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This question already has an answer here:

I have a matrix in companion form, $A=\begin{pmatrix} 0 & \cdots & 0& -a_{0} \\ 1 & \cdots & 0 & -a_{1}\\ \vdots &\ddots & \vdots &\vdots \\ 0 &\cdots & 1 & -a_{n-1} \end{pmatrix}$, $A\in M_{n}$.

I want to prove by induction that the characteristic polynomial $p_{A}=t^{n}+a_{n-1}t^{n-1}+\cdots +a_{0}$.

The part that is confusing me is if we assume this hold for $A_{1}\in M_{n-1}$ how do we transform, or "add" to $A_{1}$ to get the n by n matrix $A\in M_{n}$? I hope this makes sense.

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marked as duplicate by Marc van Leeuwen, Grigory M, Thomas Andrews, Vedran Šego, user66733 Jan 6 '14 at 16:14

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As suggested in the comment above, expand along the first row:

$$\mathrm{det}(tI_n-A) = \mathrm{det} \begin{pmatrix} t & 0 & \cdots & 0 & a_0 \\ -1 & t & \cdots & 0 & a_1 \\ \vdots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & -1 & t+a_{n-1} \end{pmatrix} = $$

$$ t \cdot \mathrm{det} \; \begin{pmatrix} t & 0 & \cdots & 0 & a_1 \\ -1 & t & \cdots & 0 & a_2 \\ \vdots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & -1 & t+a_{n-1} \end{pmatrix} + (-1)^{1+n} a_0 \cdot \mathrm{det} \begin{pmatrix} -1 & t & 0 & \cdots & 0 \\ 0 & -1 & t & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & -1 & \end{pmatrix} $$

By induction we can replace the determinant on the left by $a_1+a_2t+\cdots+a_{n-1}t^{n-2}+t^{n-1}$ and the second matrix's determinant is the product of its diagonals (since it's upper-triangular). The product of the diagonals is $(-1)^{n-1}$. Therefore, the determinant is $t(a_1+a_2t+\cdots+a_{n-1}t^{n-2}+t^{n-1})+(-1)^{n+1}(-1)^{n-1}a_0$. Which simplifies to $a_0+a_1t+\cdots+a_{n-1}t^{n-1}+t^n$.

[When writing this up, don't forget the base case for the induction.]

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  • $\begingroup$ Thank you for your explanation $\endgroup$ – Edison Oct 27 '11 at 21:32
  • $\begingroup$ Can you explain why you were able to go from line 1 to line 2 (which properties of the determinant). $\endgroup$ – IntegrateThis Nov 29 '16 at 3:58
  • $\begingroup$ As stated, I expanded along the first row. This is the standard Laplace expansion. Since all but 2 entries in the first row are zero, the expansion only has 2 nonzero terms. $\endgroup$ – Bill Cook Nov 29 '16 at 4:27

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