarily yield normalized vectors?) – Prithvidiamond Oct 05 '23 at 19:51
  • Certainly if some eigenspace $E$ has dimension greater than 1, than you can write $A = VDV^{-1}$ by taking a basis for $E$ that isn't orthonormal. But when people say "EVD" they usually implicitly take the eigenvectors to be orthonormal. The important result is that in the real case, there exists an orthonormal basis of eigenvectors of $A$ if and only if $A$ is symmetric. – Kakashi Oct 05 '23 at 23:36
  • Oh, I was trying to figure out if solving the system $A- \lambda_i I = 0$ for eigenvalues $\lambda_i$ would yield orthonormal eigenvectors directly, or not. So does this method yield such vectors and if not, what else should be done to get such vectors (I assume some form of Gram-Schmidt ensues)? – Prithvidiamond Oct 06 '23 at 00:45
  • It depends on how you solve the system. I don't know the state of the art for this. But if you are doing it by hand, you will get a basis of the $\lambda_i$ eigenspace, and you would orthonormalize it by hand using Gram-Schmidt. – Kakashi Oct 06 '23 at 00:57

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