Question

Check whether or not the matrix A=[1 1 1
0 - 2 2
0 - 2 - 3] is diagonalisable. If it is, find a matrix P, and a matrix D such that P^-1 AP=D. If A is not diagonalisable find AdjkA).

Answer 1

Answer on Question #83505 – Math – Linear Algebra

Question

Check whether or not the matrix

$$A = \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 0 & -2 & 2 \\ 0 & -2 & -3 \end{array} \right]$$

is diagonalizable. If it is, find a matrix $P$, and a matrix $D$ such that $P^{-1}AP = D$. If $A$ is not diagonalizable find $AdjA$.

Solution

Matrix $A$ ($3 \times 3$) is diagonalizable if and only if there is a basis of $R^3$ consisting of eigenvectors of $A$. So let's find the eigenvalues and eigenvectors for matrix $A$.

To define the eigenvalues, we find the values of $\lambda$ which satisfy the characteristic equation of the matrix $A$:

$$\det (A - \lambda I) = 0,$$

where $I$ is the $3 \times 3$ identity matrix.

Form the matrix $A - \lambda I$:

$$A - \lambda I = \left[ \begin{array}{ccc} 1 - \lambda & 1 & 1 \\ 0 & -2 - \lambda & 2 \\ 0 & -2 & -3 - \lambda \end{array} \right].$$

Calculate $\det (A - \lambda I)$. We define the determinant of a $3 \times 3$ matrix by the rule:

$$\left| \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right| = a_{11} \left| \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} \right| - a_{12} \left| \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} \right| + a_{13} \left| \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} \right|.$$ $$\begin{array}{l} \det (A - \lambda I) = (1 - \lambda) \left| \begin{array}{cc} -2 - \lambda & 2 \\ -2 & -3 - \lambda \end{array} \right| - 1 \left| \begin{array}{cc} 0 & 2 \\ 0 & -3 - \lambda \end{array} \right| + 1 \left| \begin{array}{cc} 0 & -2 - \lambda \\ 0 & -2 \end{array} \right| = \\ = (1 - \lambda) \left((-2 - \lambda) (-3 - \lambda) - (-2) \cdot 2\right) = (1 - \lambda) (6 + 5 \lambda + \lambda^2 + 4) \\ = (1 - \lambda) (\lambda^2 + 5 \lambda + 10) \end{array}$$

Then find solutions of the characteristic equation $\det (A - \lambda I) = 0$, i.e. solve the equation:

$$(1 - \lambda) (\lambda^2 + 5 \lambda + 10) = 0$$ $$1 - \lambda = 0 \text{ or } \lambda^2 + 5 \lambda + 10 = 0$$ $$\lambda_1 = 1$$

The quadratic equation $\lambda^2 + 5\lambda + 10 = 0$ has not any real roots, because

$$\Delta = b^2 - 4ac = 25 - 40 = 15 < 0.$$

So we can find only one eigenvector for A. But any basis for $R^3$ consists of three vectors. Therefore, there is no eigenbasis for A, and so the matrix A is not diagonalizable.

Find Adj $A$ . First, calculate the cofactors of each element.

$$c _ {1 1} = (- 1) ^ {1 + 1} \left| \begin{array}{c c} - 2 & 2 \\ - 2 & - 3 \end{array} \right| = 6 + 4 = 1 0, c _ {1 2} = (- 1) ^ {1 + 2} \left| \begin{array}{c c} 0 & 2 \\ 0 & - 3 \end{array} \right| = 0,$$ $$c _ {1 3} = (- 1) ^ {1 + 3} \left| \begin{array}{c c} 0 & - 2 \\ 0 & - 2 \end{array} \right| = 0, \quad c _ {2 1} = (- 1) ^ {2 + 1} \left| \begin{array}{c c} 1 & 1 \\ - 2 & - 3 \end{array} \right| = - 1 (- 3 + 2) = 1,$$ $$c _ {2 2} = (- 1) ^ {2 + 2} \left| \begin{array}{c c} 1 & 1 \\ 0 & - 3 \end{array} \right| = - 3, \qquad c _ {2 3} = (- 1) ^ {2 + 3} \left| \begin{array}{c c} 1 & 1 \\ 0 & - 2 \end{array} \right| = - 1 (- 2) = 2,$$ $$c _ {3 1} = (- 1) ^ {3 + 1} \left| \begin{array}{c c} 1 & 1 \\ - 2 & 2 \end{array} \right| = 2 + 2 = 4, c _ {3 2} = (- 1) ^ {3 + 2} \left| \begin{array}{c c} 1 & 1 \\ 0 & 2 \end{array} \right| = - 1 (2) = - 2,$$ $$c _ {3 3} = (- 1) ^ {3 + 3} \left| \begin{array}{c c} 1 & 1 \\ 0 & - 2 \end{array} \right| = - 2.$$

Write the cofactor matrix C:

$$C = \left[ \begin{array}{c c c} 1 0 & 0 & 0 \\ 1 & - 3 & 2 \\ 4 & - 2 & - 2 \end{array} \right], \text{adj } A = C ^ {T} = \left[ \begin{array}{c c c} 1 0 & 1 & 4 \\ 0 & - 3 & - 2 \\ 0 & 2 & - 2 \end{array} \right].$$

Answer: $A$ is not diagonalizable; $adjA = \begin{bmatrix} 10 & 1 & 4 \\ 0 & -3 & -2 \\ 0 & 2 & -2 \end{bmatrix}$ .

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