Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

1.a symmetric matrix in $\mathbb{M}_n(\mathbb{R})$ is said to be non-negative definite if $x^Tax≥0$ for all (column) vectors $x\in \mathbb{R}^n$. Which of the following statements are true?
(a) If a real symmetric $n\times n$ matrix is non-negative definite, then all of its eigenvalues are non-negative.
(b) If a real symmetric $n\times n$ matrix has all of its eigenvalues are non-negative , then it is non-negative definite.
(c) If $ A\in \mathbb{M}_n(\mathbb{R})$, then $AA^T$ is non-negetive definite.
2. only one of the following matrices is non-negetive definite. Find it.
(a) $\begin{pmatrix} 5 & -3 \\ -3 & 5 \end{pmatrix}$.
(b) $\begin{pmatrix} 1 & -3 \\ -3 & 5 \end{pmatrix}$.
(c) $\begin{pmatrix} 1 & 3 \\ 3 & 5 \end{pmatrix}$.

3.let $B$ be the real symmetric non-negative definite $2×2$ matrix such that $B^2=A$ where $A$ where is the non-negetive definite matrix in question $2$.write down the characteristic polynomial of $B$.

My thoughts..

For1. (a) & (b) are true but not sure about (c).
For 2. (a) is the correct option since it has all positive eigen values.they are $2,8$.
For 3. Eigen values of $B$ will be $\sqrt{2}$ and $\sqrt{8}$. So the answer is $x^2-3√2x+4=0$.

Can anybody help me to verify the solutions of the above problems.

share|improve this question
add comment

1 Answer 1

up vote 0 down vote accepted
  1. (c) is true: Start by proving that $AA^T$ is symmetric, otherwise you can't talk about it being non-negative definite: $(AA^T)^T=(A^T)^TA^T=AA^T$, therefore $AA^T$ is symmetric. Now take $y\in \mathbb{R}^{n\times 1}$. We wish to prove that $y^TAA^Ty\ge 0$. For the sake of a more pictorial expression, let $y=x^T$. We get $y^TAA^Ty=(x^T)^TAA^Tx^T=xAA^Tx^T=(xA)(xA)^T$. Now notice that $xA\in \mathbb{R}^{1\times n}$. The dot product $v\bullet u$ in $\mathbb{R}^{1+n}$ is given precisely by $vu^T$, plus $\sqrt{v\bullet v}=\vert \vert v \vert \vert$ is the norm induced by the dot product $\bullet$. So it follows $(xA)(xA)^T=(xA)\bullet(xA)=\vert \vert Ax \vert \vert \ge0$.

All of your answers are correct.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.