Question

For the following matrices, check whether the reexists an invertible matrix P such that P−1 A Pis diagonal. When such a P exists,find P.
i)A=( 0 1 -3)
( 2 -1 6)
( 1 -1 4)
(ii) B is equal to. ( 0 0 -2)
( 1 2 1)
( 1 0 3)

b)Find the inverse of the matrix B in part (a) by using Cayley-Hamilton theorem.(3)c)Using the fact that det(AB)=det(A)det(B)for any two matrices A and B,prove the identity (a2+b2) (c2+d2)=(ac−bd)2+(ad+bc)2

Answer 1

Answer on Question #52680 – Math – Linear Algebra

a) For the following matrices, check whether there exists an invertible matrix $P$ such that $P^{-1}AP$ is diagonal. When such a $P$ exists, find $P$.

(i) $A = \begin{pmatrix} 0 & 1 & -3 \\ 2 & -1 & 6 \\ 1 & -1 & 4 \end{pmatrix}$

(ii) $B = \begin{pmatrix} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{pmatrix}$

b) Find the inverse of the matrix $B$ in part (a) by using Cayley-Hamilton theorem.

c) Using the fact that $\det(AB) = \det(A)\det(B)$ for any two matrices $A$ and $B$, prove the identity $(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$

Solution

a) Such matrix $P$ exists if there are no zero vectors among eigenvectors of given matrix ($A$ or $B$)

(i) The columns of matrix $P$ are given by the eigenvectors of the matrix $A$. Let's first find its eigenvalues. The characteristic polynomial of $A$:

$$p(\lambda) = \det(\lambda I_3 - A) = \det \begin{pmatrix} \lambda & -1 & 3 \\ -2 & \lambda + 1 & -6 \\ -1 & 1 & \lambda - 4 \end{pmatrix} = \lambda^3 - 3\lambda^2 + 7\lambda - 5$$

Since eigenvalues satisfy the equation $p(\lambda) = 0$, we obtain the following eigenvalues

$$\lambda_{1,2,3} = 1$$

The eigenvectors are

$\lambda = 1$:

$$e_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \qquad e_2 = \begin{pmatrix} 3 \\ 0 \\ -1 \end{pmatrix}, \qquad e_3 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

Therefore there is no such matrix for $A$.

(ii) The columns of matrix $P$ are given by the eigenvectors of the matrix $B$. Let's first find its eigenvalues. The characteristic polynomial of $B$:

$$p(\lambda) = \det(\lambda I_3 - B) = \det \begin{pmatrix} \lambda & 0 & 2 \\ -1 & \lambda - 2 & -1 \\ -1 & 0 & \lambda - 3 \end{pmatrix} = \lambda^3 - 3\lambda^2 + 7\lambda - 5$$

Since eigenvalues satisfy the equation $p(\lambda) = 0$, we obtain the following eigenvalues

$$\lambda_{1,2,3} = 1, 2, 2$$

The eigenvectors are

$$\lambda = 1:$$ $$e_1 = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$$ $$\lambda = 2:$$ $$e_2 = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \qquad e_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$

Therefore

$$P = \begin{pmatrix} -2 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$$

b) The characteristic polynomial of $B$ is given by

$$p(\lambda) = \det(\lambda I_3 - B)$$

The Cayley-Hamilton theorem states that

$$p(B) = 0$$

Since

$$p(\lambda) = \begin{vmatrix} \lambda & 0 & 2 \\ -1 & \lambda - 2 & -1 \\ -1 & 0 & \lambda - 3 \end{vmatrix} = \lambda^3 - 5\lambda^2 + 8\lambda - 4,$$

we obtain

$$p(B) = B^3 - 5B^2 + 8B - 4I_3 = 0$$

Multiplying by $B^{-1}$ we obtain the following

$$B^2 - 5B + 8I_3 - 4B^{-1} = 0$$

Therefore

$$B^{-1} = \frac{1}{4}(B^2 - 5B + 8I_3)$$

Since

$$B ^ {2} = \left( \begin{array}{c c c} 0 & 0 & - 2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{array} \right) \left( \begin{array}{c c c} 0 & 0 & - 2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{array} \right) = \left( \begin{array}{c c c} - 2 & 0 & - 6 \\ 3 & 4 & 3 \\ 3 & 0 & 7 \end{array} \right),$$

we obtain

$$\begin{array}{l} B ^ {- 1} = \frac {1}{4} \left(\left( \begin{array}{c c c} - 2 & 0 & - 6 \\ 3 & 4 & 3 \\ 3 & 0 & 7 \end{array} \right) - 5 \left( \begin{array}{c c c} 0 & 0 & - 2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{array} \right) + 8 \left( \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)\right) = \\ = \frac {1}{4} \left( \begin{array}{c c c} 6 & 0 & 4 \\ - 2 & 2 & - 2 \\ - 2 & 0 & 0 \end{array} \right) = \left( \begin{array}{c c c} 3 / 2 & 0 & 1 \\ - 1 / 2 & 1 / 2 & - 1 / 2 \\ - 1 / 2 & 0 & 0 \end{array} \right) \\ \end{array}$$

c) Let's consider two matrices:

$$A = \left( \begin{array}{c c} a & b \\ d & c \end{array} \right), \qquad B = \left( \begin{array}{c c} a & d \\ b & c \end{array} \right)$$

Then

$$A B = \left( \begin{array}{c c} a & b \\ d & c \end{array} \right) \left( \begin{array}{c c} a & d \\ b & c \end{array} \right) = \left( \begin{array}{c c} a ^ {2} + b ^ {2} & a d + b c \\ a d + b c & c ^ {2} + d ^ {2} \end{array} \right)$$

Since

$$\det (A B) = \det (A) \det (B),$$

we obtain

$$\det \left( \begin{array}{c c} a ^ {2} + b ^ {2} & a d + b c \\ a d + b c & c ^ {2} + d ^ {2} \end{array} \right) = \det \left( \begin{array}{c c} a & b \\ d & c \end{array} \right) \det \left( \begin{array}{c c} a & d \\ b & c \end{array} \right)$$ $$(a ^ {2} + b ^ {2}) (c ^ {2} + d ^ {2}) - (a d + b c) ^ {2} = (a c - b d) (a c - b d)$$ $$(a ^ {2} + b ^ {2}) (c ^ {2} + d ^ {2}) = (a c - b d) ^ {2} + (a d + b c) ^ {2}$$

**Answer:**

a)

(i) Such a matrix does not exist

(ii) $\left( \begin{array}{ccc} - 2 & -1 & 0\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{array} \right)$

b) $\left( \begin{array}{ccc}3 / 2 & 0 & 1\\ -1 / 2 & 1 / 2 & -1 / 2\\ -1 / 2 & 0 & 0 \end{array} \right)$