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Problem 225

# Find the characteristic and minimum polynomials of each of the following matrices (a) $$\mid \begin{array}{cc}3 & -1 \mid \\ \mid-1 & 3 \mid\end{array}$$ (b) $$\begin{array}{cc}\mid 1 & 1 \mid \\ & \mid 0 & 2 \mid\end{array}$$ (c) \begin{aligned} \mid 1 &-2 \mid \\ & \mid 0 &-1 \mid \end{aligned} (d) $$\begin{array}{rl}\mid 1 & 1 \mid \\ & \mid 0 & 1 \mid\end{array}$$ (e) $\begin{array}{rrrr} & \mid 0 & 1 & 0 & 0 \\ & \mid 0 & 0 & 0 & 0 \\ & \mid 0 & 0 & 1 & -2 \mid \\ & \mid 0 & 0 & 1 & -1\end{array}$ (f) $\begin{array}{cccc} & \mid 3 & 1 & 0 & 0 \\ & \mid 0 & 3 & 0 & 0 \\ & \mid 0 & 0 & 2 & 1 \\ & \mid 0 & 0 & 1 & 2\end{array} \mid$

Expert verified
The characteristic and minimum polynomials of the given matrices are: (a) Matrix: $$\begin{bmatrix} 3 & -1 \\ -1 & 3 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda) = \lambda^2 - 6\lambda + 8$$ Minimum polynomial: $$m(\lambda) = (\lambda - 2)(\lambda - 4)$$ (b) Matrix: $$\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=\lambda^2 - 3\lambda + 2$$ Minimum polynomial: $$m(\lambda)=(\lambda - 1)(\lambda - 2)$$ (c) Matrix: $$\begin{bmatrix} 1 & -2 \\ 0 & -1 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=\lambda^2 +\lambda$$ Minimum polynomial: $$m(\lambda)=\lambda(\lambda + 1)$$ (d) Matrix: $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=(\lambda - 1)^2$$ Minimum polynomial: $$m(\lambda)=(\lambda - 1)^2$$ (e) Matrix: $$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 1 & -1 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=\lambda^3(\lambda - 1)$$ Minimum polynomial: $$m(\lambda)=\lambda^2(\lambda - 1)$$ (f) Matrix: $$\begin{bmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 1 & 2 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=(\lambda - 3)^2(\lambda - 2)^2$$ Minimum polynomial: $$m(\lambda)=(\lambda - 3)(\lambda - 2)$$
See the step by step solution

## Step 1: Finding the Characteristic Polynomial

Set the determinant of (A - λI) to 0: $$\begin{vmatrix} 3-\lambda & -1 \\ -1 & 3-\lambda \end{vmatrix} = 0$$ Calculate the determinant: $$(3-\lambda)(3-\lambda) - (-1)(-1) = \lambda^2 - 6\lambda + 9 - 1 = \lambda^2 - 6\lambda + 8$$ The characteristic polynomial is $$p(\lambda)=\lambda^2 - 6\lambda + 8$$.

## Step 2: Finding the Eigenvalues

Solve the characteristic polynomial equation $$p(\lambda)=0$$: $$\lambda^2 - 6\lambda + 8 = 0$$ Factoring yields: $$(\lambda - 4)(\lambda - 2) = 0$$ Eigenvalues are λ=2 and λ=4.

## Step 3: Constructing the Minimum Polynomial

Since we have two distinct eigenvalues, the minimum polynomial is formed by multiplying the factors of the characteristic polynomial corresponding to each eigenvalue: $$m(\lambda) = (\lambda - 2)(\lambda - 4)$$ Thus, the characteristic polynomial for matrix (a) is $$p(\lambda) = \lambda^2 - 6\lambda + 8$$ and the minimum polynomial is $$m(\lambda) = (\lambda - 2)(\lambda - 4)$$. These explanations are then repeated for each matrix: (b) Matrix: $$\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=\lambda^2 - 3\lambda + 2$$ Minimum polynomial: $$m(\lambda)=(\lambda - 1)(\lambda - 2)$$ (c) Matrix: $$\begin{bmatrix} 1 & -2 \\ 0 & -1 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=\lambda^2 +\lambda$$ Minimum polynomial: $$m(\lambda)=\lambda(\lambda + 1)$$ (d) Matrix: $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=(\lambda - 1)^2$$ Minimum polynomial: $$m(\lambda)=(\lambda - 1)^2$$ (e) Matrix: $$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 1 & -1 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=\lambda^3(\lambda - 1)$$ Minimum polynomial: $$m(\lambda)=\lambda^2(\lambda - 1)$$ (f) Matrix: $$\begin{bmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 1 & 2 \end{bmatrix}$$ Characteristic polynomial: $$p(\lambda)=(\lambda - 3)^2(\lambda - 2)^2$$ Minimum polynomial: $$m(\lambda)=(\lambda - 3)(\lambda - 2)$$

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