0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$
  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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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