Question

A scalar matrix is simply a matrix of the form XI, where I is the nxn identity matrix. (a) Prove that if A is similar 1 to \I

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Answer #1

A matrix A is said to be similar to B if there exists an invertible matrix such that PAP^{-1}=B

a)

Given that A is similiar to scalar matrix.

Therefore, we have

\bigskip PAP^{-1}=\lambda I\\\bigskip\Rightarrow A=P^{-1}(\lambda I)P \\\bigskip\Rightarrow A=\lambda (P^{-1}IP) \\\bigskip\Rightarrow A=\lambda I

b)

Let the eigenvalue of A be \lambda

A is said to be diagonalizable if there exists an invertible matrix U such that

U^{-1}AU=D\text{ where }D\text{ is diagonal matrix}

We know that the diagonal matrix contains the eigen values of the matrix A

But it is given that A has only one diagonal value.

Hence, by part (a), D=\lambda I

Therefore, by applying part (a) again, we have

A=\lambda I

c)

The given matrix has only one eigen value i.e. 1 is the only eigen value of the given matrix.

But the given matrix is not a scalar matrix. Hence, by part (b), the given matrix is not diagonalozable.

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