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Let

$$A =\begin{bmatrix} 1& 1& 1& 1& 1& 1& 1& 1\\ 1& 1& -1& -1& 1& -1& -1& 1\\ 1& -1& 1& -1& -1& 1& -1& 1\\ 1& -1& -1& 1& -1& -1& 1& 1\\ 1& 1& -1& -1& -1& 1& 1& -1\\ 1& -1& 1& -1& 1& -1& 1& -1\\ 1& -1& -1& 1& 1& 1& -1& -1\\ 1& 1& 1& 1& -1& -1& -1& -1 \end{bmatrix}$$

a) Find matrix equality on which you can conclude that column of matrix A make orthogonal vector system then find norm of that vectors. Find matrix equality on which you can conclude all eigenvalue then find their algebraic multiplicity.

b) From result of a) what you can say about rank of matrix A

c) Can you diagonalize this matrix? And what you can say about geometric multiplicity ?

I know matrix equality for eigenvalue,since $A=A^T$ then

$$AA^T=A^2=\begin{bmatrix} 8& 0& 0& 0& 0& 0& 0& 0\\ 0& 8& 0& 0& 0& 0& 0& 0\\ 0& 0& 8& 0& 0& 0& 0& 0\\ 0& 0& 0& 8& 0& 0& 0& 0\\ 0& 0& 0& 0& 8& 0& 0& 0\\ 0& 0& 0& 0& 0& 8& 0& 0\\ 0& 0& 0& 0& 0& 0& 8& 0\\ 0& 0& 0& 0& 0& 0& 0& 8 \end{bmatrix}$$

so spectrum of $A^2=\{8\}$, then spectrum of $A=\{-2\sqrt2,2\sqrt2\}$ and their algebraic multiplicity is 4 for $\lambda1$ and $\lambda2$. But I think that matrix multiplicity that column is orthogonal is that $AA^T=A^TA=8I$, and norm is $\|x_i\|=\sqrt8, i=1,\ldots,8$

b) $\operatorname{rank}(A)=8$

c) Yes I can diagonalize matrix since $A=A^T$ then if I use spectral theorem $A=Q\Lambda Q^T$ so geometric multiplicity is the same as algebraic multiplicity. What you say is this good?

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(a) From $A^2=8I$ you can only conclude that the only possible eigenvalues of $A$ are $\pm\sqrt{8}$, but that doesn't tell you whether $\sqrt{8}$ or $-\sqrt{8}$ are really eigenvalues of $A$, not to mention their multiplicities. To determine their algebraic multiplicities, you may use the fact that $A$ has zero trace.

(b) Yes, you are correct.

(c) Yes, that's one way to do it. An alternative way is to note that the $x^2-8$, an annihilating polynomial of $A$, is a product of two distinct linear factors over $\mathbb R$.

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  • $\begingroup$ Thank you, but if I have four $-\sqrt8$, and four $\sqrt8$ then trace=0 ,ad detA is ok, so i think that only way to prove that is to go step by step to prove their algebraic multiplicity, but i do not think that is point of this task $\endgroup$ – Marko Škorić Sep 4 '18 at 10:05
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Since $A^TA = 8I$, clearly the canonical vectors $\{e_1, \ldots, e_8\}$ are eigenvectors for $A^TA$. Your previous question then implies that $\{Ae_1, \ldots, Ae_8\}$ are orthogonal vectors and these are precisely the columns of $A$.

Since $\{Ae_1, \ldots, Ae_8\}$ is orthogonal, in particular it is linearly independent so the rank of $A$ is $$r(A) = \dim \operatorname{span} \{Ae_1, \ldots, Ae_8\} = 8$$

Direct calculation gives $\|Ae_i\| = \sqrt{8}$ for all $i = 1, \ldots, 8$.

Hence the matrix $\frac{1}{\sqrt{8}}A$ has orthonormal columns, so we conclude that $\frac{1}{\sqrt{8}}A$ is an orthogonal matrix, so its eigenvalues are on the unit circle. It is also symmetric so its eigenvalues are real. Therefore $$\sigma(A) = \sqrt{8}\sigma\left(\frac{1}{\sqrt{8}}A\right) \subseteq \sqrt{8} \cdot \{-1,1\} = \{-\sqrt8,\sqrt8\}$$

Let $a(\pm\sqrt{8})$ be the algebraic multiplicities of $\pm\sqrt8$ respectively. We have

$$0 = \operatorname{Tr} A = \sum_{\lambda \in \sigma(A)} \lambda = a(\sqrt{8})\sqrt{8} - a(-\sqrt{8})\sqrt{8}$$

Therefore $a(\sqrt{8}) = a(-\sqrt{8}) = 4$, because they are equal and $a(\sqrt{8}) + a(-\sqrt{8}) = 8$.

$A$ is of course diagonalizable because it is symmetric. Hence, geometric multiplicities of $\pm\sqrt{8}$ are equal to algebraic multiplicities, which are $4$.

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