Question

(1 point) A, P and D are n×n matrices. Check the true statements below: A. A is diagonalizable if and only if A has n eigenvalues, counting multiplicities. B. If A is diagonalizable, then A is invertible. C. A is diagonalizable if A=PDP−1 for some diagonal matrix D and some invertible matrix P. D. If there exists a basis for Rn consisting entirely of eigenvectors of A, then A is diagonalizable.

Answer 1

Answer:

Following are the answers to this questions:

A) False.

B) False.

C) True.

D) True.

Step-by-step explanation:

In option A,

Let A is $3 \times 3$ diagonal matrices that is,

$\left[\begin{array}{ccc}0&1&0\\0&0&6\\0&0&0\end{array}\right]$

here $A^3$= 0, In this A is the nilpotent matrix and A has three values that are equal to 0 but it is not a diagonalizable matrix.

In option B,

If $A =\left[\begin{array}{ccc}0&1\\0&0\end{array}\right]$ in given value A elements are 0 and 1 as a distinct eigen value, that's why A is diagonalizable  but |A| =0 its means A is not invertible, that's why it is false.

In option C,

It is given that $A= PDP^{-1}$ where A~ D, which is A is similar to diagonal matrices, that's why it is true.

In option D,

It is given that A has eigen vector, which leases from $R^n$. A has eigen vector and A is $n \times n$ matrix, the number of A eigen values = given vector of A, that's why it is correct.