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Problem 276

# Show that the matrix $$\mathrm{A}$$ is not diagonalizable where $$\mathrm{A}=\mid \begin{array}{cc} -3 & 2 \\ \mid-2 & 1 \mid \end{array}$$

Expert verified
Matrix A is given as: $$\mathrm{A}=\begin{pmatrix} -3 & 2 \\ -2 & 1 \end{pmatrix}$$ We find the characteristic equation: $$\det(\mathrm{A} - λI) = λ^2 + 2λ + 1 = (λ + 1)^2$$ Solving the characteristic equation, we find that there is only one eigenvalue, λ = -1. By calculating the eigenspace for λ = -1: $$\begin{pmatrix} -2 & 2 \\ -2 & 2 \end{pmatrix}$$ We find that there is only one linearly independent eigenvector: $$\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ Since matrix A is a 2x2 matrix and there is only one linearly independent eigenvector instead of the necessary two, the matrix A is not diagonalizable.
See the step by step solution

## Step 1: Finding the characteristic equation

Calculate the determinant of the matrix (A - λI), where λ is an eigenvalue, and I is the identity matrix. $$\det(\mathrm{A} - λI) = \det\begin{pmatrix} -3 - λ & 2 \\ -2 & 1 - λ \end{pmatrix}$$ Next, find the determinant of this matrix: $$(-3 - λ)(1 - λ) - (-2)(2) = λ^2 + 2λ + 1$$ This is our characteristic equation.

## Step 2: Solve for the eigenvalues

Solve the characteristic equation for λ to find the eigenvalues of matrix A. $$λ^2 + 2λ + 1 = (λ + 1)^2 = 0$$ This equation has only one solution: λ = -1. Since there is only one eigenvalue, the matrix A will not have enough linearly independent eigenvectors to be diagonalizable. We will now find the eigenspace corresponding to the eigenvalue λ = -1 to confirm this.

## Step 3: Finding the eigenspace for λ = -1

Substitute the eigenvalue λ = -1 back into (A - λI) and solve for the eigenvectors. $$\begin{pmatrix} -3 - (-1) & 2 \\ -2 & 1 - (-1) \end{pmatrix} = \begin{pmatrix} -2 & 2 \\ -2 & 2 \end{pmatrix}$$ Now, solve the linear system formed by this matrix by applying Gaussian elimination or another method. Row 1: $$-2X_1 + 2X_2 = 0$$ Row 2: $$-2X_1 + 2X_2 = 0$$ This linear system has only one equation after simplification. When we solve for $$X_1$$, we get: $$X_1 = X_2$$ Which means there is only one linearly independent eigenvector for λ = -1: $$\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ Since the matrix A is a 2x2 matrix and there is only one linearly independent eigenvector instead of the necessary two, the matrix A is not diagonalizable.

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