let $A\in M_{10}(\mathbb{R})$ and every row of $A$ has exactly one non zero element . find eigenvalue of $A$.

Question

let $A\in M_{10}(\mathbb{R})$ and every row of $A$ has exactly one non zero element . $$ A = \begin{bmatrix}0 & 1 &0 &\ldots & 0\\ 0 & 0 &1 &\ldots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 &0 &\ldots & 1 \\ (10)^{10} & 0 &0 &\ldots & 0\end{bmatrix} _{10 \times 10}$$

now which of following options is eigenvalue of $A$ ?

$ \\1)0 \\ 2)1 \\ 3)10 \\4) (10)^{10}$

I find $det (A)= (10)^{10}$ .but we must find $det (A- \lambda I)$ and find characteristic polynomial of $A$ and roots of characteristic polynomial is eigenvalues of $A$ .

Answer 1

Your matrix is the transpose of the companion matrix so its characteristic and minimal polynomial are equal to $x^{10} - 10^{10}$.

Therefore, the spectrum of $A$ is precisely the set of real roots of $x^{10} - 10^{10}$, which are $\pm 10$.

Answer 2

Hint: What is $A.v$ if $v=(1,10,10^2,\ldots,10^9)$?

Answer 3

The matrix $$ A-xI= \begin{bmatrix}-x & 1 &0 &\ldots & 0 &0\\ 0 & -x &1 &\ldots &0 &0 \\ \vdots & \vdots & \vdots & & \vdots &\vdots \\ 0 & 0 &0 &\ldots &-x &1 \\ 10^{10} & 0 &0 &\ldots &0 & -x\end{bmatrix} _{10 \times 10} $$ has only two generalized diagonals not containing a zero:

• The main diagonal, with signed product $x^{10}$.$\\[4pt]$
• The off diagonal (consisting of all ones), together with the entry $10^{10}$, with signed product $-10^{10}$.

It follows that the characteristic polynomial of $A$ is $x^{10}-10^{10}$, which has only two real roots, namely $\pm 10$.

Answer 4

Note that $A^{10}=10^{10}I$, which rules out most answers.