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Proof:

$Av$ = $\lambda v$

$\implies \bar{v}^{T}Av = \lambda \bar{v}^{T} v$ ------(1)

And,

$Av$ = $\lambda v \implies \bar{A}\bar{v}$=$\bar{\lambda}\bar{v} \implies \bar{v}^{T}\bar{A}^{T}=\bar{\lambda}\bar{v}^{T}$

$\implies \bar{v}^{T}\bar{A}^{T} v = \bar{\lambda}\bar{v}^{T} v$ ------(2)

Since $A$ has real eigenvalues, $\lambda = \bar{\lambda} \implies \bar{v}^{T}\bar{A}^{T} v = \lambda\bar{v}^{T} v$ ------(3)

Now, Assuming A is real ($A=\bar{A}$), and comparing equation (1) and (3):

$\bar{v}^{T}A^{T} v = \bar{v}^{T}Av$

Does that mean $A=A^{T}$?

And hence can I infer that a positive definite matrix (which of course has all eigenvalues real and positive) is symmetric? Is that the right reason behind it? If not, what makes a positive definite matrix symmetric?

EDIT: This question is not the same as "Prove that the eigenvalues of a real symmetric matrix are real.", but it actually asks about whether or not the converse it true.

Krishnkant Swarnkar
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2 Answers2

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The question has been correctly answered by Mostafa Ayaz, but, just in case you suspect that the problem arises because his example isn't diagonalizable, here's a diagonalizable example: $$ \begin{pmatrix} 2&-1\\0&1 \end{pmatrix}. $$ The eigenvalues are $1$ and $2$, with eigenvectors $\binom11$ and $\binom 10$, respectively.

More generally, notice that, for a symmetric matrix, not only are the eigenvalues real but the eigenvectors for different eigenvalues are orthogonal. So you can produce many counterexamples like mine by just choosing some different real numbers to serve as eigenvalues and some non-orthogonal vectors to serve as eigenvectors.

Andreas Blass
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    Another easy way to produce counterexamples is to take real upper triangular matrices. These will have real eigenvectors and real eigenvalues (namely the values along the diagonal), but are readily seen to be non-symmetric without citing any theorems or doing any computations beyond understanding row reduction and how to compute determinants (either via row reduction or via expansion by minors). – Aaron Aug 18 '19 at 09:02
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This is not true. The following matrix $$A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$has real eigenvalues but is asymmetric. Also it is positive definite but this too does not imply symmetry.

Mostafa Ayaz
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    Sorry this is not a good example of mon-symmetric matrix has real eigenvalues. Infact this matrix is not diagonalizable! so no eigenvalues!! – Z Ahmed Aug 18 '19 at 01:42
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    @DrZafarAhmedDSc "so no eigenvalues" The number $1$ is an eigenvalue of $A$. A corresponding eigenvector is the vector $x = \begin{bmatrix} 1 \ 0 \end{bmatrix}$. – littleO Aug 18 '19 at 06:34
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    @DrZafarAhmedDSc, do you believe that diagonalizability and existence of the eigenvalues are equivalent? In fact a matrix $A_{n\times n}$ has an eigenvalue $\lambda$ iff the equation $Av=\lambda v$ holds for some vector $v$ and is diagonalizable only if its eigenvectors span the $\Bbb R^n$ or equivalently$$A=PDP^{-1}$$for some diagonal $D$ and invertible $P$. – Mostafa Ayaz Aug 18 '19 at 06:38
  • @Mustafa Ayaz there are two things latent roots and eigenvalues they may be or may not be same. Your matrix has repeated latent roots but these are not is not eigenvalues as your matrix is not diagonalizable. may. I have edited my solution in this regard. – Z Ahmed Aug 18 '19 at 07:01