I know that we can say if $A$ is real then it has real eigenvalues, but is the opposite always true (i.e. if $A$ has real eigenvalues then it is Hermitian)?
123546
Let us say I know that a given $N\times N$ matrix has all its eigenvalues as real, does it mean, it is hermitian. How do I prove (or disprove) that?
62243
Alastair Rae states that there are 4 postulates of Quantum Mechanics in his text on the subject matter.  The first part of his second postulate can be stated as:      Every dynamical variable may be represented by a Hermitian operator whose eigenvalues represent the result of carrying out a measurement of the value of the dynamical variable...   My question is: just how much can we deduce from saying that the some of the operators involved in Quantum Mechanics are Hermitian?  When I use the word Hermitian, I am referring to the property that $A=A^+$ where $A$ is an operator and $A^+$ is the adjoint of $A$.  Does an operator being Hermitian automatically imply that there will exist eigenfunctions and eigenvalues corresponding to this operator?  Does being Hermitian imply that, should there exist eigenfunctions corresponding to this operator, they will form a complete and orthonormal set?  That any wavefunction can be expanded in terms of these eigenfunction?  Just how much is hidden in the word "Hermitian"?
33340
If $\lambda$ is a real eigenvalue of a real matrix $M$, does there necessarily exist a real eigenvector of $M$ corresponding to $\lambda$?  Edit: Never mind. I figured it out.. If $\lambda$ is a real eigenvalue, then $\det(\lambda I-M)=0$, which means there exists a real vector $x$ such that $Mx=\lambda x$. Right?
78529
what do it mean by "A real nxn matrix always has a real eigenvalue when n is odd?" Shouldnt it be that regardless of odd or even there will always have a real root?  hope to hear from you guys soon:)
117842
As the title says. Is the matrices $A^*A$ and $AA^*$ hermitian (symmetric if $A$ is real)?
104732
This is a problem I've been stuck on for a while. I know that this statement is false if it had $A$ instead of $A^2$, for example, the matrix $A = \begin{bmatrix} 0&amp;1\\0&amp;0 \end{bmatrix}$ can't be written as $X^3$, but since $A^2$ equals $ \begin{bmatrix} 0&amp;0\\0&amp;0 \end{bmatrix}$, it can be written as $X^3$ with $X$ being$ \begin{bmatrix} 0&amp;0\\0&amp;0 \end{bmatrix}$ for example. Other than this, I have no insight into why this statement could be true: nothing can be assumed about diagonalizability, invertibility or anything else. I might be overlooking something simple. Thanks in advance.
10749
I am a physicist. I always heard physicists used the terminology "", "Hermitian", "", and "essentially self-adjoint" operators interchangeably.  Actually what is the difference between all those operators? Presumably understandable by a physicist.
12785
Let $A$ belongs to $M_3(\mathbb R)$ which is not a diagonal matrix. Pick out the cases  when $A$ is diagonalizable over $\mathbb R$:  a. when $A^2 = A$; b. when $(A - 3I)^2 = 0$; c. when $A^2 + I = 0$  (a) is certainly true as eigen values are distinct. (c) is certainly not true as eigen values are imaginary not real. but i am not sure about (b). can anybody help? thanks.
56098
This can be found at , but I don't know how to prove it:  $A$ is normal if and only if there's a unitary matrix V,s.t. $A^*=AV$
114097
I understand the proof of the identity in the title for $A$ Hermitian. One uses that any Hermitian matrix can be diagonalized as $A = X \Lambda X^{-1}$, such that $$ \det{A} = \prod_i \lambda_i, $$ and we have $$ \exp(Tr(\log(A)) = \exp(Tr(X\log\Lambda X^ {-1}) = \exp(\sum_i\log(\lambda_i)) = \prod_i \lambda_i. $$  However, is it possible to show the identity for $A$ not Hermitian? My motivation for this question is that in physics the identity is often used without it being clear that $A$ is Hermitian.
73092
These facts are taken for granted in a QM text I read. The purportedly guaranteed non-normalizability of eigenfunctions which correspond to a continuous eigenvalue spectrum is only partly justified by the author, who merely states that the non-normalizability is linked to the fact that such eigenfunctions do not tend to zero at infinity.  Not a very satisfying answer. What I'm really after is an explanation based in functional analysis. I believe there is a generalized result about inner products being finite for discrete spectra but infinite for continuous spectra.  Can anyone shed some light on this?
60640
Let $\mathbf A, \mathbf B$ be Hermitian matrices of the same size. What is the characterization of $\mathbf A, \mathbf B$ such that $p(\lambda)=\det(\mathbf A-\lambda\mathbf B)=0$ has only real roots?   If $\mathbf B$ is positive definite (I corrected this), it is easy to see $p(\lambda)$ has only real roots.
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Clearly if $\lambda$ is an eigenvalue of $A$ then $p(\lambda)$ is an eigenvalue of $p(A)$ where $p$ is a polynomial. And there are cases where $A$ may have eigenvalues other than these.   Is there a general rule for finding all eigenvalues of $p(A)$? Does it depend on the field from which the elements of $A$ are taken? Are their any partial solutions to this problem?
43751
A symmetric operator has only real eigenvalues and different eigenvectors corresponding to different eigenvalues are orthogonal.   These are exactly what we want for a physical observable. I think this is the all required.   So why is quantum mechanics not satisfied with symmetric operators, but wants more?
91498
So I was working on a specific problem related to Hermitian matrices. If we let $H_n$ denote the set of n x n Hermitian matrices. We're told that $H_n$ is a real vector space under matrix addition and scalar multiplication by a real number. I don't understand why though, because what would prevent you from adding two imaginary numbers?   Then we're told to write a basis for $H_2$. Now for a Hermitian matrix, it's equal to its conjugate transpose, so a basis would be:  $$\left(\begin{array}{ccc} 0 &amp; 1 \\ 1 &amp; 0\\ \end{array}\right) , \left(\begin{array}{ccc} 0 &amp; 0 \\ 0 &amp; 1\\ \end{array}\right), \left(\begin{array}{ccc} 1 &amp; 0 \\ 0 &amp; 0\\ \end{array}\right), \left(\begin{array}{ccc} 0 &amp; i \\ -i &amp; 0\\ \end{array}\right)$$  Which should be the same as the basis for a symmetric matrix, correct?   From there though, we have $sH_n$ represent skew-Hermitian matrices. I believe the basis for $sH_n$ is:  Edit: updated my basis here.   $$\left(\begin{array}{ccc} 0 &amp; -1 \\ 1 &amp; 0\\ \end{array}\right), \left(\begin{array}{ccc} 0 &amp; i \\ i &amp; 0\\ \end{array}\right)$$  because the diagonals would have to be zero right, to make the conjugate transpose of A be equal to -A?  From there though, they ask if $sH_n$ is a real vector space, and I'm not sure what to answer. It's real if we only use reals?  I'm asked the same thing for $U_n$, where $U_n$ represents n x n unitary matrices. I'm not even sure how to come up with a basis for this set??  Edit: Technically, the question says "Is $U_n$ a real vector space? Write a basis if possible."  I suppose I couldn't write a basis because $U_n$ is not a real vector space then. How would I show that though?   Thanks in advance for your help.
108527
Let $A$ be an $n \times n$ matrix with real enteries. Define $\langle x,y\rangle_A = \langle Ax,Ay\rangle, \quad x,y \in \mathbb R^n$ , where $\langle,\rangle$ is a standard inner product on $\mathbb R^n$. Then  $\langle x,y\rangle_A$ is an inner product iff  1)  $\ker A = \{0\}$.  2) $\operatorname{rank} A= n$.  3) all eigenvalues of $A$ are positive.  4) all eigen value of $A$ are non negative.  Clearly if $\ker A \neq \{0\}$, then there exist $x \neq 0$ such that $Ax =0$, then $\langle Ax, Ax\rangle_A = 0$, then $\langle ,\rangle_A$ is not inner product, so $\ker A = \{0\}$, thus rank of $A$ is $n$.  I am unable to check the 3) and 4) option. Please help to check the other option. Thank you
99852
I'm reading about linear algebra and I came across with the following theorem where I have a problem convincing myself:      Theorem 2.1 $\,$ Every linear operator on a finite-dimensional complex vector space has an eigenvalue.      Proof:       To show that $T$ (our linear operator on $V$) has an   eigenvalue, fix any nonzero vector $v \in V$. The vectors $v, Tv,  T^2v,..., T^nv$ cannot be linearly independent, because $V$ has   dimension $n$ and we have $n + 1$ vectors. Thus there exist complex   numbers $a_0,...,a_n$, not all $0$, such that      $$a_0v + a_1Tv + ··· + a_nT^nv = 0.$$ Make the $a$’s the coefficients   of a polynomial, which can be written in factored form as $$a_0 + a_1z  + ··· + a_nz^n = c(z − r_1)\cdots(z − r_m),$$ where $c$ is a non-zero complex number, each $r_j$ is complex, and the equation holds for all   complex $z$. We then have      $${\color{red}{ 0=(a_0I + a_1T + ··· + a_nT^n)v= c(T − r_1I)\cdots(T −  r_mI)v}},$$ which means that $T − r_j$ I is not injective for at least   one $j$. In other words, $T$ has an eigenvalue.$\;\blacksquare$   I'm having trouble with the factorized form of the matrix polynomial (in red). I understood that the factorization holds for a polynomial by the fundamental theorem of algebra but why does this also hold for matrices?  In other words, why is the the part I highlighted true? Does such an factorization always exist?    Could I have some help to see this? Thank you =)   P.S.  is my reference (page 3).  UPDATE:   Someone else also has asked the  before it seems.