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I was confused in a proof of: every symmetric matrix $A\in\mathbb{R}^{n\times n}$ has eigen vectors which form the orthonormal basis of $\mathbb{R}^n$. The proof that I was looking into procceded like this:

  1. Since $A$ is symmetric, it is self-adjoint and hence all eigen-values of $A$ are real.
  2. There is a unitary matrix $U\in \mathbb{C}^{n\times n}$ (matrix may have complex numbers as its entries) such that $U^{-1}AU$ is diagonal.
  3. We use 'Gauss Elimination Method' followed by 'Gram-Schmidt Orthogonalization Process' to find orthonormal basis of $A-\lambda I$ for each eigen-value $\lambda \in \mathbb R$.
  4. We get $n$ orthonormal eigen-vector, all in real space $\mathbb{R}^n$.

From the steps of the proof, I am stuck in step 3, where I am not sure that solutions of $A-\lambda I$ will give us orthogonal real eigen-vectors, which form the basis of eigen-vector of $A-\lambda I$. I am not sure how the jump from complex to real is made here. Can someone please explain? Or give an alternate first-principle's solution? Thanks in advance!

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  • $\begingroup$ Maybe I am mistaken, but since $A$ is a real matrix, Gauss Elimination and Gram-Schmidt should yield real vectors in any case. If one looks at the formula for Gram-Schmidt it should be clear that 'orthogonalizing' real vectors yields real vectors and Gauss Elimination just solves the Equation $(A - \lambda I) x = 0$. Which only has real solutions as well since its a (real) linear equation. $\endgroup$ Aug 30, 2023 at 9:45
  • $\begingroup$ Yes, i think it is correct that Gauss Elimination can solve the Equation (𝐴−𝜆𝐼)𝑥=0, such that there is atleast one real eigen-vector of $A$, but i am not so sure about why the dimension of the null space ( of $(A-\lambda I)x$ ) in $\mathbb R^n$ is same as that $\mathbb C^n$. $\endgroup$
    – A J
    Aug 31, 2023 at 7:01
  • $\begingroup$ Well the dimension of the solution of $(A- \lambda I) x = 0$ (regarded as a real equation) is equal to the dimension of the eigenspace of $\lambda$. For this one does not need to consider the complex solutions what so ever. I might not be understanding you correctly I feel like. $\endgroup$ Aug 31, 2023 at 7:59
  • $\begingroup$ Yes I agree in their respective spaces but not sure whether it can be translated from one space to another. For example, let the dimension of solution of $(A-\lambda I)x=0$ where $x\in\mathbb{R}^n$ be $d_1$ and the dimension of solution of the same equation where $x\in\mathbb{C}^n$ be $d_2$. It is known that $A\in\mathbb{R}^{n\times n}$ and $\lambda \in \mathbb{R}$. I am not sure whether it will always be the case that $d_1=d_2$? I think this is exactly what @ronno has answered, but it would be great to look at the proof of the claim. $\endgroup$
    – A J
    Sep 1, 2023 at 16:50
  • $\begingroup$ That's a fair point. It is basically answered here although without an explanation I guess. $\endgroup$ Sep 1, 2023 at 16:56

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A real matrix (the relevant one here is $A - \lambda I$) has the same nullity over $\mathbb{R}$ and $\mathbb{C}$, eg by rank-nullity and the determinant definition of rank. Alternatively, once you have one real eigenvector $x$, you can restrict $A$ to $\{y \mid y^tx = 0\}$ and induct on dimension.

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  • $\begingroup$ Is there any proof of the claim? It would be great to look into it from the first principles if possible. $\endgroup$
    – A J
    Sep 1, 2023 at 16:51
  • $\begingroup$ @AJ Which claim? That the nullity is the same? A proof is: "rank--nullity and the determinant definition of rank". $\endgroup$
    – ronno
    Sep 1, 2023 at 17:53
  • $\begingroup$ Yes that nullity is the same (a more detailed proof is preferable). Does this also hold in general subspaces? $\endgroup$
    – A J
    Sep 1, 2023 at 18:04
  • $\begingroup$ There aren't any more details, the rank is the same over $\mathbb{R}$ and $\mathbb{C}$ because it's the size of the largest non-vanishing subdeterminant, and then the nullity is $n - \text{rank}$. What is the statement for general subspaces that you want to know about? $\endgroup$
    – ronno
    Sep 1, 2023 at 19:01
  • $\begingroup$ I think the statement is false, but it is as follows. Let $\mathbb{V}\subseteq \mathbb{C}^{n}$ be a subspace, and let $S\subseteq \mathbb{C}^{n}$ be the subspace of solution of $(A-\lambda I)x=0$ where $A\in \mathbb{R}^{n\times n}$, $\lambda \in \mathbb{R}$, and $x\in \mathbb{C}^{n}$. Then the dimension of $S\cap \mathbb{V}$ is same as dimension of $S$? This may be true when $\mathbb{V}=\mathbb{R}$, is it possible to put a characterization on $\mathbb{V}$ when the statement is true? $\endgroup$
    – A J
    Sep 1, 2023 at 21:23

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