Find the characteristic polynomial |$\lambda - AI $| for this $5 \times 5$ matrix

Question

Find the characteristic polynomial |$\lambda - AI $| for the $5\times 5$ matrix

$$A=\left(\begin{matrix} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0& 0 & 1 & 0\\ 0& 0 & 0 & 0 & 1\\ 10^{10} & 0 & 0 & 0 & 0\\ \end{matrix}\right)$$

My attempts:

$$A - \lambda I = \left(\begin{matrix} -\lambda & 1 & 0 & 0 & 0\\ 0 & -\lambda & 1 & 0 & 0\\ 0 & 0& -\lambda & 1 & 0\\ 0& 0 & 0 & -\lambda & 1\\ 10^{10} & 0 & 0 & 0 & -\lambda \\ \end{matrix}\right)$$

From my thinking, the characteristic polynomial |$\lambda - AI $| for the $5\times 5$ matrix is -$\lambda ^5$.

Is it true or false? Any hints/solution will be appreciated. Please help me. Thanks you.

Answer 1

Doing a Laplace expansion along the first column, you get that the determinant of that matrix is equal to\begin{align}-\lambda\begin{vmatrix}-\lambda&1&0&0\\0&-\lambda&1&0\\0&0&-\lambda&1\\0&0&0&-\lambda\end{vmatrix}+10^{10}\begin{vmatrix}1&0&0&0\\-\lambda&1&0&0\\0&-\lambda&1&0\\0&0&-\lambda&1\end{vmatrix}=-\lambda^5+10^{10}.\end{align}

Answer 2

If the characteristic polynomial is $-\lambda^5$, you are suggesting that $0$ is an eigenvalue and the matrix is singular. However, it should be clear that the rows are linearly independent and hence the matrix is nonsingular.

To find it characteristic polynomial, try to expand along the first column.

Note that characteristic polynomial is $|\lambda I - A|$.

Answer 3

Expand on the last row $$\det (A - \lambda I)=det \left(\begin{matrix} -\lambda & 1 & 0 & 0 & 0\\ 0 & -\lambda & 1 & 0 & 0\\ 0 & 0& -\lambda & 1 & 0\\ 0& 0 & 0 & -\lambda & 1\\ 10^{10} & 0 & 0 & 0 & -\lambda \\ \end{matrix}\right)$$

$$ = 10^{10}-\lambda ^5$$

Answer 4

Consider the effect of $A$ on the canonical basis: $$ Ae_1 = 10^{10} e_5,\quad Ae_2 = e_1,\quad Ae_3 = e_2,\quad Ae_4 = e_3,\quad Ae_5 = e_4 $$ Then $A^5 e_i = 10^{10} e_i$ and so $A^5 = 10^{10} I$.