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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.

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  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.

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