Let $A\in\mathbb{C}^{m,n}$ be a complex matrix. Then define $$ \tilde{A}=\begin{pmatrix} 0 &amp; A\\ A^*&amp; 0 \end{pmatrix} \in \mathbb{C}^{m+n,m+n}. $$ Order the eigenvalues of $\tilde A \ \lambda_1 \geq\dots\geq\lambda_{m+n}$ and the singular values of $A$ $s_1\geq \dots \geq s_n$.  I'm trying to prove$$ \lambda_i=s_i \ \forall 1\leq i\leq n.$$ Does the fact that $AB$ and $BA$ have the same eigenvalues (except zeros) suffice How do I go about showing this?
121270
$$B = \begin{bmatrix} 0 &amp; A^* \\ A&amp;0 \end{bmatrix}$$  I think that $\det(B) = \det(A) * \det(A^*)$ and probably eigen values just get squared. What is the right answer?  EDIT: $\operatorname{rank}(B) = 2 * \operatorname{rank}(A)$, so there are twice more eigenvalue or their multiples. And the additional eigenvalues are just negatives of the eigenvalues of $A$ (Thanks to Simon).  Still can't figure out.
53742
Let $A \in \mathcal{M}_{n}(\mathbb{Q})$ such that $A^{2017}=I_n$. Is it always true that $$A=I_n?$$   When $n \leq 2015$, I proved that this result must hold using the minimal polynomial and the fact that $2017$ is prime. I suspect that when $n \geq 2016$, this equality is not always true, but I can't find any matrix other than $I_n$ with that property, which also has only rational entries.  I also thought of an approach with eigenvalues, but since they are all complex roots of $2017$, I am not sure how to make use of them here.
112517
Let $A$ be a $n\times n$ complex matrix. Assume that $A$ is self-adjoint and $B$ denote the inverse of $A+iI$, where $I$ is identity matrix of order $n\times n$.  Then all eigenvalues of $(A-iI)B$ are   purely imaginary real of modulus one of modulus less than one   My attempt: I discarded options $1,4$ by taking $A=O$ $($null matrix$)$. Let '$a$' be an eigenvalue of $A$ then the eigenvalues of $(A-iI)B$ are $(a-i)(a+i)^{-1}$. But I'm not sure about it. Please help me.
91087
How to show that the relationship $\left | \lambda_{min} \right | \geq \sigma_{min}$ holds between the minimum eigenvalue and singular value of a square matrix $A \in \mathbb{C}^{n \times n}$?
18160
Let $A$ be a Hermitian square complex matrix of size $n$ then $\langle Ax,x \rangle=0$,   $\forall x \in \Bbb C^n$ if and only if $A=0$   I want to know if the condition that $A$ is Hermitian is required for the statement to still hold, please?  Thank you
627
If $B\succeq A$, show that $\lambda_n(B)\ge\lambda_n(A)$, where $\lambda_n$ is the $n$th largest eigenvalue.  It is Theorem 6 in  but the proof is only given as "by characterization." Do they mean the min/max characterization of eigenvalues, $$\lambda_k=\min\max\frac{x^TAx}{x^Tx}$$ and if so, how is that applied?  Thanks.
71606
Let $A \in {M_n}$ is nonsingular and each eigenvalue of $A$ is either $+1$ or $-1$.Why $A$ is similar to ${A^{ - 1}}$?
42734
Let $A$ be an $n \times n$ matrix and let $\Lambda$ be an $n \times n$ diagonal matrix. Is it always the case that $A\Lambda = \Lambda A$? If not, when is it the case that $A \Lambda = \Lambda A$?  If we restrict the diagonal entries of $\Lambda$ to being the equal (i.e. $\Lambda = \text{drag}(a, a, \dots, a)$), then it is clear that $A\Lambda = AaI = aIA = \Lambda A$. However, I can't seem to come up with an argument for the general case.
58764
Let $A$ be a nonsingular $n \times n$ matrix, $\|\cdot\|$ be any natural norm, and $K_{p}(A)=$ $\|A\|_{p}\left\|A^{-1}\right\|_{p} .$ Let $\lambda_{1}$ be the smallest and $\lambda_{n}$ be the largest eigenvalues of the matrix $A^{t} A$ (b) Show that $K_{2}(A)=\sqrt{\frac{\lambda_{n}}{\lambda_{1}}}$. (Hint: One should use the $\|A\|=\sqrt{\rho\left(A^{t} A\right)}=\sqrt{\rho\left(A A^{t}\right)}$ relation) For part $(b)$, it seems minimal eigenvalue of matrix $A$ will be the largest eigenvalue of the inverse of matrix $A$, so we have to use this idea but could not understand formally why square roots are needed for $K_{2}(A)$ and how spectral radius is necessary for this part?!
107774
I have tried this problem and I'm unsuccessful.  Let $T: M(n,n)\rightarrow M(n,n)$ be the mapping defined by $ T(A) = A^T$. Show that the only eigenvalues of $T$ are $\pm 1$.
94085
Let $A$ be an n×n matrix with eigenvalues $\lambda_i, i=1,2,\dots,n$. Then $\lambda_1^k,\dots,\lambda_n^k$ are eigenvalues of $A^k$.   I was wondering if $\lambda_1^k,\dots,\lambda_n^k$ are all the eigenvalues of $A^k$? Are the algebraic and geometric multiplicities of $\lambda_i^k$ for $A^k$ same as those of $\lambda_i$ for $A$ respectively?   Thanks!
97035
Given the matrices $A,B,C\in M_n(\mathbb R)$ which satisfy the condiditions:  $A+B+C+\lambda ΑΒ=0 \quad(1)$  $Α+Β+C+\lambda BC=0 \quad(2)$  $A+B+C+\lambda CA=0 \quad(3)$  for some $\lambda\neq0$.  (a) Prove that $I+\lambda Α$, $Ι+\lambda Β$, $I+\lambda C$ are invertible and $AB=BC=CA$.  (b) Prove that $A=B=C$.  For (a) i understand that the matrices are invertible, because we have non zero eigenvalue. In addition we have AB=BC=CA, because the (1)-(2) and the (2)-(3) give us these equalities. I have been stucked in (b). I tried to use the equality $AB=BC=CA$ and the fact that $I+\lambda Α,Ι+\lambda Β,Ι+\lambda C$ are invertible, but i can't prove that $A=B=C$. I thought that A, B and C are null matrix, but i don't find a way to prove that. I want help for (b).
94909
Title restated: Show that $A\in\mathbb{C}_n$ is normal $\iff$ $tr(A^*A) = \sum_{i = 1}^n|\lambda_i|^2$, where $\lambda_1,...,\lambda_n$ are the eigenvalues of $A$.  This question comes from "Matrices and Linear Transformations" by Charles Cullen. I'm studying for an exam and am trying to do some problems but I'm having trouble with this one. Any help would be appreciated.  Thank you.
38811
If $\lambda_1$, $\lambda_2$, $\lambda_3$ are the eigenvalues of the matrix : $$         \begin{pmatrix}         26 &amp; -2 &amp; 2 \\         2 &amp; 21 &amp; 4 \\         4 &amp; 2 &amp; 28 \\         \end{pmatrix} $$ Show that : $$\sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2} \le \sqrt{1949}$$   I found the characterstic equation as $$\lambda^3 - 75\lambda^2 + 1850\lambda - 15000 = 0$$ This gave $$\lambda_1 + \lambda_2 + \lambda_3 = 75$$ and $$\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_1\lambda_3 = 1850$$ Also $${(\lambda_1 + \lambda_2 + \lambda_3)}^2 = \lambda_1^2 + \lambda_2^2 + \lambda_3^2 + 2{(\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_1\lambda_3)}$$ From this I found out $$\sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2} = \sqrt{1925}$$ But am not satisfied with the way I have approached the problem. Does anyone know of an easier way of doing it?
95694
If $A \in \mathbb{C}^{n \times n}$ is a non-zero Hermitian matrix, I need to show that  $$\text{rank}(A) \geq \frac{(\text{tr}(A))^2}{\text{tr}(A^2)}$$ and reason out when the equality is attained?    Can anyone help me in showing this result?
113002
Let $a$ and $b$ be distinct nonzero real numbers and let $A\in \mathbb{R}^{n\times n}(n\geq 2)$ with each diagonal entry equal to $a$ and each off-diagonal entry equal to $b$.     Determine all eigenvalues and eigenvectors of $A$ together with their algebraic multiplicities. Is $A$ diagonalizable? Why? Determine the minimal polynomial of $A$.   My idea: Considering a $4\times 4$ matrix for simplicity:  $\left( \begin{array}{ccccc} a&amp; b&amp;b &amp; b \\ b &amp;a&amp;b &amp;b \\ b &amp; b&amp;a&amp; b\\ b&amp;b&amp;b&amp;a \end{array} \right)$ now we perform row operations to transform this matrix into a simple one like this: $\left( \begin{array}{ccccc} a-b&amp; b-a&amp; 0 &amp; 0 \\ 0 &amp;a-b&amp; b-a&amp;0 \\ 0 &amp; 0&amp;a-b &amp; b-a\\ b&amp;b&amp;b&amp;a \end{array} \right)$ Then find the eigen values and from there eigen vectors? I think eigen values are $a\pm b$ but not sure how to prove it.  Thanks Marvis, your answer is very helpful.   Can you also comment on my ideas about the rest of the problem: to find eigen vectors we need $x$ such that $(A-\lambda I)x=0.$ So, $\lambda =a-b$ gives me that $(A-\lambda I)=(b)$ a matrix with all entries $b$, thus eigen vector in this case is a $n\times 1$ vector with first entry $n-1$ and the rest of the entries as 1's. So, using the eigen vectors I can find an invertible matrix $P$ and find out if $P^{-1}AP=D$ where $D$ is a diagonal matrix? Then I can get minimal polynomial from the diagonal matrix.
116317
Let $A$ be a complex matrix such that $A^n = I$, show that $A$ is diagonalisable.  How do I do this?    I would hazard a guess that since $A^n = I$ that means that $A^n$ is obvious diagonal, and I suppose that a diagonal matrix can only be formed by the square of another diagonal matrix, but I am not sure if that is the case. Also I think this neglects the complex cases.  Is there a Jordan block argument that I should use?