In a text book I am following, there is a definition of positive definite matrices, which I did not see before:  $x^TAx \geq \alpha x^Tx$   where $\alpha$ is a positive scalar and $x \in \mathbb{R^n}$.  The usual definition I know is for every non-zero vector $x \in \mathbb{R^n}$, it is $x^TAx &gt; 0$.  These definitions must be equivalent. It is trivial to show that the first definition implies $x^TAx &gt; 0$. But I failed to see a way to show the other way around of the equivalence. I thought of applying SVD to the matrix $A$ and  tried to find a way to show that $x^TAx$ is always lower bounded by a $\alpha x^Tx$ with $\alpha$ being positive, using singular values and vectors of $A$ but couldn't move forward. What is the correct way of approach here?  (The other question in the board assumes symmetric matrices. This one does not).
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Is there a theorem saying ${\bf x^TAx} \ge { a \bf x^Tx}$ for some $a&gt;0$ not dependent on $\bf x$ if $\bf A$ is positive definite matrix? If there is, can some help provide a reference or some explanation? Thank you!
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I recently show that every polynomial with real coefficient and $P$ is always is a sum of two squares of polynomials, therefore positive. These questions also appear often in arithmetic.  What if we change polynomials by matrices?  I ask my professor and unfortunately he had no idea if this result is already shown or not. I tried, unsuccessfully, several tracks : First denote that $$\forall M\in M_n(\mathbb R) , M=1/2[(A+A{}^t)+(A-A{}^t)].$$ and perhaps used the following result :  Any symmetric positive matrix admits a square root  Proof. By spectral theorem, there exist $\Omega \in \mathbb O_n(\mathbb R)$ and $(\lambda_1,...\lambda_n)\in (\mathbb R^{+})^n$ such that $S=\Omega D \Omega^{-1}$ with $D$ the diagonal matrix formed by the positive eigenvalues​​. Let $D'$ the diagonal matrix formed of the roots of the eigenvalues​​. Then , $D'²=S$ and  $D'{}^t=D'$ thus she is a symmetric positive matrix.  Question : Is it possible to believe that all matrices (say in $M_2(\mathbb R)$) are the sum of two squares?
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Assume that $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{A-B}$ are all positive definite Hermitian symmetric matrices of same dimensions. Prove that $\mathbf{B}^{-1}-\mathbf{A}^{-1}$ is positive definite.
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Let $A$ and $B$ be two $n \times n$ (symmetric) positive definite matrices, and denote the $k$th smallest eigenvalue of a general $n \times n$ matrix by $\lambda_k(X)$, $k = 1, 2, \ldots, n$ so that $$\lambda_1(X) \leq \lambda_2(X) \leq \cdots \leq \lambda_n(X).$$ I guess the following relation holds: $$\lambda_k(A + B) &gt; \max\{\lambda_k(A), \lambda_k(B)\}, \; k = 1, 2, \ldots, n.$$ This looks intuitive but I have difficulty to prove it, any hints?
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This is a homework problem, but I feel like I'm struggling with not knowing facts from linear algebra.  Apparently this is supposed to be an easy question but I hit a brick wall at the following point.  Can anyone give me a tip or tell me if I'm going down the wrong road?  My approach:  Let $A$ be a symmetric $n\times n$ matrix, and then suppose $x$ is a real-valued column vector of dimension $n$ with not all entries equal to $0$.  I need to show that $x^Te^{A}x&gt; 0$.  (based on looking up the definition of positive definite on wikipedia)  Then   $\begin{eqnarray*} x^{T}(e^{A})x &amp;=&amp; x^{T}(\sum_{n=0}^{\infty}\frac{A^{n}}{n!})x\\ &amp;=&amp; \sum_{n=0}^{\infty}\frac{x^{T}A^{n}x}{n!}\\ &amp;=&amp; \sum_{n=0}^{\infty}\frac{x^{T}A^{n}x}{n!})\\ \end{eqnarray*}$  As noted below, from here on is incorrect:  $\begin{eqnarray*} &amp;=&amp; \sum_{n=0}^{\infty}\frac{(x^{T}Ax)^{n}}{n!\|x\|^{2(n-1)}})\\ &amp;=&amp; \sum_{n=0}^{\infty}\frac{(x^{T}A^{T}x)^{n}}{n!\|x\|^{2(n-1)}}\\ &amp;=&amp; \sum_{n=0}^{\infty}\frac{((Ax)^{T}x)^{n}}{n!\|x\|^{2(n-1)}}\\ &amp;=&amp; \sum_{n=0}^{\infty}\frac{((Ax)^{T}(x^{T})^{T})^{n}}{n!\|x\|^{2(n-1)}}\\ &amp;=&amp; \sum_{n=0}^{\infty}\frac{((x^{T}Ax)^{T})^{n}}{n!\|x\|^{2(n-1)}}\\ \end{eqnarray*}$  As you can see this brings me no closer to getting $x^Tx$ somewhere, which I may assume is greater than $0$.  So my conclusion is that using only the fact that $A$ is symmetric is not enough.  Is there some result about symmetric matrices that I should use?
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Which of the following statements are true?   $1.$ If $A ∈M_n(C)$ is such that $(Ax,x) = 0$ for all $x ∈C^n$, then $A = 0$.   $2.$ If $A ∈M_n(C)$ is such that $(Ax,x) ≥ 0$ for all $x ∈C^n$, then $A = A^∗$.  I know that $A$ is skew symmetric Matrix iff $x^tAx = 0$ for all $x ∈R^n$ and $A ∈M_n(R)$.  The question posted by Neha Gupta had an WRONG answer. That's why I posted it again.
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I have problems to show that if $A$ is a $2 \times 2$ matrix and if there exists some positive integer such that $A^{n}=0$ then $A^{2}=0$. I only showed that $A$ is a singular matrix but nothing else. Thanks any help will be appreciated.
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Let $A$ be a positive semidefinite matrix of rank $1$. Let $B$ be a general Hermitian matrix.  Under what conditions on $B$ (probably in terms of $A$) is $A-B$ positive semidefinite?  I was thinking that it may be along the lines of the generalized eigenvalue problem but can't quite see how.
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For random variables $X \in \mathbb{R}^h$, and a positive semi-definite matrix $A$: Is there a simplified expression for the expected value, $\mathop {\mathbb E}[Tr(X^TAX)]$ and variance, $Var[Tr(X^TAX)]$? Please note that $A$ is not a random variable.
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Is there a simple way to show that a positive definite matrix must be Hermitian?  I feel there is a long drawn out proof of this to be had by taking unit vectors and applying the positive definiteness property, and brute forcing it.  But is there some simple clever proof why a positive definite matrix is necessarily Hermitian?
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I am confused about the difference between positive semi-definite and positive definite.  May I understand that positive semi-definite means symmetric and $x'Ax \ge 0$, while positive definite means symmetric and $x'Ax \gt 0$?
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If I have a positive semidefinite matrix $A_{n\times n}$, I want to transform it to positive definite matrix. I plug the following operation to the original matrix: $$B=A+\lambda I_{n}$$ where $\lambda &gt; 0$. I want to know whether B is positive definite matrix. Could anyone proof it?
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Let $A \in \mathbb{R}^{n\times n}$ be a positive definite matrix, $x \in \mathbb{R}^n$ and $c \in \mathbb{R} \setminus \{0\}$. I got to determine the maximum  $$\max\{c^Ty:y\in \mathcal{E} (A,x)\}$$  where $\mathcal{E} (A,x)$ is an ellipsoid defined by $A$ and $x$.   How can I determine this maximum? Can anyone give me a hint?
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Exercise 12 from SEC. 82 of Finite-Dimensional Vector Spaces - 2nd Edition by Paul R. Halmos.  If $A$ and $B$ are linear transformations on a finite-dimensional inner product space, and if $\textbf{0} \le A \leq B$, then $\det A \leq \det B$.  Notation: $0 \le A$ means that $A$ is positive.  If $\det B = 0$, choose $x \in \ker B$, then $0 \le ((B-A)x,x) = (Bx, x) - (Ax, x) = -(Ax, x) \le 0$. Therefore $\ker B \subseteq \ker A$ and $0 = det A \le det B = 0$. How to prove for the case when $B$ is invertible? Intuitively, I feel that since both $A$ and $B$ are positive their proper values are also positive and $B$'s proper values should be greater the $A$'s and same for determinants.
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Suppose that $A \in \mathbb{R}^{n\times n }$, and $x,y\in \mathbb{R}^n$.    Show that there exist numbers $a,b$ so that $\det{(A + sxy^*)}=a+bs$   Show that if $\det{(A)}\neq 0$ then $a = \det{(A)}$ and $b = \det{(A)} \cdot y^*A^{-1}x$.   Is it true that if $\det{(A)}=0$ then $a=0$?    My answer so far: The third question seems trivial and silly, and I don't understand why it's on there. I have solved the first one as follows: take $\beta$ to be any basis whose first element is $x$. Then with respect to basis $\beta$, $A+ sxy^*$ has the form   $$\left(  \begin{matrix} \alpha_{11} + sy_1 &amp; \alpha_{12} + sy_2  &amp; \dotsb &amp; \alpha_{1n}+sy_n \\ \alpha_{21}&amp;\dotsb&amp;&amp;\alpha_{2n} \\ \vdots &amp; \ddots &amp;&amp;\vdots\\ \alpha_{n1} &amp; \dotsb &amp;&amp;\alpha_{nn} \end{matrix} \right),$$  from which it's clear that the determinant is a polynomial in $s$ of degree $\leq 1$.    I am less clear on the second question.   For a limited case, note that if $A$ is diagonalizable and $x$ is an eigenvector of $A$ with eigenvalue $\lambda_x \neq 0$, then adding $sxy^*$ to $A$ has the effect of preserving all the eigenvectors, bumping up $\lambda_x$ to $\lambda_x + sy^*x$, and since the determinant is the product of all the eigenvalues, the product of $\lambda_1 \lambda_2  \dotsb (\lambda_x + sy^*x)\dotsb \lambda_n $ is   $$\det{(A)} + \det{(A)}\frac{sy^*x}{\lambda_x}=\det{(A)}\cdot sy^*(\frac{x}{\lambda_x}) = \det{(A)}\cdot s y^*(A^{-1}x).$$  Note that $y^*A^{-1}x$ is $\alpha_1 y_1 + \dotsb + \alpha_n y_n$, for $\alpha_i$ such that $x = \alpha_1 A_1 + \dotsb + \alpha_n A_n$, where $A_i$ are the columns of $A$.  But this is all I've got. Can someone give me a hint? I'm not looking for you to feed me the answer. Thanks as always
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Can we have the following?     For matrices $A$ and $B$, if $A \succeq B \implies \overline\sigma(A) \ge \overline\sigma(B)$?    where $\overline\sigma(\cdot)$ means the largest singular value.
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I am having trouble with understanding the following definition while studying some basic things related with matrix norms:  For every matrix $A\in M_n(\mathbb{R})$  $$\sup_{x\neq 0}{} \frac{\|Ax\|}{\|x\|} = \sup_{\|x\| = 1}{\|Ax\|},\; x \in\mathbb{R}^n$$  Why we are taking $\|x\| = 1$?  Is there any proof of above statement?  Added: Why to take sup in the definition of norm?  I need help to understand above mentioned fact.   Thank you very much.