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$A\in \mathbb{C}^{n,n}$ is nonsingular, then:
a. $P_A(0)\neq 0$, where $P_A$ is characteristic polynomial matrix $A$
b. $\ker(AA^T)=\{0\}$
c. matrix $I_n-A$ is also nonsingular.

a. is not true. $A=\left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$ is nonsingular and $P_A(\lambda)=(1-\lambda)^2$
b. it is true:
$AA^Tx=0\Leftrightarrow A^Tx=0 \Leftrightarrow x =0$. Here, I use the fact that $\det A=\det A^T$ so both matrixes are reverable.
c. it is not true, $A=\left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$, then $I_n-A=\left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)=\left( \begin{array}{ccc} 0 & 0 \\ 0 & 0 \\ \end{array} \right)$ is not nonsingular.

Am I right ?

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    $\begingroup$ Your counter-example to a. is not right, since $P_A(0)=1$ in your case. $\endgroup$ – Luiz Cordeiro Aug 6 '16 at 23:05
  • $\begingroup$ Ok, I will try to repair it. What about b. and c. ? $\endgroup$ – user343207 Aug 6 '16 at 23:06
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a. is true, since $P_A(0)=\det(A-0I)=\det(A)\neq 0$, because $A$ is nonsingular.

b. and c. are ok.

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  • $\begingroup$ Is it ok to say that nonsingular matrix has no $0$ on principial diagonal ? $\endgroup$ – user343207 Aug 6 '16 at 23:11
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    $\begingroup$ @HaskellFun No. For example $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ is nonsingular and only has zeroes in the principal diagonal. $\endgroup$ – Luiz Cordeiro Aug 6 '16 at 23:13

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