Question

Suppose \(A = PD{P^{ - 1}}\), where \(P\) is \(2 \times 2\) and \(D = \left( {\begin{array}{*{20}{l}}2&0\\0&7\end{array}} \right)\)

a. Let \(B = 5I - 3A + {A^2}\). Show that \(B\) is diagonalizable by finding a suitable factorization of \(B\).

b. Given \(p\left( t \right)\) and \(p\left( A \right)\) as in Exercise 5 , show that \(p\left( A \right)\) is diagonalizable.

Short answer

a. It is proved that \(B\) is diagonalizable by finding a suitable factorization of \(B\).

b. Since \(I,D,{D^2},.......\) are diagonal matrices, the matrix in the middle is also diagonal as a linear combination of diagonal matrices, which proves that \(p\left( A \right)\) is diagonalizable.

Step-by-step solution

Definition of Diagonalizable Matrix

A diagonal matrixis a matrix in which non-zero values appear only on its main diagonal. In other words, every entry not on the diagonal is zero. Diagonalization is the process of transforming a matrix into a diagonal form.

If \(A\) is a matrix of order \(n\), then \(A\) is said to be diagonalizable if there exists an invertible matrix \(P\) such that

\(A = PD{P^{ - 1}}\)

Where \(D\) is the diagonal matrix with eigenvaluesof \(A\) as the diagonal entries.

Given that if A = PDPˉ¹

First of all, note that if\(A = PD{P^{ - 1}}\), then for any natural\(k\).

\(\begin{aligned}{}{A^k} &= A \cdot A \cdot A \cdot \ldots \cdot A\\ &= \left( {PD{P^{ - 1}}} \right)\left( {PD{P^1}} \right)\left( {PD{P^{ - 1}}} \right) \ldots \left( {PD{P^{ - 1}}} \right)\\ &= PD\left( {{P^{ - 1}}P} \right)D\left( {{P^{ - 1}}P} \right)D\left( {{P^{ - 1}} \cdot \ldots \cdot P} \right)D{P^{ - 1}}\\ &= P{D^k}{P^{ - 1}}\end{aligned}\)

Calculating the matrix is diagonal

It is given that\(A = PD{P^{ - 1}}\), so we can write

\(\begin{aligned}{}B &= 5I - 3A + {A^2}\\ &= 5PI{P^{ - 1}} - 3PD{P^{ - 1}} + P{D^2}{P^{ - 1}}\\ &= P\left( {5I - 3D + {D^2}} \right){P^{ - 1}}\end{aligned}\)

The matrix in the middle is,

\(\begin{aligned}{}5I - 3D + {D^2} &= \left( {\begin{aligned}{*{20}{l}}5&0\\0&5\end{aligned}} \right) - 3\left( {\begin{aligned}{*{20}{l}}2&0\\0&7\end{aligned}} \right) + {\left( {\begin{aligned}{*{20}{l}}2&0\\0&7\end{aligned}} \right)^2}\\ &= \left( {\begin{aligned}{*{20}{l}}5&0\\0&5\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{}}6&0\\0&{21}\end{aligned}} \right) + \left( {\begin{aligned}{*{20}{}}4&0\\0&{49}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}3&0\\0&{32}\end{aligned}} \right)\end{aligned}\)

Since this matrix is diagonal, we conclude that \(B\) is diagonalizable (with the same \(P\) as \(A\) )

(b) Step 4: Proving that \(p\left( A \right)\) is diagonalizable 

We have\(p\left( A \right) = {c_0} + {c_1}A + {c_2}{A^2} + ...... + {c_n}{A^n}\)

Now using\(A = PD{P^{ - 1}}\)we get,

\(\begin{aligned}{}p\left( A \right) &= {c_0}PI{P^{ - 1}} + {c_1}PD{P^{ - 1}} + {c_2}{\left( {PD{P^{ - 1}}} \right)^2} + ...... + {c_n}{\left( {PD{P^{ - 1}}} \right)^n}\\p\left( A \right) &= {c_0}PI{P^{ - 1}} + {c_1}PD{P^{ - 1}} + {c_2}P{D^2}{P^{ - 1}} + .... + {c_n}P{D^n}{P^{ - 1}}\end{aligned}\)

Finally, we get:

\(p\left( A \right) = P\left( {{c_0}I + {c_1}D + {c_2}{D^2} + \ldots + {c_n}{D^n}} \right){P^{ - 1}}\)

Since\(I,D,{D^2}, \ldots ,{D^n}\)are diagonal matrices, the matrix in the middle is also diagonal as a linear combination of diagonal matrices, which proves that\(p\left( A \right)\)is diagonalizable (with the same\(P\)as\(A\)).