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Let $M$ be a real $n\times n$ matrix such that for$ every$ non-zero vector $x\in \mathbb{R}^{n},$ we have $x^{T}M x> 0.$ Then

  1. Such an $M$ cannot exist
  2. Such $Ms$ exist and their rank is always $n$
  3. Such $Ms$ exist, but their eigenvalues are always real
  4. No eigenvalue of any such $M$ can be real
  5. None of the above
in Linear Algebra edited by
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-- For $a)$ and $d),$ counter-examples are identity matrices like $M= \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$ for $n=2.$

Because, considering a non-zero vector  $x$  as $= \begin{bmatrix} a\\b \end{bmatrix}$ where, $a >0, b>0$

Now, $x^TMx = \begin{bmatrix} a &b \end{bmatrix} \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a \\b \end{bmatrix} = a^2 + b^2 > 0.$ So, such $Ms$ exist and eigen values can be real.

-- For $(c),$ counter-example is : $M= \begin{bmatrix} 2k &2k \\ -2k & 2k \end{bmatrix}$ where $k \geq 1$

Here, also on considering a non-zero vector  $x$  as $= \begin{bmatrix} a\\b \end{bmatrix}$ where, $a >0, b>0$

Now, $x^TMx = \begin{bmatrix} a &b \end{bmatrix} \begin{bmatrix} 2k &2k \\-2k & 2k \end{bmatrix} \begin{bmatrix} a \\b \end{bmatrix} = 2k(a^2 + b^2) > 0.$ So, such $Ms$ exist and eigen values can be complex here.

-- Now, for $(b),$ Suppose, $x \in \mathbb{R}^n$ is a non-zero eigen vector associated with matrix $M$.

Then, $Mx = \lambda x$

$\Rightarrow x^TMx = x^T\lambda x$

$\Rightarrow x^TMx = \lambda x^Tx$

Since, $x^Tx > 0 $   and If $\lambda > 0$ then $x^TMx > 0$

So, For positive eigenvalues $\lambda$ and a non-zero eigen-vectors $x$, $x^TMx > 0$ is always true.

Since, determinant is the product of eigen values. So, if all eigen values are positive then determinant will be positive and matrix $M$ will be non-singular and has full rank $n$. So, $(b)$ is true.
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