# Do real eigenvalues $\implies$ symmetric matrix? And why is a positive definite matrix symmetric?

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.

• @RodrigoPizarro This is not a duplicate of that question which is symmetric $\implies$ real eigenvalues. The OP is asking about the converse: does real eigenvalues $\implies$ symmetric. – 0XLR Aug 17 at 23:29
• Note that the term "positive definite" is often (but not always) used to mean "symmetric positive definite". – littleO Aug 17 at 23:32
• Possible duplicate of Prove that the eigenvalues of a real symmetric matrix are real. – Shailesh Aug 18 at 0:04
• @littleO Thanks. That was helpful. Actually I thought the same but just recently referred this link which confused me (page 2 para 1 line 2): math.utah.edu/~zwick/Classes/Fall2012_2270/Lectures/… But now what u say is clear :-) – Krishnkant Swarnkar Aug 18 at 0:28

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.
• @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 at 6:34
• @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 at 6:38
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.